SOLUTIONS TO EXERCISES

CHEMISTRY

THE CENTRAL SCIENCE

14 TH EDITION


 

SOLUTIONS TO EXERCISES

Roxy Wilson

University of Illinois, Urbana-Champaign

CHEMISTRY

THE CENTRAL SCIENCE

14 TH EDITION

BROWN | L E MAY

BURSTEN | MURPHY

WOODWARD | STOLTZFUS

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1 17

ISBN-10: 0-13-455224-5

ISBN-13: 978-0-13-455224-8


 

Contents

Chapter 1 Introduction: Matter, Energy, and Measurement

Chapter 2 Atoms, Molecules, and Ions

Chapter 3 Chemical Reactions and Reaction Stoichiometry

Chapter 4 Reactions in Aqueous Solution

Chapter 5 Thermochemistry

Chapter 6 Electronic Structure of Atoms

Chapter 7 Periodic Properties of the Elements

Chapter 8 Basic Concepts of Chemical Bonding

Chapter 9 Molecular Geometry and Bonding Theories

Chapter 10 Gases

Chapter 11 Liquids and Intermolecular Forces

Chapter 12 Solids and Modern Materials

Chapter 13 Properties of Solutions

Chapter 14 Chemical Kinetics

Chapter 15 Chemical Equilibrium

Chapter 16 Acid–Base Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 18 Chemistry of the Environment

Chapter 19 Chemical Thermodynamics

Chapter 20 Electrochemistry

Chapter 21 Nuclear Chemistry

Chapter 22 Chemistry of the Nonmetals

Chapter 23 Transition Metals and Coordination Chemistry

Chapter 24 The Chemistry of Life: Organic and Biological Chemistry


 

Introduction

Chemistry: The Central Science, 14th edition , contains over 2700 end-of-chapter exercises. Considerable attention has been given to these exercises because one of the best ways for students to master chemistry is by solving problems. Grouping the exercises according to subject matter is intended to aid the student in selecting and recognizing particular types of problems. Within each subject matter group, similar problems are arranged in pairs. This provides the student with an opportunity to reinforce a particular kind of problem. There are also a substantial number of general exercises in each chapter to supplement those grouped by topic. Integrative exercises, which require students to integrate concepts from several chapters, are a continuing feature of the 14th edition. Answers to the odd-numbered topical exercises plus selected general and integrative exercises, about 1150 in all, are provided in the textbook. These appendix answers help to make the textbook a useful self-contained vehicle for learning.

This manual, Solutions to Exercises in Chemistry: The Central Science, 14th edition , was written to enhance the end-of-chapter exercises by providing documented solutions. The manual assists the instructor by saving time spent generating solutions for assigned problem sets and aids the student by offering a convenient independent source to check their understanding of the material. Most solutions have been worked in the same detail as the in-chapter sample exercises to help guide students in their studies.

To reinforce the Analyze, Plan, Solve, Check problem-solving method used extensively in the text, this strategy has also been incorporated into the Solutions Manual. Solutions to most red topical exercises and selected Additional and Integrative exercises feature this four-step approach. We strongly encourage students to master this powerful and totally general method.

When using this manual, keep in mind that the numerical result of any calculation is influenced by the precision of the numbers used in the calculation. In this manual, for example, atomic masses and physical constants are typically expressed to four significant figures, or at least as precisely as the data given in the problem. If students use slightly different values to solve problems, their answers will differ slightly from those listed in the appendix of the text or this manual. This is a normal and a common occurrence when comparing results from different calculations or experiments.

Rounding methods are another source of differences between calculated values. In this manual, when a solution is given in steps, intermediate results will be rounded to the correct number of significant figures; however, unrounded numbers will be used in subsequent calculations. By following this scheme, calculators need not be cleared to re-enter rounded intermediate results in the middle of a calculation sequence. The final answer will appear with the correct number of significant figures. This may result in a small discrepancy in the last significant digit between student-calculated answers and those given in this manual. Variations due to rounding can occur in any analysis of numerical data.


The first step in checking your solution and resolving differences between your answer and the listed value is to look for similarities and differences in problem-solving methods. Ultimately, resolving the small numerical differences described above is less important than understanding the general method for solving a problem. The goal of this manual is to provide a reference for sound and consistent problem-solving methods in addition to accurate answers to text exercises.

Extraordinary efforts have been made to keep this manual as error-free as possible. All exercises were worked and proofread by at least three chemists to ensure clarity in methods and accuracy in mathematics. The ongoing work and advice of Dr. Richard Helmich, Ms. Rene Musto, and Dr. Christopher Musto continue to be invaluable to this project. In any written work as technically challenging as this manual, typos and errors inevitably creep in, despite our combined efforts. Please help us find and eliminate them. We hope that both instructors and students will find this manual accurate, helpful, and instructive.

Roxy B. Wilson, Ph.D.

1829 Maynard Dr.

Champaign, IL 61822

rbwilson@illinois.edu


 

1 Introduction: Matter, Energy, and Measurement

Visualizing Concepts

1.1 Pure elements contain only one kind of atom. Atoms can be present singly or as tightly bound groups called molecules. Compounds contain two or more kinds of atoms bound tightly into molecules. Mixtures contain more than one kind of atom and/or molecule, not bound into discrete particles.

  1. pure element: i
  2. mixture of elements: v, vi
  3. pure compound: iv
  4. mixture of an element and a compound: ii, iii

1.2   After a physical change , the identities of the substances involved are the same as their identities before the change. That is, molecules retain their original composition. During a chemical change , at least one new substance is produced; rearrangement of atoms into new molecules occurs.

Diagram (b) represents a chemical change because the molecules after the change are different than the molecules before the change.

1.3

  1. Brass is composed of two different kinds of atoms, so it is a mixture. The mixture appears homogeneous under an optical microscope, so it is a homogeneous mixture.
  2. Because brass is a homogeneous mixture, it is a solution. We usually think of solutions as liquids, but they can be solids, liquids, or gases.

1.4   Refer to Sample Exercise 1.5.

  1. density = mass volume ; mass = density × volume mass(Al) = 2.70 g cm 3 × 196 cm 3 × 1 kg 1000 g = 0.5292 = 0.529 kg mass(Ag) = 10.49 g cm 3 × 196 cm 3 × 1 kg 1000 g = 2.056 = 2.06 kg
  2. The amount of work done to lift each sphere to a height of 2.2 m can be calculated using Equation 1.1, w = F × d = m × g × d. Use mass in kg and distance in meters so the final unit for work is joules, J.

  1. w ( Al ) = 0.5292 kg Al × 9.8 m s 2 × 2.2 m = 11.41 = 11 kg-m 2 1 s 2 = 11 J w(Ag) = 2.056 kg Ag × 9.8 m s 2 × 2.2 m = 44.33 = 44 kg-m 2 s 2 = 44 J

    (The results have two significant figures because the height had only two significant figures.)

    Note: This is the first exercise where “intermediate rounding” occurs. In this manual, when a solution is given in steps, the intermediate result will be rounded to the correct number of significant figures. However, the unrounded number will be used in subsequent calculations. The final answer will appear with the correct number of significant figures. That is, calculators need not be cleared and new numbers entered in the middle of a calculation sequence. This may result in a small discrepancy in the last significant digit between student-calculated answers and those given in the manual. These variations occur in any analysis of numerical data.

    For example, in this exercise the mass of the Ag sphere, 2.056 × 10 3 g, is rounded to 2.06 × 10 3 g, but 2.056 × 10 3 g is retained in the subsequent calculation of work, 44 J. In this case, the unrounded and rounded masses both lead to the same rounded value for work. In other exercises, the correctly rounded results of the two methods may not be identical.

  1. Less work is done on the Al sphere, so its potential energy increases by a smaller amount than the potential energy of the Ag sphere.
  1. No. The two spheres do not have the same kinetic energy when they touch the ground. Because the Ag sphere has greater mass, it gains more potential energy when it is lifted. At the point of impact, all potential energy has been converted to kinetic energy. Because of its greater potential energy, the Ag sphere has greater kinetic energy when it hits the floor.

1.5   Filtration. When brewing a cup of coffee, hot water contacts the coffee grounds and dissolves components of the coffee bean that are water-soluble. This creates a heterogeneous mixture of undissolved coffee bean solids and liquid coffee solution; this mixture is separated by filtration. Undissolved grounds remain on the filter paper and liquid coffee drips into the container below.

1.6

  1. time
  2. mass
  3. temperature
  4. area
  5. length
  6. area
  7. temperature
  8. density
  9. volume

1.7   Density is the ratio of mass to volume. For a sphere, size is like volume; both are determined by the radius of the sphere.

  1. For spheres of the same size or volume, the denominator of the density relationship is the same. The denser the sphere, the heavier it is. A list from lightest to heaviest is in order of increasing density and mass. The aluminum sphere (density = 2.70 g/cm 3 ) is lightest, then nickel (density = 8.90 g/cm 3 ), then silver (density = 10.49 g/cm 3 ).

  1. For cubes of equal mass, the numerator of the density relationship is the same. The denser the sphere, the smaller its volume or size. A list from smallest to largest is in order of decreasing density. The platinum sphere (density = 21.45 g/cm 3 ) is smallest, then gold (density = 19.30 g/cm 3 ), then lead (density = 11.35 g/cm 3 ).

1.8   Results (bullet holes) that are close to each other are precise . Results that are close to the “true value” (the bull’s-eye) are accurate .

  1. The results on target A are precise but not accurate. The individual shots landed close to each other, but not close to the bull’s-eye. The results on target B are both precise and accurate. The individual shots landed close together and in the center of the target. The results on target C are neither precise nor accurate. The individual shots are scattered around the target.
  2. The precise grouping on A is high and to the right of center. To improve accuracy, the sighting mechanism on the gun should be adjusted down and slightly left from its current position.

    To improve the results on C, someone who can produce a precise grouping must shoot the student’s target rifle. Then, the position of the sighting mechanism can be adjusted to produce an accurate shot. After adjusting the sight, the student needs more practice to improve precision.

1.9

  1. 7.5 cm. There are two significant figures in this measurement; the number of cm can be read precisely, but there is some estimating required to read tenths of a centimeter. Listing two significant figures is consistent with the convention that measured quantities are reported so that there is uncertainty in only the last digit.
  2. The speed is 72 mi/hr (inner scale, two significant figures) or 115 km/hr (outer scale, three significant figures). Both scales are read with certainty in the “hundreds” and “tens” place, and some uncertainty in the “ones” place. The km/hr speed has one more significant figure because its magnitude is in the hundreds.

1.10

  1. Volume = length × width × height. Because the operation is multiplication, the dimension with fewest significant figures (sig figs) determines the number of sig figs in the result. The dimension “2.5 cm” has 2 sig figs, so the volume is reported with 2 sig figs.
  2. Density = mass/volume. Because the operation is division, again the datum with fewer significant figures determines the number of sig figs in the result. While mass, 104.72 g, has 5 sig figs, volume [from (a)] has 2 sig figs, so density is also reported to 2 sig figs.

1.11  Given: masses of six jelly beans, mass of full jar, mass of empty jar. Find: number of jelly beans in the jar. The total mass of jelly beans is the mass of the jar full minus the mass of the jar empty. The mass of an “average” jelly bean is the average of the six masses. Then, the number of jelly beans is the total mass of beans divided by the average mass of a single bean.

Total mass of beans = 2082 g – 653 g = 1429 g

Average mass of a bean = (3.15 + 3.12 + 2.98 + 3.14 + 3.02 + 3.09) / 6 = 3.08 g

Number of beans = 1429 g total / 3.08 g per bean = 463.96 = 464 beans


By applying the significant figure rules for addition and subtraction, the total mass of beans has 0 decimal places and thus 4 significant figures. The mass of an average bean has 2 decimal places and 3 sig figs. The number of beans then has 3 sig figs, by the rules for multiplication and division. This makes sense, because we expect an integer number of beans in the jar. (Note that using an unrounded average bean mass of 3.0833 g predicts the number of beans to be 463.46, which rounds to 463 beans. The difference in these two values shows uncertainty in the last significant figure of the number of beans, as we expect in an experimental result.)

1.12  Compounds are pure substances, they have constant composition and properties throughout. The agate stone cannot be a compound, because materials with different properties appear as irregular rings in the stone. Ellen is correct.

Classification and Properties of Matter (Sections 1.2 and 1.3)

1.13

  1. heterogeneous mixture
  2. homogeneous mixture (If there are undissolved particles, such as sand or decaying plants, the mixture is heterogeneous.)
  3. pure substance
  4. pure substance

1.14

  1. homogeneous mixture
  2. heterogeneous mixture (particles in liquid)
  3. pure substance
  4. heterogeneous mixture

1.15

  1. S
  2. Au
  3. K
  4. Cl
  5. Cu
  6. uranium
  7. nickel
  8. sodium
  9. aluminum
  10. silicon

1.16

  1. C
  2. N
  3. Ti
  4. Zn
  5. Fe
  6. phosphorus
  7. calcium
  8. helium
  9. lead
  10. silver

1.17  A(s) → B(s) + C(g)

Substances A and C are definitely compounds; B is probably a compound. When solid carbon is burned in excess oxygen gas, the two elements combine to form a gaseous compound, carbon dioxide. Clearly substance C is this compound. Because C is produced when A is heated in the absence of oxygen (from air), both the carbon and the oxygen in C must have been present in A originally. A is, therefore, a compound composed of two or more elements chemically combined. Without more information on the chemical or physical properties of B, we cannot determine absolutely whether it is an element or a compound. However, few if any elements exist as white solids, so B is probably also a compound.

1.18  Gold, Au, and “fool’s gold,” FeS 2 , are similar in appearance and physical state, so these two properties are not useful for discriminating between the two substances. Density and melting point would both help determine if the nugget is gold. (For these two substances, density is easiest to measure and is definitive.)


1.19 Physical properties : silvery white (color); lustrous; melting point = 649 °C; boiling point = 1105 °C; density at 20 °C = 1.738 g/cm 3 ; pounded into sheets (malleable); drawn into wires (ductile); good conductor. Chemical properties : burns in air to give intense white light; reacts with Cl 2 to produce brittle white solid.

1.20

  1. Physical properties : melting point = 420 °C; hardness = 2.5 Mohs; density = 7.13 g/cm 3 at 25 °C. Chemical properties : granules react with dilute sulfuric acid to produce hydrogen gas; at elevated temperatures, reacts slowly with oxygen gas to produce ZnO.
  2. From the photo, we see that zinc is a shiny dark gray solid at atmospheric conditions, (physical state, color). These are physical properties.

1.21

  1. chemical
  2. physical
  3. physical
  4. chemical
  5. chemical

1.22

  1. chemical
  2. physical
  3. physical (The production of H 2 O is a chemical change, but its condensation is a physical change.)
  4. physical (The production of soot is a chemical change, but its deposition is a physical change.)

1.23  Distillation. A solution of sugar and water is a homogeneous mixture that cannot be separated by filtration. Distillation takes advantage of the much lower boiling point of water.

1.24

  1. Chemical. A substance with new chemical properties is produced when the two clear liquids are mixed. The product of physical mixing would be another colorless solution.
  2. Of the three methods, filtration is the most convenient way to separate an insoluble solid from a liquid.

The Nature of Energy (Section 1.4)

1.25

  1. Plan .  E k = 1/2 mv 2 ; m = 1200 kg; v = 18 m/s; 1 kg-m 2 /s 2 = 1 J

    Solve .  E k = 1/2 × 1200 kg × (18) 2 m 2 /s 2 = 1.944 × 10 5 = 1.9 × 10 5 J

  2. 1 cal = 4.184 J; 1.944 × 10 5 J × 1 cal 4.184 J = 4.646 × 10 4 cal = 4.6 × 10 4 cal
  3. As the automobile brakes to a stop, its speed (and hence its kinetic energy) drops to 0. This “lost” kinetic energy is mostly converted to heat. (The heat shows up in the brake parts, tires, and road.) It is not converted to some form of potential energy.

1.26

  1. Analyze .  Given: mass and speed of ball. Find: kinetic energy.

    Plan . Because 1 J = 1 kg-m 2 /s 2 convert oz to kg and mph to m/s to obtain E k in J.

    Solve . 5.13 oz × 1 lb 16 oz × 1 kg 2.205 lb = 0.14541 = 0.145 kg


  1. 95.0 mi 1 hr × 1.6093 km 1 mi × 1000 m 1 km × 1 hr 60 min × 1 min 60 sec = 42.468 = 42.5 m/s E k = 1 / 2 mv 2 = 1 / 2 × 0.14541 kg × ( 42.468 m 1 s ) 2 = 131 kg-m 2 1 s 2 = 131 J

    Check .  1/2(0.15 × 1600) ≈ 1/2(160+80) ≈ 120 J

  1. Kinetic energy is related to velocity squared (v 2 ); if the speed of the ball decreases to 55.0 mph, the kinetic energy of the ball will decrease by a factor of (55.0/95.0) 2 . (The conversion factors to m/s apply to both speeds and will cancel in the ratio.) The numerical multiplier is (55/95) 2 = 0.335. The kinetic energy decreases by approximately a factor of 3.
  1. As the ball hits the catcher’s glove, its speed (and hence its kinetic energy) drops to 0. Most of the energy is converted to heat, which is transferred to the glove and the hand inside. If the catcher’s arm (and shoulder) is considered a spring, some kineic energy is transferred to potential energy of the arm assembly, which recoils while catching the ball. [The event can also be considered an inelastic collision between the ball and the arm assembly, with the combined masses (arm+ ball) having the same kinetic energy and the arm doing work to decelerate the (arm + ball).]

1.27

  1. Kinetic energy; the particles move apart.
  2. Potential energy decreases. The greater the separation prior to release, the smaller the electrostatic repulsion and potential energy.

1.28

  1. Increases. The magnitude of the electrostatic attractive potential energy gets smaller as the distance between the particles increases, but the sign is negative. The potential energy increases because it becomes less negative. From another perspective, energy must be supplied to separate particles that are attracted to each other. This energy leads to an increase in the potential energy of the particles.
  2. Increases. Work must be done to pump water in opposition to the force of gravity. This work increases the potential energy of the water.
  3. Increases. Energy must be supplied to break a chemical bond. The potential energy of the atoms increases.

1.29 Analyze/Plan. Use results from Solution 1.4 and energy relationships discussed in Section 1.4 to solve for kinetic energy and velocity.

Solve. From Solution 1.4 we have the work required to lift the spheres. This is equal to the potential energy of the spheres at 2.2 m. As a sphere hits the floor, all potential energy is changed to kinetic energy. The kinetic energy of the Al sphere is then 11 J.

E k = 11.41 = 11 J; mass(Al) = 0.5292 = 0.529 kg; E k = 1/2 mv 2 ; v = (2 E k /m) 1/2 ; v(Al) = ( 2 × 11 .41 kg-m 2 s 2 × 1 0 .5292 kg ) 1 / 2 = 6.567 = 6.6 m/s


1.30  E k = 44.33 = 44 J; mass(Ag) = 2.056 kg = 2.06 kg; E k = 1/2 mv 2 ; v = (2 E k /m) 1/2 ; v(Ag) = ( 2 × 44.33 kg-m 2 s 2 × 1 2.056 kg ) 1 / 2 = 6.567 = 6.6 m/s

Note that calculation shows the speeds of the Al and Ag spheres at impact to be equal, as stated in Exercise 1.4(d).

Units and Measurement (Section 1.5)

1.31

  1. 1 × 10 −1
  2. 1 × 10 −2
  3. 1 × 10 −15
  4. 1 × 10 −6
  5. 1 × 10 6
  6. 1 × 10 3
  7. 1 × 10 −9
  8. 1 × 10 −3
  9. 1 × 10 −12

1.32

  1. 2.3 × 10 10 L × 1 nL 1 × 10 9 L = 0.23 nL
  2. 4.7 × 10 6 g × 1 μ g 1 × 10 6 g = 4.7 μ g
  3. 1.85 × 10 12 m × 1 pm 1 × 10 12 m = 1.85 pm
  4. 16.7 × 10 6 s × 1 Ms 1 × 10 6 s = 16.7 Ms
  5. 15.7 × 10 3 g × 1 kg 1 × 10 3 g = 15.7 kg
  6. 1.34 × 10 3 m × 1 mm 1 × 10 3 m = 1.34 mm
  7. 1.84 × 10 2 cm × 1 m 1 × 10 2 cm = 1.84 m

1.33

  1. °C = 5/9 (°F – 32 °); 5/9 (72 – 32) = 22 °C
  2. °F = 9/5 (°C) + 32 °; 9/5 (216.7) + 32 = 422.1 °F
  3. K = °C + 273.15; 233 °C + 273.15 = 506 K
  4. °C = 315 K – 273.15 = 41.85 = 42 °C; °F = 9/5 (41.85 °C) + 32 = 107 °F
  5. °C = 5/9 (°F – 32); 5/9 (2500 – 32) = 1371 = 1400 °C

    K = 1371 °C + 273.15 = 1644 = 1600 K

  6. °C = 0 K – 273.15 = –273.15 °C; °F = 9/5 (–273.15 °C) + 32 = –459.67 °F (assuming 0 K has infinite sig figs)

1.34

  1. °C = 5/9 (87 °F – 32 °) = 31 °C
  2. K = 25 °C + 273.15 = 298 K; °F = 9/5 (25 °C) + 32 = 77 °F
  3. °C = 5/9 (400 °F – 32 °) = 204.444 = 200 °C

    K = °C + 273.15 = 204.444 °C + 273.15 = 477.59 = 500 K

  4. °C = 77 K – 273.15 = –196.15 = –196 °C

    °F = 9/5 (–196.15 °C) + 32 = –321.07 = –321 °F


1.35

  1. density = mass volume = 40.55 g 25.0 mL = 1.62 g/mL or 1.62 g/cm 3

    (The units cm 3 and mL will be used interchangeably in this manual.)

    Tetrachloroethylene, 1.62 g/mL, is more dense than water, 1.00 g/mL; tetrachloroethylene will sink rather than float on water.

  2. 25.0 cm 3 × 0.469 g cm 3 = 11.7 g

1.36

  1. volume = length 3 (cm 3 ); density = mass/volume (g/cm 3 )

    volume = (1.500) 3 cm 3 = 3.375 cm 3

    density = 76.31 g 3.375 cm 3 = 22.61 g/cm 3 osmium
  2. 125.0 mL × 1 cm 3 1 mL × 4.51 g 1 cm 3 = 563.75 = 564 g titanium
  3. 0.1500 L × 1 mL 1 × 10 3 L × 0.8787 g 1 mL = 131.8 g benzene

1.37

  1. calculated density = 38.5 g 45 mL = 0.86 g/mL

    The substance is probably toluene, density = 0.866 g/mL.

  2. 45.0 g × 1 mL 1.114 g = 40.4 mL ethylene glycol
  3. A 100 mL graduated cylinder such as the one in Figure 1.21 usually has 1-mL markings. One can read the volume with certainty to the nearest mL, and estimate tenths of a mL. The volume calculated in has uncertainty in the tenths place, so a graduated cylinder like this will provide the appropriate accuracy of measurement.
  4. ( 5.00 ) 3 cm 3 × 8.90 g 1 cm 3 = 1.11 × 10 3 g ( 1.11 kg ) nickel

1.38

  1. 21 .95 g 25 .0 mL = 0 .878 g/mL

    The tabulated value has four significant figures, while the experimental value has three. The tabulated value rounded to three figures is 0.879. The values agree within one in the last significant figure of the experimental value; the two results agree. The liquid could be benzene.

  2. 15.0 g × 1 mL 0.7781 g = 19.3 mL cyclohexane

  1. r = d/2 = 5.0 cm/2 = 2.5 cm

    V = 4/3 π r 3 = 4/3 × π × (2.5) 3 cm 3 = 65.4498 = 65 cm 3

    65.4498 cm 3 × 11.34 g cm 3 = 7.4 × 10 2 g

    (The answer has two significant figures because the diameter had only two significant figures.)

1.39 36 billion metric tons × 1 × 10 9 metric tons 1 billion metric tons × 1000 kg 1 metric ton × 1000 g 1 kg = 3.6 × 10 16 g

The metric prefix for 1 × 10 15 is peta, abbreviated P.

3.6 × 10 16 g × 1 Pg 1 × 10 15 g = 36 Pg

1.40

  1. The wafers have the same diameter as the boule, so the question becomes “how many 0.75 mm wafers can be cut from the 2 m boule?”

    2.0 m boule × 1 mm 1 × 10 3 m × 1 wafer 0.75 mm = 2667 = 2.7 × 10 3 wafers

    As a practical matter, the thickness of the cutting blade reduces the actual number of disks that can be produced, so the real number is something less than 2667 wafers. Perhaps a more realistic answer, to 2 sig figs, is 2.6 × 10 3 wafers.

  2. Calculate the volume of the wafer in cm 3 . V = r 2 h

    r = d 2 = 300 mm 2 × 1 cm 10 mm = 15 cm ; h = 0.75 mm × 1 cm 10 mm = 7.5 × 10 2 cm

    V = π r 2 h = π ( 15 cm ) 2 ( 7.5 × 10 2 cm ) = 53.0144 = 53 cm 3

    Density = mass / V;    mass = density × V

    2.33 g cm 3 × 53.0144 cm 3 = 123.52 = 1.2 × 10 2 g

1.41 Analyze .  Given: heat capacity of water = 1 Btu/lb-°F Find: J/Btu

Plan . heat capacity of water = 1 cal g- C o ; cal g- C o J g- C o J lb- F o J Btu

This strategy requires changing °F to °C. Because this involves the magnitude of a degree on each scale, rather than a specific temperature, the 32 in the temperature relationship is not needed. 100 °C = 180 °F; 5 °C = 9 °F

Solve . 1 cal g- C o × 4 .184 J cal × 453 .6 g lb × 5 C o 9 F o × 1 lb- F o 1 Btu = 1054 J/Btu


1.42

  1. Analyze .  Given: 1 kwh; 1 watt = 1 J/s; 1 watt-s = 1 J. Find: conversion factor for joules and kwh.

    Plan .  kwh → wh → ws → J

    Solve . 1 kwh × 1000 w 1 kw × 60 min h × 60 s min × 1 J 1 w-s = 3.6 × 10 6 J

    1 kwh = 3.6 × 10 6 J

  2. Analyze .  Given: 100 watt bulb. Find: heat in kcal radiated by bulb or person in 24 hr.

    Plan .  1 watt = 1 J/s; 1 kcal = 4.184 × 10 3 J; watt → J/s → J → kcal. Solve .

    100 watt = 100 J 1 s × 60 sec min × 60 min hr × 24 hr × 1 kcal 4 .184 × 10 3 J = 2065 = 2.1 × 10 3 kcal

    24 hr has 2 sig figs, but 100 watt is ambiguous. The answer to 1 sig fig would be 2 × 10 3 kcal.

Uncertainty in Measurement (Section 1.6)

1.43  Exact: (b), (d), and (f) (All others depend on measurements and standards that have margins of error, e.g., the length of a week as defined by the Earth’s rotation.)

1.44  Exact: (b), (e) (The number of students is exact on any given day.)

1.45

  1. 3
  2. 2
  3. 5
  4. 3
  5. 5
  6. 1 [See Sample Exercise 1.7 (c)]

1.46

  1. 4
  2. 3
  3. 4
  4. 5
  5. 6
  6. 2

1.47

  1. 1.025 × 10 2
  2. 6.570 × 10 2
  3. 8.543 × 10 −3
  4. 2.579 × 10 −4
  5. –3.572 × 10 −2

1.48

  1. 7.93 × 10 3 mi
  2. 4.001 × 10 4 km

1.49

  1. 14.3505 + 2.65 = 17.0005 = 17.00 (For addition and subtraction, the minimum number of decimal places, here two, determines decimal places in the result.)
  2. 952.7 – 140.7389 = 812.0
  3. (3.29 × 10 4 )(0.2501) = 8.23 × 10 3 (For multiplication and division, the minimum number of significant figures, here three, determines sig figs in the result.)
  4. 0.0588/0.677 = 8.69 × 10 −2

1.50

  1. [320.5 – 6104.5/2.3] = –2.3 × 10 3 (The intermediate result has two significant figures, so only the thousand and hundred places in the answer are significant.)
  2. [285.3 × 10 5 – 0.01200 × 10 5 ] × 2.8954 = 8.260 × 10 7 (Because subtraction depends on decimal places, both numbers must have the same exponent to determine decimal places/sig figs. The intermediate result has 1 decimal place and 4 sig figs, so the answer has 4 sig figs.)

  1. ( 0.0045 × 20,000 .0 ) + ( 2813 × 12 ) = 3.4 × 10 4 2 sig figs /0 dec pl 2 sig figs/first 2 digits
  1. 863 × [ 1255 ( 3.45 × 108 ) ] = 7.60 × 10 5 ( 3 sig figs /0 dec pl) 3 sig figs × [0 dec pl/3 sig figs] = 3 sig figs

1.51  The mass 21.427 g has 5 significant figures.

1.52  The volume in the graduated cylinder is 19.5 mL. Liquid volumes are read at the bottom of the meniscus, so the volume is slightly less than 20 mL. Volumes in this cylinder can be read with certainty to 1 mL, and with some uncertainty to 0.1 mL, so this measurement has 3 sig figs.

Dimensional Analysis (Section 1.7)

1.53  In each conversion factor, the old unit appears in the denominator, so it cancels, and the new unit appears in the numerator.

  1. mm nm : 1 × 10 3 m 1 mm × 1 nm 1 × 10 9 m = 1 × 10 6 nm/mm
  2. mg kg : 1 × 10 3 g 1 mg × 1 kg 1000 g = 1 × 10 6 kg/mg
  3. km ft : 1000 m 1 km × 1 cm 1 × 10 2 m × 1 in 2.54 cm × 1 ft 12 in = 3.28 × 10 3 km/ft
  4. in 3 cm 3 : ( 2.54 ) 3 cm 3 1 3 in 3 = 16.4 cm 3 /in 3

1.54  In each conversion factor, the old unit appears in the denominator, so it cancels, and the new unit appears in the numerator.

  1. μm mm : 1 × 10 6 m 1 μm × 1 mm 1 × 10 3 m = 1 × 10 3 mm/ μm
  2. ms ns : 1 × 10 3 s 1 ms × 1 ns 1 × 10 9 s = 1 × 10 6 ns/ms
  3. mi → km: 1.6093 km/mi
  4. ft 3 L : ( 1 2 ) 3 in 3 1 ft 3 × ( 2 . 5 4 ) 3 cm 3 1 in 3 × 1 L 1 0 0 0 cm 3 = 2 8 . 3 L/ ft 3

1.55

  1. 15.2 m s × 1 km 1000 m × 60 s 1 min × 60 min 1 hr = 54.7 km/hr
  2. 5.0 × 10 3 L × 1 gal 3.7854 L = 1.3 × 10 3 gal

  1. 151 ft × 1 yd 3 ft × 1 m 1.0936 yd = 46.025 = 46.0 m
  1. 60.0 cm d × 1 in 2.54 cm × 1 d 24 hr = 0.984 in/hr

1.56

  1. 2.998 × 10 8 m s × 1 km 1000 m × 1 mi 1.6093 km × 60 s 1 min × 60 min 1 hr = 6.707 × 10 8 mi/hr
  2. 1454 ft × 1 yd 3 ft × 1 m 1.0936 yd = 443.18 = 443.2 m
  3. 3 , 666 , 500 m 3 × 1 3 dm 3 ( 1 × 10 1 ) 3 m 3 × 1 L 1 dm 3 = 3.6665 × 10 9 L
  4. 242 mg cholesterol 100 mL blood × 1 mL 1 × 10 3 L × 5.2 L × 1 × 10 3 g 1 mg = 12.58 = 13 g cholesterol

1.57

  1. 5.00 days × 24 hr 1 day × 60 min 1 hr × 60 s 1 min = 4.32 × 10 5 s
  2. 0.0550 mi × 1.6093 km mi × 1000 m 1 km = 88.5 m
  3. $ 1.89 gal × 1 gal 3.7854 L = $ 0.499 L
  4. 0.510 in ms × 2.54 cm 1 in × 1 × 10 2 m 1 cm × 1 km 1000 m × 1 ms 1 × 10 3 s × 60 s 1 min × 60 min 1 hr = 46.6 km hr

    Estimate: 0.5 × 2.5 = 1.25; 1.25 × 0.01 ≈ 0.01; 0.01 × 60 × 60 ≈ 36 km/hr

  5. 22.50 gal min × 3.7854 L gal × 1 min 60 s = 1.41953 = 1.420 L/s

    Estimate: 20 × 4 = 80; 80/60 ≈ 1.3 L/s

  6. 0.02500 ft 3 × 12 3 in 3 1 ft 3 × 2.54 3 cm 3 1 in 3 = 707.9 cm 3

    Estimate: 10 3 = 1000; 3 3 = 27; 1000 × 27 = 27,000; 27,000/0.04 ≈ 700 cm 3

1.58

  1. 0.105 in × 2.54 cm in × 1 × 10 2 m cm × 1 mm 1 × 10 3 m = 2.667 = 2.67 mm
  2. 0.650 qt × 1 L 1.057 qt × 1 mL 1 × 10 3 L = 614.94 = 615 mL

  1. 8.75 μm s × 1 × 10 6 m 1 μm × 1 km 1 × 10 3 m × 60 s 1 min × 60 min 1 hr = 3.15 × 10 5 km/hr
  1. 1.955 m 3 × ( 1.0936 ) 3 yd 3 1 m 3 = 2.55695 = 2.557 yd 3
  1. $ 3.99 lb × 2.205 lb 1 kg = 8.798 = $ 8.80 /kg
  1. 8.75 lb ft 3 × 453.59 g 1 lb × 1 ft 3 12 3 in 3 × 1 in 3 2.54 3 cm 3 × 1 cm 3 1 mL = 0.140 g/mL

1.59

  1. 31 gal × 4 qt 1 gal × 1 L 1.057 qt = 1.2 × 10 2 L

    Estimate: (30 × 4)/1 ≈ 120 L

  2. 6 mg kg (body) × 1 kg 2.205 lb × 185 lb = 5 × 10 2 mg

    Estimate: 6/2 = 3; 3 × 180 = 540 mg

  3. 400 km 47.3 L × 1 mi 1.6093 km × 1 L 1.057 qt × 4 qt 1 gal = 19.9 mi gal

    (2 × 10 1 mi/gal for 1 sig fig)

    Estimate: 400/50 = 8; 8/1.6 = 5; 5/1 = 5; 5 × 4 ≈ 20 mi/gal

  4. 200 cups × 1 lb 50 cups × 1 kg 2.205 lb = 1.81 kg

    (2 kg for 1 sig fig)

    Estimate: 1 lb = 50 cups, 4 lb = 200 cups; 4 lb ≈ 2 kg

1.60

  1. 1 2 5 7 m i × 1 k m 0 . 6 2 1 3 7 m i × c h a r g e 2 2 5 k m = 8 . 9 9 c h a r g e s

    Because charges are integral events, 9 total charges are required. The trip begins with a full charge, so 8 additional charges during the trip are needed.

  2. 14 m s × 1 km 1 × 10 3 m × 1 mi 1.6093 km × 60 s 1 min × 60 min 1 hr = 31 mi/hr
  3. 450 in 3 × ( 2.54 ) 3 cm 3 1 in 3 × 1 mL 1 cm 3 × 1 × 10 3 L 1 mL = 7.37 L
  4. 2.4 × 10 5 barrels × 42 gal 1 barrel × 4 qt 1 gal × 1 L 1.057 qt = 3.8 × 10 7 L

1.61  14.5 ft × 16.5 ft × 8.0 ft = 1914 = 1.9 × 10 3 ft 3 (2 sig figs)

1914 ft 3 × ( 1 yd ) 3 ( 3 ft ) 3 × ( 1 m ) 3 ( 1.0936 ) 3 yd 3 × 10 3 dm 3 1 m 3 × 1 L 1 dm 3 × 1.19 g L × 1 kg 1000 g = 64.4985 = 64 kg air

Estimate: 1900/27 ≈ 60; (60 × 1)/1 ≈ 60 kg

1.62  10.6 ft × 14.8 ft × 20.5 ft = 3216.04 = 3.22 × 10 3 ft 3

3216.04 ft 3 × ( 1 yd ) 3 ( 3 ft ) 3 × ( 1 m ) 3 ( 1.0936 yd ) 3 × 48 μg Co 1 m 3 × 1 × 10 6 g 1 μg = 4.4 × 10 3 g Co

1.63  Strategy:

  1. Calculate volume of gold (Au) in cm 3 in the sheet
  2. Mass = density × volume
  3. Change g troy oz and $

100 ft × 82 ft × ( 12 ) 2 in 2 1 ft 2 × 5 × 10 6 in × ( 2.54 ) 3 cm 3 1 in 3 = 96.75 = 1 × 10 2 cm 3 Au

96.75 cm 3 Au × 19.32 g 1 cm 3 × 1 troy oz 31.1034768 g × $ 1654 troy oz = $ 99 , 399 = $ 1 × 10 5

(Strictly speaking, the datum 100 ft has 1 sig fig, so the result has 1 sig fig.)

1.64  A wire is a very long, thin cylinder of volume, V = π r 2 h, where h is the length of the wire and π r 2 is the cross-sectional area of the wire.

Strategy:

  1. Calculate total volume of copper in cm 3 from mass and density
  2. h (length in cm) = V π r 2
  3. Change cm → ft

150 lb Cu × 453.6 g 1 lb Cu × 1 cm 3 8.94 g = 7610.7 = 7.61 × 10 3 cm 3

r = d/ 2 = 7.50 mm × 1 cm 10 mm × 1 2 = 0.375 cm

h = V π r 2 = 7610.7 cm 3 π ( 0.375 ) 2 cm 2 = 1.7227 × 10 4 = 1.72 × 10 4 cm

1.7227 × 10 4 cm × 1 in 2.54 cm × 1 ft 12 in = 565 ft

(too difficult to estimate)


Additional Exercises

1.65

  1. A gold ingot is pure metallic gold, a pure substance.
  2. A cup of coffee is a solution if there are no suspended solids (coffee grounds). It is a heterogeneous mixture if there are grounds. If cream or sugar is added, the homogeneity of the mixture depends on how thoroughly the components are mixed.
  3. A wood plank is a heterogeneous mixture of various cellulose components. The different domains in the mixture are visible as wood grain or knots.

1.66

  1. A hypothesis is more likely to eventually be shown to be incorrect. A hypothesis is a possible explanation based on preliminary experimental data. A theory may be more general, and has a significant body of experimental evidence to support it.
  2. A law reliably predicts the behavior of matter, while a theory provides an explanation for that behavior.

1.67  According to the law of constant composition, any sample of vitamin C has the same relative amount of carbon and oxygen; the ratio of oxygen to carbon in the isolated sample is the same as the ratio in synthesized vitamin C.

2 .00 g O 1 .50 g C = x g O 6 .35 g C ; x = (2 .00 g O) (6 .35 g C) 1 .50 g C = 8.47 g O

1.68

  1. (l) → (g)
  2. °F = 9/5 (°C) + 32 °; 9/5 (12) + 32 = 53.6 = 54 °F
  3. 0.765 g/mL × 103.5 mL = 79.178 = 79.2 g ethyl chloride

1.69

    1. (22.52 + 22.48 + 22.54)/3 = 22.51
    2. (22.64 + 22.58 + 22.62)/3 = 22.61

    Based on the average, set I is more accurate. That is, it is closer to the true value of 22.52%.

  1. Average deviation = Σ | v a l u e - a v e r a g e | / 3
    1. | 22.52 – 22.51 | + |22.48 – 22.51 | + |22.54 – 22.51 |/3 = 0.02
    2. | 22.64 – 22.61 | + |22.58 – 22.61 | + |22.62 – 22.61 |/3 = 0.02

    Based on average deviations, the two sets display the same precision, even though set I is more accurate. [According to Section 1.5, standard deviation is a measure that is often used to determine precision. Using the formula for calculating standard deviation given in Appendix A.5, the values for the two sets are 0.03 and 0.03, respectively. The standard deviations of the two sets are also the same, confirming that the two sets are equally precise.]

1.70

  1. Inappropriate. The circulation number indicates a precision of one part per million, 1 ppm. First, the term “circulation” is not specific, and can include paid subscriptions, news stand sales and other means of distribution. Even if we consider paid subscriptions only, it is unlikely that the number of paid subscriptions is know to a precision of 1 ppm.

  1. Inappropriate. The population number implies a precision of 1 part per million, 1 ppm. Census data are taken only once every ten years and cannot directly measure every citizen even in the year the census is taken. Population fluctuates daily, so this precision is not reasonable.
  1. Appropriate. The percentage has three significant figures. In a population as large as that of the United States, the number of people named Brown can surely be counted by census data or otherwise to a precision of three significant figures.
  1. Inappropriate. Letter grades are posted at most to two decimal places and three significant figures (if plus and minus modifiers are quantified). The grade-point-average, obtained by addition and division, cannot have more decimal places or significant figures than the numbers being averaged.

1.71

  1. volume
  2. area
  3. volume
  4. density
  5. time
  6. length
  7. temperature

1.72

  1. m s 2
  2. k g - m s 2
  3. k g - m s 2 × m = k g - m 2 s 2
  4. kg-m s 2 × 1 m 2 = kg m-s 2
  5. kg-m 2 s 2 × 1 s = kg-m 2 s 3
  6. m s
  7. kg × ( m s ) 2 = kg-m 2 s 2

1.73

  1. 2.4 × 10 5 mi × 1 .609 km 1 mi × 1000 m 1 km = 3.862 × 10 8 = 3.9 × 10 8 m
  2. 2.4 × 10 5 mi × 1.609 km 1 mi × 1 hr 350 km × 60 min 1 hr × 60 s 1 min = 4.0 × 10 6 s
  3. 3.862 × 10 8 m × 2 × 1 s 3.00 × 10 8 m = 2.574 = 2.6 s
  4. 29 .783 km s × 1 mi 1 .6093 km × 60 s 1 min × 60 min 1 hr = 6.6624 × 10 4 mi/hr

1.74

  1. Baking powder is not a pure substance. It is a mixture of basic and acidic compounds that, in the presence of water, react to form CO 2 which causes baked goods to rise.
  2. Lemon juice is not a pure substance. It is a mixture of water, citric acid, and other natural flavors. Its exact composition depends on the characteristics of the lemon that produces the juice.
  3. Propane is a nearly pure substance. Propane itself is odorless, but tank propane contains trace amounts of an odor-producing substance so that leaks can easily be detected.
  4. Aluminum foil is a nearly pure substance.
  5. Ibuprofen itself is a pure substance. Ibuprofen tablets probably contain the active ingredient and other binders, drying agents and coatings.

  1. Bourbon whiskey is not a pure substance. It is a mixture of water, alcohol, and flavoring compounds.
  1. Helium gas is a pure substance.
  1. Clear water pumped from a deep aquifer is a nearly pure substance. All natural water (not distilled or deionized) contains trace minerals.

1.75

  1. 575 ft × 12 in 1 ft × 2 .54 cm 1 in × 10 mm 1 cm × 1 quarter 1 .55 mm = 1.1307 × 10 5 = 1.13 × 10 5 quarters
  2. 1.1307 × 10 5 quarters × 5.67 g 1 quarter = 6.41 × 10 5 g (641 kg)
  3. 1.1307 × 10 5 quarters × 1 dollar 4 quarters = $ 28 , 268 = $ 2.83 × 10 4
  4. $ 16 , 213 , 166 , 914 , 811.11 × 1 stack $28,268 = 5.7355 × 10 8 = 5.74 × 10 8 stacks

1.76

  1. $1950 acre-ft × 1 acre 4840 yd 2 × 3 ft 1 yd × (1 .094 yd) 3 (1 m) 3 × (1 m) 3 (10 dm) 3 × (1 dm) 3 1 L = $ 1.583 × 10 3 / L or 0.1583 ¢ /L   (0 .158 ¢ /L to 3 sig figs)
  2. $1950 acre-ft × 1 acre-ft 2 households-year × 1 year 365 days × 1 household = $2 .671 day = $2 .67 day

1.77  Select a common unit for comparison, in this case the kg.

1 kg > 2 lb, 1 L ≈ 1 qt

5 lb potatoes < 2.5 kg

5 kg sugar = 5 kg

1 gal = 4 qt ≈ 4 L; 1 mL H 2 O = 1 g H 2 O; 1 L = 1000 g, 4 L = 4000 g = 4 kg

The order of mass from lightest to heaviest is 5 lb potatoes < 1 gal water < 5 kg sugar.

1.78  There are 347 degrees between the freezing and boiling points on the oleic acid (O) scale and 100 degrees on the celsius (C) scale. Also, 13 °C = 0 °O. By analogy with °F and °C,

O o = 100 347 ( C o 13 ) or C o = 347 100 ( O o ) + 13

These equations correctly relate the freezing point (and boiling point) of oleic acid on the two scales.

f .p . of H 2 O: O o = 100 347 ( 0 C o 13 ) = 3.746 = 4 O o


1.79  The most dense liquid, Hg, will sink; the least dense, cyclohexane, will float; H 2 O will be in the middle.

image

1.80  Density is the ratio of mass and volume. For samples with the same volume, in this case spheres with the same diameter, the denser ball will have a greater mass. The heavier ball, the red one on the right in the diagram is more dense.

1.81  The mass of water in the bottle does not change with temperature, but the density (ratio of mass to volume) does. That is, the amount of volume occupied by a certain mass of water changes with temperature. Calculate the mass of water in the bottle at 25 °C, and then the volume occupied by this mass at −10 °C.

  1. 25 o C : 1.50 L H 2 O × 1000 cm 3 1 L × 0 .997 g H 2 O 1 cm 3 = 1.4955 × 10 3 = 1.50 × 10 3 g H 2 O

    −10 °C: 1.4955 × 10 3 g H 2 O × 1 c m 3 0 . 9 1 7 g H 2 O × 1 L 1 0 0 0 c m 3 = 1.6309 = 1.63 L

  2. No. If the soft-drink bottle is completely filled with 1.50 L of water, the 1.63 L of ice cannot be contained in the bottle. The extra volume of ice will push through any opening in the bottle, or crack the bottle to create an opening.

1.82  mass of toluene = 58.58 g – 32.65 g = 25.93 g

volume of toluene = 25.93 g × 1 mL 0 .864 g = 30.0116 = 30.0 mL

volume of solid = 50.00 mL – 30.0116 mL = 19.9884 = 20.0 mL

density of solid = 32 .65 g 19 .9884 mL = 1.63 g/mL

1.83  V = 4/3 π r 3 = 4/3 π (28.9 cm) 3 = 1.0111 × 10 5 = 1.01 × 10 5 = 1.01 × 10 5 cm 3

1.0111 × 10 5 cm 3 × 19 .3 g cm 3 × 1 lb 453 .59 g = 4302 = 4.30 × 10 3 lb

No, the thief is unlikely to be able to carry the sphere without assistance. It weighs 4300 pounds, more than two tons.


1.84 1.00 gal battery acid × 4 qt 1 gal × 1000 mL 1 .0567 qt × 1 .28 g mL = 4845.3 = 4.85 × 10 3 g battery acid

4.8453 × 10 3 g battery acid × 38 .1 g sulfuric acid 100 g battery acid = 1846 = 1.85 × 10 3 g sulfuric acid

1.85

  1. 40 lb peat 14 × 20 × 30 in 3 × 1 in 3 ( 2.54 ) 3 cm 3 × 453.6 g 1 lb = 0.13 g/cm 3 peat

    40 lb soil 1 .9 gal × 1 gal 4 qt × 1 .057 qt 1 L × 1 × 10 3 L 1 mL × 1 mL 1 cm 3 × 453.6 g 1 lb = 2.5 g/cm 3 soil

    No. Volume must be specified to compare mass. The densities tell us that a certain volume of peat moss is “lighter” (weighs less) than the same volume of top soil.

  2. 1 bag peat = 14 × 20 × 30 = 8.4 × 10 3 in 3

    15.0 ft × 20.0 ft × 3.0 in × 12 2 in 2 ft 2 = 129 , 600 = 1.3 × 10 5 in 3 peat needed

    129 , 600 in 3 × 1 bag 8.4 × 10 3 in 3 = 15.4 = 15 bags (Buy 16 bags of peat .)

1.86 8.0 oz × 1 lb 16 oz × 453.6 g lb × 1 cm 3 2.70 g = 84.00 = 84 cm 3

84 cm 3 50 ft 2 × 1 2 ft 2 12 2 in 2 × 1 2 in 2 2 .54 2 cm 2 × 10 mm 1 cm = 0.018 mm

1.87  The total solar flux, the power provided by the sun, is the average solar flux times the area of the disc in sunlight at any instant. However, because the earth rotates, each disc is in sunlight for only half of the day, so only half of this flux can be collected.

680 W m 2 × 1.28 × 10 14 m 2 × 1/ 2 = 4.352 × 10 16 = 4.4 × 10 16 W

Collection is 10% efficient, so only 10% of this power is available, 4.4 × 10 15 W

15 TW × 1 × 10 12 W 1 TW = 15 × 10 12 W needed

15 × 10 12 W needed 4.352 × 10 15 W available × 100 = 0.3447 = 0.34 % Earth’s surface needed

1.88  Warren’s first hypothesis was that the bacterium he observed was involved in causing ulcers. This was a radical hypothesis in 1979, when it was commonly accepted that the acidic environment of the stomach would not support microorganisms. Strong evidence was needed to change the opinion and practice of the medical community.


To test his hypothesis, Warren first collected many more tissue samples, from patients with and without ulcers. He established that healthy tissue did not contain the bacterium, while many of the ulcerated samples did. From there, with Marshall and other collaborators, he established two different ways to diagnose the infection (determine the presence of the bacterium in patients with potential ulcers), and learned to culture (grow in the laboratory) the new bacterium so that its effects could be studied. Finally, in patients diagnosed and treated for bacterial ulcers, he demonstrated that the recurrence of ulcers was rare.

This information was obtained from the autobiography of J. Robin Warren, http://www.nobelprize.org/nobel_prizes/medicine/laureates/2005/warren-autobio.php

1.89 45.23 g ethanol × 1 cm 3 0.789 g ethanol = 57.3257 = 57.33 cm 3 , volume of cylinder

V = πr 2 h ; r = ( V / πh ) 1 / 2 = [ 57.3257 cm 3 π × 25.0 cm ] = 0.854338 = 0.854 cm d = 2 r = 1.71 cm

1.90

  1. Let x = mass of Au in jewelry

    9.85 – x = mass of Ag in jewelry

    The total volume of jewelry = volume of Au + volume of Ag

    0.675 cm 3 = x g × 1 cm 3 19.3 g + ( 9.85 x ) g × 1 cm 3 10.5 g

    0.675 = x 19 .3 + 9.85 x 10.5 (To solve, multiply both sides by ( 19.3 ) ( 10.5 ) )

    0.675 ( 19.3 ) ( 10.5 ) = 10.5 x + ( 9.85 x ) ( 19.3 ) 136.79 = 10.5 x + 190.105 19.3 x 53.315 = 8.8 x

    x = 6.06 g Au; 9.85 g total – 6.06 g Au = 3.79 g Ag

    mass % Au = 6.06 g Au 9.85 g jewelry × 100 = 61.5 % Au

  2. 24 carats × 0.615 = 15 carat gold

1.91  The separation with distinctly separated red and blue spots is more successful. The procedure that produced the purple blur did not separate the two dyes. To quantify the characteristics of the separation, calculate a reference value for each spot that is

distance traveled by spot distance traveled by solvent

If the values for the two spots are fairly different, the separation is successful. (One could measure the distance between the spots, but this would depend on the length of paper used and be different for each experiment. The values suggested above are independent of the length of paper.)


1.92

  1. False. Air and water are not elements. Air is a homogeneous mixture of gases and pure water is a compound.
  2. False. Mixtures can contain any number of pure substances, either elements or compounds or both.
  3. True.
  4. True.
  5. False. When yellow stains in a kitchen sink are treated with bleach water, a chemical change occurs.
  6. True.
  7. False. The number 0.0033 has the same number of significant figures as 0.033.
  8. True. (In a conversion factor, the quantity in the numerator is equal to the quantity in the denominator, so the overall numerical value is one.)
  9. True.

1.93  The densities are:

carbon tetrachloride (methane, tetrachloro) – 1.5940 g/cm 3

hexane – 0.6603 g/cm 3

benzene – 0.87654 g/cm 3

methylene iodide (methane, diiodo) – 3.3254 g/cm 3

Only methylene iodide will separate the two granular solids. The undesirable solid (2.04 g/cm 3 ) is less dense than methylene iodide and will float; the desired material is more dense than methylene iodide and will sink. The other three liquids are less dense than both solids and will not produce separation.

1.94

  1. 10.0 mg × 1 × 10 3 g 1 mg × 1 cm 3 0.20 g = 0.050 cm 3 = 0.050 mL volume
  2. 10.0 mg × 1 × 10 3 g 1 mg × 1242 m 2 1 g = 12.42 = 12.4 m 2 surface area
  3. 7.748 mg Hg initial – 0.001 mg Hg remain = 7.747 mg Hg removed

    7.747 mg Hg removed 7.748 mg Hg initial × 100 = 99.99 % Hg removed

  4. 10.0 mg “spongy” initial + 7.747 mg Hg removed = 17.747 = 17.7 mg after exposure

 

2 Atoms, Molecules, and Ions

Visualizing Concepts

2.1

  1. Like charges repel and opposite charges attract, so the sign of the electrical charge on the particle is negative.
  2. The greater the magnitude of the charges, the greater the electrostatic repulsion or attraction. As the charge on the plates is increased, the bending will increase.
  3. As the mass of the particle increases and speed stays the same, linear momentum (mv) of the particle increases and bending decreases. (See A Closer Look : The Mass Spectrometer.)

2.2

  1. % abundance = # of mass number × particles total number of particles × 100

    12 red 293 Nv particles

    8 blue 295 Nv particles

    20 total particles

    % abundance 293 Nv = 12 20 × 100 = 60 %

    % abundance 295 Nv = 8 20 × 100 = 40 %

  2. Atomic weight (AW) is the same as average atomic mass.

    Atomic weight (average atomic mass) = Σ fractional abundance mass of isotope

    AW of Nv = 0.60(293.15) + 0.40(295.15) = 293.95 amu

    (Because % abundance was calculated by counting exact numbers of particles, assume % abundance is an exact number. Then, the number of significant figures in the AW is determined by the number of sig figs in the masses of the isotopes.)

2.3   In general, metals occupy the left side of the chart, and nonmetals the right side.

metals : red and green nonmetals : blue and yellow

alkaline earth metal : red noble gas : yellow


2.4   Because the number of electrons (negatively charged particles) does not equal the number of protons (positively charged particles), the particle is an ion. The charge on the ion is 2–.

Atomic number = number of protons = 16. The element is S, sulfur.

Mass number = protons + neutrons = 32

16 32 S 2

2.5   In a solid, particles are close together and their relative positions are fixed. In a liquid, particles are close but moving relative to each other. In a gas, particles are far apart and moving. Most ionic compounds are solids because of the strong forces among charged particles. Molecular compounds can exist in any state: solid, liquid, or gas.

Because the molecules in ii are far apart, ii must be a molecular compound. The particles in i are near each other and exist in a regular, ordered arrangement, so i is likely to be an ionic compound.

2.6   Formula: IF 5 Name: iodine pentafluoride

Because the compound is composed of elements that are all nonmetals, it is molecular.

2.7   See Figure 2.17. yellow box: 1+ (group 1A); blue box: 2+ (group 2A)

black box: 3+ (a metal in Group 3A); red box: 2– (a nonmetal in group 6A);

green box: 1– (a nonmetal in group 7A)

2.8   Cations (red spheres) have positive charges; anions (blue spheres) have negative charges. There are twice as many anions as cations, so the formula has the general form CA 2 . Only Ca(NO 3 ) 2 , calcium nitrate, is consistent with the diagram.

2.9   These two compounds are isomers. They have the same chemical formula, C 4 H 9 Cl, but different arrangements of atoms. That is, they have different chemical structures. In the first isomer, the Cl atom is bound to the second C atom from the left. In the second isomer, the Cl atom is bound to the right-most C atom.

2.10

  1. In the absence of an electric field, there is no electrostatic interaction between the oil drops and the apparatus, so the rate of fall of the oil drops is determined solely by the force of gravity. In the presence of an electric field, there is electrostatic attraction between the negatively charged oil drops and the positively charged plate, as well as electrostatic repulsion between the negatively charged oil drops and the negative plate. These electrostatic forces oppose the force of gravity and change the rate of fall of the drops.
  2. Each individual drop has a different number of electrons associated with it. The greater the accumulated negative charge on the drop, the greater the electrostatic forces between the oil drop and the plates. If the combined electrostatic forces are greater than the force of gravity, the drop moves up.

The Atomic Theory of Matter and the Discovery of Atomic Structure (Sections 2.1 and 2.2)

2.11

  1. ratio of masses = 0.727 g O 0.273 g C = 2.663 = 2.66
  2. ratio of masses = 0.571 g O 0.429 g C = 1.331 = 1.33
  3. The two mass ratios are related by a factor of 2. In the first compound, CO 2 , twice as much O is bound to one gram of C as in the second compound. The empirical formula of the second compound is then CO.

2.12

  1. 6.500 g compound – 0.384 g hydrogen = 6.116 g sulfur
  2. Conservation of mass

2.13

  1. 17 .60 g oxygen 30 .82 g nitrogen = 0 .5711 g O 1 g N ; 0 .5711/0 .5711 = 1 .0

    35 .20 g oxygen 30 .82 g nitrogen = 1 .142 g O 1 g N ; 1 .142/0 .5711 = 2 .0

    70 .40 g oxygen 30 .82 g nitrogen = 2 .284 g O 1 g N ; 2 .284/0 .5711 = 4 .0

    88 .00 g oxygen 30 .82 g nitrogen = 2 .855 g O 1 g N ; 2 .855/0 .5711 = 5 .0

  2. These masses of oxygen per one gram nitrogen are in the ratio of 1:2:4:5 and thus obey the law of multiple proportions . Multiple proportions arise because atoms are the indivisible entities combining, as stated in Dalton’s theory. Because atoms are indivisible, they must combine in ratios of small whole numbers.

2.14

  1. 1: 3 .56 g fluorine 4 .75 g iodine = 0 .749 g fluorine/1 g iodine

    2: 3 .43 g fluorine 7 .64 g iodine = 0 .449 g fluorine/1 g iodine

    3: 9 .86 g fluorine 9 .41 g iodine = 1 .05 g fluorine/1 g iodine

  2. To look for integer relationships among these values, divide each one by the smallest. If the quotients aren’t all integers, multiply by a common factor to obtain all integers.

    1: 0.749/0.449 = 1.67; 1.67 × 3 = 5

    2: 0.449/0.449 = 1.00; 1.00 × 3 = 3

    3: 1.05/0.449 = 2.34; 2.34 × 3 = 7

    The ratio of g fluorine to g iodine in the three compounds is 5:3:7. These are in the ratio of small whole numbers and, therefore, obey the law of multiple proportions . This integer ratio indicates that the combining fluorine “units” (atoms) are indivisible entities.


2.15   Electrons, in the form of cathode rays, were discovered first. Neutrons were discovered last.

2.16

  1. Beta rays have the mass and charge of an electron. If the unknown particle is a proton, it will be deflected in the opposite direction as a beta ray because protons and beta rays have opposite electrical charges.
  2. In an electric field, lighter particles are deflected by a greater amount than heavier ones. Protons have larger mass than beta rays (electrons) so they would be deflected by a smaller amount.

2.17 Analyze . We are given the diameters of a gold atom and its nucleus, and a gold foil that is two atoms thick. What fraction of alpha particles in Rutherford’s experiment are deflected at large angles?

Plan . In order to be deflected at a large angle, an alpha particle must directly strike a gold nucleus. Assume that the gold atoms in a single row touch. Consider the cross-sectional area of the gold foil exposed to the beam of alpha particles. Calculate the percentage of this area occupied by the nucleus. But, there are two rows of gold particles, offset relative to one another (Figure 2.9). Assume each alpha particle has two chances to hit a gold nucleus, so the fraction deflected at large angles is twice the ratio of areas. [This approach ignores empty space in the arrangement of gold atoms, which is about 9% of the total cross-sectional area.]

Solve .

fraction of alpha particles deflected at large angles = area of Au nucleus area of Au atom × 2

The cross-sectional area of a spherical atom is a circle. Area = πr 2

fraction deflected at large angles = π [ r (nucleus)] 2 π [ r (atom) ] 2 × 2

fraction deflected at large angles = ( 1.0 × 10 4 Å ) 2 ( 2.7 Å ) 2 × 2 = 2.7 × 10 9

That is, 1 out of approximately 365 million alpha particles is deflected at a large angle.

2.18

  1. The droplets carry different total charges because there may be 1, 2, 3, or more electrons on the droplet.
  2. The electronic charge is likely to be the lowest common factor in all the observed charges.
  3. Assuming this is so, we calculate the apparent electronic charge from each drop as follows:

    A:  1.60 × 10 –19 / 1 = 1.60 × 10 –19 C

    B:  3.15 × 10 –19 / 2 = 1.58 × 10 –19 C

    C:  4.81 × 10 –19 / 3 = 1.60 × 10 –19 C

    D:  6.31 × 10 –19 / 4 = 1.58 × 10 –19 C

    The reported value is the average of these four values. Because each calculated charge has three significant figures, the average will also have three significant figures.

    (1.60 × 10 –19 C + 1.58 × 10 –19 C + 1.60 × 10 –19 C + 1.58 × 10 –19 C) / 4 = 1.59 × 10 –19 C


The Modern View of Atomic Structure; Atomic Weights (Sections 2.3 and 2.4)

2.19

  1. 1.35 Å × 1 × 10 10 m 1 Å × 1 nm 1 × 10 9 m = 0.135 nm 1.35 Å × 1 × 10 10 m 1 Å × 1 pm 1 × 10 12 m = 1.35 × 10 2 or 135pm (1 Å = 100 pm)
  2. Aligned Au atoms have diameters touching. d = 2r = 2(1.35 Å) = 2.70 Å

    1.0 mm × 1 m 1000 mm × 1 Å 1 × 10 10 m × 1 Au atom 2.70 Å = 3.70 × 10 6 Au atoms

  3. V = 4 / 3 πr 3 · r = 1 . 35 Å × 1 × 10 - 10 m 1 Å × 100 cm m = 1 . 35 × 10 - 8 cm V = ( 4 / 3 ) ( π ) ( 1 . 35 × 10 - 8 ) 3 cm 3 = 1 . 03 × 10 - 23 cm 3

2.20

  1. r = d/2;r = 2.7 × 10 8 cm 2 × 1 A 1 × 10 8 cm = 1.35 = 1.4 A r = 2.7 × 10 8 cm 2 × 1 m 100 cm = 1.35 × 10 10 = 1.4 × 10 10 m
  2. Aligned Rh atoms have diameters touching. d = 2.7 × 10 –8 cm = 2.7 × 10 –10 m

    6.0 μm × 1 × 10 6 m 1 μm × 1 Rh atom 2.7 × 10 10 m = 2.2 × 10 4 Rh atom

  3. V = 4/3 π r 3 ; r = 1.35 × 10 –10 = 1.4 × 10 –10 m

    V = (4/3)[(π(1.35 × 10 –10 ) 3 ] m 3 = 1.031 × 10 –29 = 1.0 × 10 –29 m 3

2.21

  1. proton, neutron, electron
  2. proton = +1, neutron = 0, electron = –1
  3. The neutron is most massive. (The neutron and proton have very similar masses.)
  4. The electron is least massive.

2.22

  1. False. The nucleus has most of the mass but occupies very little of the volume of an atom.
  2. True.
  3. False. The number of electrons in a neutral atom is equal to the number of protons in the atom.
  4. True.

2.23

  1. 5 protons, 5 neutrons, 5 electrons. Every neutral 10 B atom has 5 protons and 5 neutrons. The mass number of this B atom is 10, so it has (10 – 5) = 5 neutrons.
  2. 6 11 C . Adding a proton increases the atomic number of the atom to 6 and the mass number to 11. The element with atomic number 6 is carbon.

  1. 5 11 B Adding a neutron increases the mass number by 1, but does not change the identity of the atom.
  1. The atom in part (c) is an isotope of B 10 . Isotopes are atoms of the same element with different masses.

2.24

  1. 29 protons, 34 neutrons, 29 electrons. Every neutral Cu atom has 29 protons and 29 electrons. The mass number of this Cu atom is 63, so it has (63 – 29) = 34 neutrons.
  2. 29 63 Cu 2 + . Removing electrons produces a positively charged ion; it does not change the atomic number or mass number of the atom.
  3. 29 63 Cu . A copper atom with 36 neutrons has a mass number of (29 + 36) = 65.

2.25

  1. Atomic number is the number of protons in the nucleus of an atom. Mass number is the total number of nuclear particles, protons plus neutrons, in an atom.
  2. The mass number can vary without changing the identity of the atom, but the atomic number of every atom of a given element is the same.

2.26

  1. 31 16 X and 32 16 X are isotopes of the same element, because they have identical atomic numbers.
  2. These are isotopes of the element sulfur, S, atomic number = 16.

2.27   p = protons, n = neutrons, e = electrons

  1. 40 Ar has 18 p, 22 n, 18 e
  2. 65 Zn has 30 p, 35 n, 30 e
  3. 70 Ga has 31 p, 39 n, 31 e
  4. 80 Br has 35 p, 45 n, 35 e
  5. 184 W has 74 p, 110 n, 74 e
  6. 243 Am has 95 p, 148 n, 95 e

2.28

  1. 32 P has 15 p, 17 n
  2. 51 Cr has 24 p, 27 n
  3. 60 Co has 27 p, 33 n
  4. 99 Tc has 43 p, 56 n
  5. 131 I has 53 p, 78 n
  6. 201 TI has 81 p, 120 n

2.29

Symbol 79 Br 55 Mn 112 Cd 222 Rn 207 Pb
Protons 35 25 48 86 82
Neutrons 44 30 64 136 125
Electrons 35 25 48 86 82
Mass no. 79 55 112 222 207

2.30

Symbol 112 Cd 96 Sr 87 Sr 81 Kr 235 U
Protons 48 38 38 36 92
Neutrons 64 58 49 45 143
Electrons 48 38 38 36 92
Mass No. 112 96 87 81 235

2.31

  1. 78 196 PT
  2. 36 84 Kr
  3. 33 75 As
  4. 24 12 Mg

2.32  Because the two nuclides are atoms of the same element, by definition they have the same number of protons, 54. They differ in mass number (and mass) because they have different numbers of neutrons. 129 Xe has 75 neutrons and 130 Xe has 76 neutrons.

2.33

  1. 6 12 C
  2. Atomic weights are really average atomic masses, the sum of the mass of each naturally occurring isotope of an element times its fractional abundance. Each B atom will have the mass of one of the naturally occurring isotopes, whereas the “atomic weight” is an average value. The naturally occurring isotopes of B, their atomic masses, and relative abundances are:

    10 B, 10.012937, 19.9%; 11 B, 11.009305, 80.1%.

2.34

  1. 12 amu
  2. The atomic weight of carbon reported on the front-inside cover of the text is the abundance-weighted average of the atomic masses of the two naturally occurring isotopes of carbon, 12 C and 13 C. The mass of a 12 C atom is exactly 12 amu, but the atomic weight of 12.011 takes into account the presence of some 13 C atoms in every natural sample of the element.

2.35   Atomic weight (average atomic mass) = Σ fractional abundance × mass of isotope

Atomic weight = 0.6917(62.9296) + 0.3083(64.9278) = 63.5456 = 63.55 amu

2.36   Atomic weight (average atomic mass) = Σ fractional abundance × mass of isotope

Atomic weight = 0.7215(84.9118) + 0.2785(86.9092) = 85.4681 = 85.47 amu

(The result has 2 decimal places and 4 sig figs because each term in the sum has 4 sig figs and 2 decimal places.)

2.37

  1. In Thomson’s cathode ray tube, the charged particles are electrons. In a mass spectrometer, the charged particles are positively charged ions (cations).
  2. The x-axis label (independent variable) is atomic mass (or particle mass) and the y-axis label (dependent variable) is signal intensity.
  3. The Cl 2+ ion will be deflected more. The greater the charge on the positive ion, the larger its interaction with the electric and magnetic fields. (For this reason, the x-axis label of a mass spectrum is usually mass-to-charge ratio of the particles.)

2.38

  1. True.
  2. False. The height of each peak in the mass spectrum is directly proportional to the relative abundance of the isotope.
  3. True.

2.39

  1. Average atomic mass = 0 .7899(23 .98504) + 0 .1000(24 .98584) + 0 .1101(25 .98259) = 24 .31 amu
  2. image

    The relative intensities of the peaks in the mass spectrum are the same as the relative abundances of the isotopes. The abundances and peak heights are in the ratio 24 Mg: 25 Mg: 26 Mg as 7.8 : 1.0 : 1.1.

2.40

  1. Three peaks: 1 H – 1 H, 1 H – 2 H, 2 H – 2 H
  2. 1 H – 1 H = 2(1.00783) = 2.01566 amu

    1 H – 2 H = 1.00783 + 2.01410 = 3.02193 amu

    2 H – 2 H = 2(2.01410) = 4.02820 amu

    The mass ratios are 1 : 1.49923 : 1.99845 or 1 : 1.5 : 2.

  3. 1 H – 1 H is largest, because there is the greatest chance that two atoms of the more abundant isotope will combine.

    2 H – 2 H is the smallest, because there is the least chance that two atoms of the less abundant isotope will combine.

The Periodic Table, Molecules and Molecular Compounds, and Ions and Ionic Compounds (Sections 2.5, 2.6, and 2.7)

2.41

  1. Cr, 24 (metal)
  2. He, 2 (nonmetal)
  3. P, 15 (nonmetal)
  4. Zn, 30 (metal)
  5. Mg, 12 (metal)
  6. Br, 35 (nonmetal)
  7. As, 33 (metalloid)

2.42

  1. lithium, 3 (metal)
  2. scandium, 21 (metal)
  3. germanium, 32 (metalloid)
  4. ytterbium, 70 (metal)
  5. manganese, 25 (metal)
  6. antimony, 51 (metalloid)
  7. xenon, 54 (nonmetal)

2.43

  1. K, alkali metals (metal)
  2. I, halogens (nonmetal)
  3. Mg, alkaline earth metals (metal)
  4. Ar, noble gases (nonmetal)
  5. S, chalcogens (nonmetal)

2.44   C, carbon, nonmetal; Si, silicon, metalloid; Ge, germanium, metalloid; Sn, tin, metal; Pb, lead, metal

2.45

  1. C 4 H 10 is the molecular formula for both compounds. For the molecular formula, count the total number of each kind of atom in the structural formula.
  2. C 2 H 5 . Starting with the molecular formula, divide subscripts by any common factors to determine the simplest ratio of atom types in the molecule. In this example the common factor for both molecules is 2.
  3. Structural. In this example, the molecules are structural isomers and only the structural formulas allow us to determine that the molecules are different.

2.46

  1. Benzene, C 6 H 6 ; acetylene, C 2 H 2 . For the molecular formula, count the total number of each kind of atom in the ball and stick representations.
  2. Benzene, CH; acetylene, CH. Starting with the molecular formula, divide subscripts by any common factors to determine the simplest ratio of atom types in the molecule. In this example the common factor for benzene is 6 and for acetylene is 2.

2.47   From center-text to right, the molecular and empirical formulas are: N 2 H 4 , NH 2 ; N 2 H 2 , NH; NH 3 , NH 3

2.48   No. Two substances with the same molecular and empirical formulas can be isomers. They are not necessarily the same compound.

2.49

  1. AlBr 3
  2. C 4 H 5
  3. C 2 H 4 O
  4. P 2 O 5
  5. C 3 H 2 Cl
  6. BNH 2

2.50   A molecular formula contains all atoms in a molecule. An empirical formula shows the simplest ratio of atoms in a molecule or elements in a compound.

  1. molecular formula: C 6 H 6 ; empirical formula: CH
  2. molecular formula: SiCl 4 ; empirical formula: SiCl 4 (1:4 is the simplest ratio)
  3. molecular: B 2 H 6 ; empirical: BH 3
  4. molecular: C 6 H 12 O 6 ; empirical: CH 2 O

2.51

  1. 6
  2. 10
  3. 12

2.52

  1. 6
  2. 6
  3. 9

2.53

  1. image
  2. image
  3. image
  4. image

2.54

  1. image
  2. image
  3. image
  4. image

2.55

Symbol 59 Co 3+ 80 Se 2– 192 Os 2+ 200 Hg 2+
Protons 27 34 76 80
Neutrons 32 46 116 120
Electrons 24 36 74 78
Net Charge 3+ 2– 2+ 2+

2.56

Symbol 31 P 3– 79 Se 2– 119 Sn 4+ 197 Au 3+
Protons 15 34 50 79
Neutrons 16 45 69 118
Electrons 18 36 46 76
Net Charge 3– 2– 4+ 3+

2.57

  1. Mg 2+
  2. Al 3+
  3. K +
  4. S 2–
  5. F–

2.58

  1. Ga 3+
  2. Sr 2+
  3. As 3–
  4. Br–
  5. Se 2–

2.59

  1. GaF 3 , gallium(III) fluoride
  2. LiH, lithium hydride
  3. AlI 3 , aluminum iodide
  4. K 2 S, potassium sulfide

2.60

  1. ScI 3
  2. Sc 2 S 3
  3. ScN

2.61

  1. CaBr 2
  2. K 2 CO 3
  3. Al(CH 3 COO) 3
  4. (NH 4 ) 2 SO 4
  5. Mg 3 (PO 4 ) 2

2.62

  1. CrBr 3
  2. Fe 2 O 3
  3. Hg 2 CO 3
  4. Ca(ClO 3 ) 2
  5. (NH 4 ) 3 PO 4

2.63

Ion K + NH 4 + Mg 2+ Fe 3+
Cl KCl NH 4 Cl MgCl 2 FeCl 3
OH KOH NH 4 OH * Mg(OH) 2 Fe(OH) 3
CO 3 2– K 2 CO 3 (NH 4 ) 2 CO 3 MgCO 3 Fe 2 (CO 3 ) 3
PO 4 3– K 3 PO 4 (NH 4 ) 3 PO 4 Mg 3 (PO 4 ) 2 FePO 4

* Equivalent to NH 3 (aq).

2.64

Ion Na + Ca 2+ Fe 2+ Al 3+
O 2– Na 2 O CaO FeO Al 2 O 3
NO 3 NaNO 3 Ca(NO 3 ) 2 Fe(NO 3 ) 2 Al(NO 3 ) 3
SO 4 2– Na 2 SO 4 CaSO 4 FeSO 4 Al 2 (SO 4 ) 3
AsO 4 3– Na 3 AsO 4 Ca 3 (AsO 4 ) 2 Fe 3 (AsO 4 ) 2 AlAsO 4

2.65   Molecular (all elements are nonmetals):

  1. B 2 H 6
  2. CH 3 OH
  3. NOCl
  4. NF 3

Ionic (formed by a cation and an anion, usually contains a metal cation):

  1. LiNO 3
  2. Sc 2 O 3
  3. CsBr
  4. Ag 2 SO 4

2.66   Molecular (all elements are nonmetals):

  1. PF 5
  2. SCl 2
  3. N 2 O 4

Ionic (formed from ions, usually contains a metal cation):

  1. NaI
  2. Ca(NO 3 ) 2
  3. FeCl 3
  4. LaP
  5. CoCO 3

Naming Inorganic Compounds; Some Simple Organic Compounds (Sections 2.8 and 2.9)

2.67

  1. ClO 2
  2. Cl
  3. ClO 3
  4. ClO 4
  5. ClO

2.68

  1. selenate
  2. selenide
  3. hydrogen selenide (biselenide)
  4. hydrogen selenite (biselenite)

2.69

  1. calcium, 2+; oxide, 2−
  2. sodium, 1+; sulfate, 2−
  3. potassium, 1+; perchlorate, 1−
  4. iron, 2+; nitrate, 1−
  5. chromium, 3+; hydroxide, 1−

2.70

  1. copper, 2+; sulfide, 2−
  2. silver, 1+; sulfate, 2−
  3. aluminum, 3+; chlorate, 1−
  4. cobalt, 2+; hydroxide, 1−
  5. lead, 2+; carbonate, 2−

2.71

  1. lithium oxide
  2. iron(III) chloride (ferric chloride)
  3. sodium hypochlorite
  4. calcium sulfite
  5. copper(II) hydroxide (cupric hydroxide)
  6. iron(II) nitrate (ferrous nitrate)
  7. calcium acetate
  8. chromium(III) carbonate (chromic carbonate)
  9. potassium chromate
  10. ammonium sulfate

2.72

  1. potassium cyanide
  2. sodium bromite
  3. strontium hydroxide
  4. cobalt(II) telluride (cobaltous telluride)
  5. iron(III) carbonate (ferric carbonate)
  6. chromium(III) nitrate (chromic nitrate)
  7. ammonium sulfite
  8. sodium dihydrogen phosphate
  9. potassium permanganate
  10. silver dichromate

2.73

  1. Al(OH) 3
  2. K 2 SO 4
  3. Cu 2 O
  4. Zn(NO 3 ) 2
  5. HgBr 2
  6. Fe 2 (CO 3 ) 3
  7. NaBrO

2.74

  1. Na 3 PO 4
  2. Zn(NO 3 ) 2
  3. Ba(BrO 3 ) 2
  4. Fe(ClO 4 ) 2
  5. Co(HCO 3 ) 2
  6. Cr(CH 3 COO) 3
  7. K 2 Cr 2 O 7

2.75

  1. bromic acid
  2. hydrobromic acid
  3. phosphoric acid
  4. HClO
  5. HIO 3
  6. H 2 SO 3

2.76

  1. HI
  2. HClO 3
  3. HNO 2
  4. carbonic acid
  5. perchloric acid
  6. acetic acid

2.77

  1. sulfur hexafluoride
  2. iodine pentafluoride
  3. xenon trioxide
  4. N 2 O 4
  5. HCN
  6. P 4 S 6

2.78

  1. dinitrogen monoxide
  2. nitrogen monoxide
  3. nitrogen dioxide
  4. dinitrogen pentoxide
  5. dinitrogen tetroxide

2.79

  1. ZnCO 3 , ZnO, CO 2
  2. HF, SiO 2 , SiF 4 , H 2 O
  3. SO 2 , H 2 O, H 2 SO 3
  4. PH 3
  5. HClO 4 , Cd, Cd(ClO 4 ) 2
  6. VBr 3

2.80

  1. NaHCO 3
  2. Ca(ClO) 2
  3. HCN
  4. Mg(OH) 2
  5. SnF 2
  6. CdS, H 2 SO 4 , H 2 S

2.81

  1. A hydrocarbon is a compound composed of the elements hydrogen and carbon only.
  2. image

    molecular and empirical formulas: C 5 H 12

2.82

  1. Isomers are molecules with the same molecular formula, but different structural formulas. Isomers have the same number and kinds of atoms, but these atoms are arranged in different ways.
  2. Butane and pentane are both capable of existing in isomeric forms. There is more than one way to arrange the four C atoms and ten H atoms of butane, and more than one way to arrange the five C atoms and twelve H atoms of pentane. There is only one way to arrange the two C atoms and six H atoms of ethane and only one way to arrange the three C atoms and eight H atoms of propane.

2.83

  1. A functional group is a group of specific atoms that are constant (arranged the same way) from one molecule to the next.
  2. The characteristic alcohol functional group is an –OH. Another way to say this is that whenever a molecule is called an alcohol, it contains the –OH group.
  3. image

2.84

  1. The suffix -ol indicates that an organic compound contains an –OH group. Organic compounds that contain the –OH functional group are called alcohols. Ethanol and propanol contain –OH groups (and are alcohols.)
  2. The root prop- indicates that a molecule contains three carbon atoms. Propane and propanol contain three carbon atoms.

2.85

  1. image
  2. 1-chloropropane        2-chloropropane

2.86

image

Additional Exercises

2.87

  1. Droplet D would fall most slowly. It carries the most negative charge, so it would be most strongly attracted to the upper (+) plate and most strongly repelled by the lower (–) plate. These electrostatic forces would provide the greatest opposition to gravity.
  2. Calculate the lowest common factor.

    A: 3.84 × 10 –8 / 2.88 × 10 –8 = 1.33; 1.33 × 3 = 4

    B: 4.80 × 10 –8 / 2.88 × 10 –8 = 1.67; 1.67 × 3 = 5

    C: 2.88 × 10 –8 / 2.88 × 10 –8 = 1.00; 1.00 × 3 = 3

    D: 8.64 × 10 –8 / 2.88 × 10 –8 = 3.00; 3.00 × 3 = 9

    The total charge on the drops is in the ratio of 4:5:3:9. Divide the total charge on each drop by the appropriate integer and average the four values to get the charge of an electron in warmombs.

    A: 3.84 × 10 –8 / 4 = 9.60 × 10 –9 wa

    B: 4.80 × 10 –8 / 5 = 9.60 × 10 –9 wa

    C: 2.88 × 10 –8 / 3 = 9.60 × 10 –9 wa

    D: 8.64 × 10 –8 / 9 = 9.60 × 10 –9 wa

    The charge on an electron is 9.60 × 10 –9 wa

  3. The number of electrons on each drop are the integers calculated in part (b). A has 4 e , B has 5 e , C has 3 e and D has 9 e .
  4. 9.60 × 10 9 wa 1 e × 1 e 1.60 × 10 16 C = 6.00 × 10 7 wa/C

2.88

  1. 3 He has 2 protons, 1 neutron, and 2 electrons.
  2. 3 H has 1 proton, 2 neutrons, and 1 electron.

    3 He: 2(1.672621673 × 10 –24 g) + 1.674927211 × 10 –24 g + 2(9.10938215 × 10 –28 g) = 5.021992 × 10 –24 g

    3 H: 1.672621673 × 10 –24 g + 2(1.674927211 × 10 –24 g) + 9.10938215 × 10 –28 g = 5.023387 × 10 –24 g

    Tritium, 3 H, is more massive.

  3. The masses of the two particles differ by 0.0014 × 10 –24 g. Each particle loses 1 electron to form the +1 ion, so the difference in the masses of the ions is still 1.4 × 10 –27 . A mass spectrometer would need precision to 1 × 10 –27 g to differentiate 3 He + and 3 H.

2.89

  1. Calculate the mass of a single gold atom, then divide the mass of the cube by the mass of the gold atom.

    197.0 amu gold atom × 1 g 6.022 × 10 23 amu = 3.2713 × 10 22 = 3.271 × 10 22 g/gold atom 19.3 g cube × 1 gold atom 3.271 × 10 22 g = 5.90 × 10 22 Au atoms in the cube

  2. The shape of atoms is spherical; spheres cannot be arranged into a cube so that there is no empty space. The question is, how much empty space is there? We can calculate the two limiting cases, no empty space and maximum empty space. The true diameter will be somewhere in this range.

    No empty space: volume cube/number of atoms = volume of one atom

    V = 4/3π r 3 ; r = (3 V/4 π) 1/3 ; d = 2r

    volume of cube = (1 .0 × 1 .0 × 1 .0)= 1.0 cm 3 5.90 × 10 22 Au atoms = 1.695 × 10 23 = 1.7 × 10 23 cm 3

    r = [3 (1.695 × 10 –23 cm 3 )/4 π] 1/3 = 1.6 × 10 –8 cm; d = 2r = 3.2 × 10 –8 cm

    Maximum empty space: Assume atoms are arranged in rows in all three directions so they are touching across their diameters. That is, each atom occupies the volume of a cube, with the atomic diameter as the length of the side of the cube. The number of atoms along one edge of the gold cube is then

    (5.90 × 10 22 ) 1/3 = 3.893 × 10 7 = 3.89 × 10 7 atoms/1.0 cm.

    The diameter of a single atom is 1.0 cm/3.89 × 10 7 atoms = 2.569 × 10 –8

    = 2.6 × 10 –8 cm.

    The diameter of a gold atom is between 2.6 × 10 –8 cm and 3.2 × 10 –8 cm

    (2.6 – 3.2 Å).


  1. Some atomic arrangement must be assumed, because none is specified. The solid state is characterized by an orderly arrangement of particles, so it isn’t surprising that atomic arrangement is required to calculate the density of a solid. A more detailed discussion of solid-state structure and density appears in Chapter 11 .

2.90

  1. In arrangement A, the number of atoms in 1 cm 2 is just the square of the number that fit linearly in 1 cm.

    1.0 cm × 1 atom 4.95 Å × 1 × 10 10 Å 1 m × 1 m 100 cm = 2.02 × 10 7 = 2.0 × 10 7 atoms/cm

    1.0 cm 2 = (2.02 × 10 7 ) 2 = 4.081 × 10 14 = 4.1 × 10 14 atoms/cm 2

  2. In arrangement B, the atoms in the horizontal rows are touching along their diameters, as in arrangement A. The number of Rb atoms in a 1.0 cm row is then 2.0 × 10 7 Rb atoms. Relative to arrangement A, the vertical rows are offset by 1/2 of an atom. Atoms in a “column” are no longer touching along their vertical diameter. We must calculate the vertical distance occupied by a row of atoms, which is now less than the diameter of one Rb atom.

    Consider the triangle shown below. This is an isosceles triangle (equal side lengths, equal interior angles) with a side-length of 2d and an angle of 60°. Drop a bisector to the uppermost angle so that it bisects the opposite side.

    image

    The result is a right triangle with two known side lengths. The length of the unknown side (the angle bisector) is 2h, two times the vertical distance occupied by a row of atoms. Solve for h, the “height” of one row of atoms.

    (2h) 2 + d 2 = (2d) 2 ; 4h 2 = 4d 2 – d 2 = 3d 2 ; h 2 = 3d 2 /4

    h = (3d 2 /4) 1/2 = (3(4.95 Å) 2 /4) 1/2 = 4.2868 = 4.29 Å

    The number of rows of atoms in 1 cm is then

    1.0 cm × 1 row 4.2868 Å × 1 × 10 10 Å 1 m × 1 m 100 cm = 2.333 × 10 7 = 2.3 × 10 7

    The number of atoms in a 1.0 cm 2 square area is then

    2.020 × 10 7 atoms 1 row × 2.333 × 10 7 rows = 4.713 × 10 14 = 4.7 × 10 14 atoms

    Note that we have ignored the loss of “1/2” atom at the end of each horizontal row. Out of 2.0 × 10 7 atoms per row, one atom is not significant.

  3. The ratio of atoms in arrangement B to arrangement A is then 4.713 × 10 14 atoms/4.081 × 10 14 = 1.155 = 1.2:1. Clearly, arrangement B results in less empty space per unit area or volume. If extended to three dimensions, arrangement B would lead to a greater density for Rb metal.

2.91

  1. diameter of nucleus = 1 × 10 –4 Å; diameter of atom = 1 Å

    V = 4/3 π r 3 ; r = d/2; r n = 0.5 × 10 –4 Å; r a = 0.5 Å

    volume of nucleus = 4/3 π (0.5 × 10 –4 ) 3 Å 3

    volume of atom = 4/3 π (0.5) 3 Å 3

    volume fraction of nucleus = volume of nucleus volume of atom = 4 / 3 π ( 0.5 × 10 4 ) 3 Å 3 4 / 3 π ( 0.5 ) 3 Å 3 = 1 × 10 12

    diameter of atom = 5 Å, r a = 2.5 Å

    volume fraction of nucleus = 4 / 3 π ( 0.5 × 10 4 ) 3 Å 3 4 / 3 π ( 2.5 ) 3 Å 3 = 8 × 10 15

    Depending on the radius of the atom, the volume fraction of the nucleus is between 1 × 10 –12 and 8 × 10 –15 , that is, between 1 part in 10 12 and 8 parts in 10 15 .

  2. mass of proton = 1.0073 amu

    1.0073 amu × 1.66054 × 10 –24 g/amu = 1.6727 × 10 –24 g

    diameter = 1.0 × 10 15 m , radius = 0.50 × 10 15 m × 100 cm 1 m = 5.0 × 10 14 cm

    Assuming a proton is a sphere, V = 4/3 π r 3 .

    density = g cm 3 = 1.6727 × 10 24 4 / 3 π ( 5.0 × 10 14 ) 3 cm 3 = 3.2 × 10 15 g/cm 3

2.92   The integer on the lower left of a nuclide is the atomic number; it is the number of protons in any atom of the element and gives the element’s identity. The number of neutrons is the mass number (upper left) minus atomic number.

  1. As, 33 protons, 41 neutrons
  2. I, 53 protons, 74 neutrons
  3. Eu, 63 protons, 89 neutrons
  4. Bi, 83 protons, 126 neutrons

2.93

  1. 6 Li, 3 protons, 3 neutrons; 7 Li, 3 protons, 4 neutrons
  2. Average atomic weight of sample = Σ fractional abundance × mass of isotope

    Av. atomic weight = 0.01442(6.015122) + 0.98558(7.016004) = 7.001571 = 7.002 amu

2.94

  1. 8 16 O , 8 17 O , 8 18 O ,
  2. All isotopes are atoms of the same element, oxygen, with the same atomic number (Z = 8), 8 protons in the nucleus and 8 electrons. Elements with similar electron arrangements have similar chemical properties (Section 2.5). Because the 3 isotopes all have 8 electrons, we expect their electron arrangements to be the same and their chemical properties to be very similar, perhaps identical. Each has a different number of neutrons (8, 9, or 10), a different mass number (A = 16, 17, or 18) and thus a different atomic mass.

2.95   Atomic weight (average atomic mass) = Σ fractional abundance × mass of isotope

Atomic weight = 0.014(203.97302) + 0.241(205.97444) + 0.221(206.97587) +

0.524(207.97663) = 207.22 = 207 amu

(The result has 0 decimal places and 3 sig figs because the fourth term in the sum has 3 sig figs and 0 decimal places.)

2.96

  1. The 68.926 amu isotope has a mass number of 69, with 31 protons, 38 neutrons and the symbol G 31 69 a . The 70.925 amu isotope has a mass number of 71, 31 protons, 40 neutrons, and symbol G 31 71 a . (All Ga atoms have 31 protons.)
  2. The average mass of a Ga atom is 69.72 amu. Let x = abundance of the lighter isotope, 1–x = abundance of the heavier isotope. Then x(68.926) + (1–x)(70.925) = 69.72; x = 0.6028 = 0.603, 69 Ga = 60.3%, 71 Ga = 39.7%.

2.97

  1. There are 24 known isotopes of Ni, from 51 Ni to 74 Ni.

(b,c) The five most abundant isotopes (b) and their natural abundances (c) are

  1. 58 Ni, 57.935346 amu, 68.077%

    60 Ni, 59.930788 amu, 26.223%

    62 Ni, 61.928346 amu, 3.634%

    61 Ni, 60.931058 amu, 1.140%

    64 Ni, 63.927968 amu, 0.926%

    Data from Handbook of Chemistry and Physics , 74th edition [Data may differ slightly in other editions.]

2.98

  1. A Br 2 molecule could consist of two atoms of the same isotope or one atom of each of the two different isotopes. This second possibility is twice as likely as the first. Therefore, the second peak (twice as large as peaks 1 and 3) represents a Br 2 molecule containing different isotopes. The mass numbers of the two isotopes are determined from the masses of the two smaller peaks. Because 157.836 ≈ 158, the first peak represents a 79 Br— 79 Br molecule. Peak 3, 161.832 ≈ 162, represents a 81 Br— 81 Br molecule. Peak 2 then contains one atom of each isotope, 79 Br— 81 Br, with an approximate mass of 160 amu.
  2. The mass of the lighter isotope is 157.836 amu/2 atoms, or 78.918 amu/atom. For the heavier one, 161.832 amu/2 atoms = 80.916 amu/atom.
  3. The relative size of the three peaks in the mass spectrum of Br 2 indicates their relative abundance. The average mass of a Br 2 molecule is

    0.2569(157.836) + 0.4999(159.834) + 0.2431(161.832) = 159.79 amu.

    (Each product has four significant figures and two decimal places, so the answer has two decimal places.)


  1. 159.79 amu avg . Br 2 molecule × 1 Br 2 molecule 2 Br atoms = 79.895 amu
  1. Let x = the abundance of 79 Br, 1 – x = abundance of 81 Br. From (b), the masses of the two isotopes are 78.918 amu and 80.916 amu, respectively. From (d), the mass of an average Br atom is 79.895 amu.

    x(78.918) + (1 – x)(80.916) = 79.895, x = 0.5110

    79 Br = 51.10%, 81 Br = 48.90%

2.99

  1. Five significant figures. 1 H + is a bare proton with mass 1.0073 amu. 1 H is a hydrogen atom, with 1 proton and 1 electron. The mass of the electron is 5.486 × 10 –4 or 0.0005486 amu. Thus the mass of the electron is significant in the fourth decimal place or fifth significant figure in the mass of 1 H.
  2. Mass of 1 H = 1.0073 amu (proton) 0.0005486 amu ¯ (electron) 1.0078 amu (We have not rounded up to 1.0079 because 49 < 50 in the final sum .)

    Mass % of electron= mass of e mass of 1 H × 100 = 5.486 × 10 4 amu 1.0078 amu × 100 = 0.05444 %

2.100

  1. an alkali metal: K
  2. an alkaline earth metal: Ca
  3. a noble gas: Ar
  4. a halogen: Br
  5. a metalloid: Ge
  6. a nonmetal in 1A: H
  7. a metal that forms a 3+ ion: Al
  8. a nonmetal that forms a 2– ion: O
  9. an element that resembles Al: Ga

2.101

  1. S 106 266 g has 106 protons, 160 neutrons and 106 electrons
  2. Sg is in Group 6B (or 6) and immediately below tungsten, W. We expect the chemical properties of Sg to most closely resemble those of W.

2.102   Strontium is an alkaline earth metal, similar in chemical properties to calcium and magnesium. Calcium is ubiquitous in biological organisms, humans included. It is a vital nutrient required for formation and maintenance of healthy bones and teeth. As such, there are efficient pathways for calcium uptake and distribution in the body, pathways that are also available to chemically similar strontium. Harmful strontium imitates calcium and then behaves badly when the body tries to use it as it uses calcium.

2.103   Calculate the volume of the penny, use density to calculate mass and price to calculate the value of copper in the penny.

V = π r 2 h; d = 19 mm, r = d/2 = 9.5 mm; h = 1.5 mm

V = π × ( 9.5 mm ) 2 × 1.5 mm × 1 cm 3 ( 10 ) 3 mm 3 = 0.4253 = 0.43 cm 3 0.4253 cm 3 × 8.9 g cm 3 × 1 lb 453.6 g × $ 2.40 lb = $ 0.02003 = $ 0.020

That is, the copper in each penny is worth two pennies!


2.104   Calculate the volume of the coin, use density to calculate mass and price to calculate the value of silver in the coin.

V = π r 2 h ; d = 41 mm, r = d/2 = 20.5 mm; h = 2.5 mm

V = π × ( 20.5 mm ) 2 × 2.5 mm × 1 cm 3 ( 10 ) 3 mm 3 = 3.3006 = 3.3 cm 3 3.3006 cm 3 × 10.5 g cm 3 × $ 0.51 g = $ 17.675 = $ 18

Wow! The silver in each Silver Eagle dollar coin is worth $18.

2.105

  1. chlorine gas, Cl 2 : ii
  2. propane, C 3 H 8 : v
  3. nitrate ion, NO 3 : i
  4. sulfur trioxide, SO 3 : iii
  5. methylchloride, CH 3 Cl: iv

2.106

  1. nickel(II) oxide, 2+
  2. manganese(IV) oxide, 4+
  3. chromium(III) oxide, 3+
  4. molybdenium(VI) oxide, 6+

2.107

Cation Anion Formula Name
Li + O 2– Li 2 O Lithium oxide
Fe 2+ PO 4 3– Fe 3 (PO 4 ) 2 Iron(II) phosphate
Al 3+ SO 4 2– Al 2 (SO 4 ) 3 Aluminum sulfate
Cu 2+ NO 3 Cu(NO 3 ) 2 Copper(II) nitrate
Cr 3+ I CrI 3 Chromium(III) iodide
Mn + ClO 2 MnClO 2 Manganese(I) chlorite
NH 4 + CO 3 2– (NH 4 ) 2 CO 3 Ammonium carbonate
Zn 2+ ClO 4 Zn(ClO 4 ) 2 Zinc perchlorate

2.108

  1. Empirical formula, CH 3

    The empirical and molecular formulas of propane are C 3 H 8 . Propane has two more H atoms than cyclopropane, so the empirical and molecular formulas are different.

  2. The solid wedges indicate bonds from C atoms to H atoms that are above the plane of the page; the dashed wedges show bonds from C atoms to H atoms that are behind the plane of the page.
  3. To illustrate chlorocyclopropane, replace any one of the H atoms on cyclopropane with a Cl atom. There are no isomers of chlorocyclopropane, because a structure with a Cl atom at any one of the six positions can be rotated into the original structure.

2.109

  1. perbromate ion
  2. selenite ion
  3. AsO 4 3–
  4. HTeO 4

2.110   Carbonic acid: H 2 CO 3 ; the cation is H + because it is an acid; the anion is carbonate because the acid reacts with lithium hydroxide to form lithium carbonate.

Lithium hydroxide: LiOH; lithium carbonate: Li 2 CO 3

2.111

  1. sodium chloride
  2. sodium bicarbonate (or sodium hydrogen carbonate)
  3. sodium hypochlorite
  4. sodium hydroxide
  5. ammonium carbonate
  6. calcium sulfate

2.112

  1. potassium nitrate
  2. sodium carbonate
  3. calcium oxide
  4. hydrochloric acid
  5. magnesium sulfate
  6. magnesium hydroxide

2.113

  1. CaS, Ca(HS) 2
  2. HBr, HBrO 3
  3. AlN, Al(NO 2 ) 3
  4. FeO, Fe 2 O 3
  5. NH 3 , NH 4 +
  6. K 2 SO 3 , KHSO 3
  7. Hg 2 Cl 2 , HgCl 2
  8. HClO 3 , HClO 4

2.114   In the nucleus. The strong force holds the protons together against the repulsive electrostatic force.


 

3 Chemical Reactions and Reaction Stoichiometry

Visualizing Concepts

3.1   Reactant A = blue, reactant B = red

Overall, 4 blue A 2 molecules + 4 red B atoms → 4 A 2 B molecules

Because 4 is a common factor, this equation reduces to equation (a).

3.2

  1. There are four CH 3 OH molecules in the products box. CO is the only source of C atoms for the reaction, so there must be four CO molecules in the reactants box.
  2. CO + 2 H 2 → CH 3 OH

3.3

  1. There are twice as many O atoms as N atoms, so the empirical formula of the original compound is NO 2 .
  2. No, because we have no way of knowing whether the empirical and molecular formulas are the same. NO 2 represents the simplest ratio of atoms in a molecule but not the only possible molecular formula.

3.4   The box contains 4 C atoms and 10 H atoms, so the empirical formula of the hydrocarbon is C 2 H 5 .

3.5

  1. Analyze .  Given the molecular model, write the molecular formula.

    Plan .  Use the colors of the atoms (spheres) in the model to determine the number of atoms of each element.

    Solve .  Observe 2 gray C atoms, 5 white H atoms, 1 blue N atom, 2 red O atoms. C 2 H 5 NO 2

  2. Plan .  Follow the method in Sample Exercise 3.9. Calculate formula weight in amu and molar mass in grams.

    2 C atoms = 2(12.0 amu) = 24.0 amu

    5 H atoms = 5(1.0 amu)  = 5.0 amu

    1 N atoms = 1(14.0 amu) = 14.0 amu

    2 O atoms = 2(16.0 amu) = 32.0 amu

    75.0 amu

    Formula weight = 75.0 amu, molar mass = 75.0 g/mol


  1. Plan . The molar mass of a substance provides the factor for converting grams to moles (or moles to grams).

    Solve .

    Because the mass of glycine has 4 significant figures, use a molar mass of glycine that has at least 4 significant figures. Using molar masses of the component elements from the periodic chart, the molar mass of glycine is

    [2(12.0107) + 5(1.00794) + 14.0067 + 2(15.9994)] = 75.0666 = 75.067 g/mol

    1 0 0 . 0 g glycine × 1 mol glycine 7 5 . 0 6 7 g glycine = 1 . 3 3 2 1 = 1 . 3 3 2 mol glycine

  1. Plan .  Use the definition of mass % and the results from parts (a) and (b) above to find mass % N in glycine.

    Solve . mass % N = g N g C 2 H 5 N O 2 × 1 0 0

    Assume 1 mol C 2 H 5 NO 2 . From the molecular formula of glycine [part (a)], there is 1 mol N/mol glycine.

    mass % N = 1 × (molar mass N) molar mass glycine × 100 = 14.0 g 75 .0 g × 100 = 18.7 %

3.6 Analyze .  Given: 4.0 mol CH 4 . Find: mol CO and mol H 2

Plan .  Examine the boxes to determine the CH 4 :CO mol ratio and CH 4 :H 2 O mole ratio.

Solve .  There are 2 CH 4 molecules in the reactant box and 2 CO molecules in the product box. The mole ratio is 2:2 or 1:1. Therefore, 4.0 mol CH 4 can produce 4.0 mol CO. There are 2 CH 4 molecules in the reactant box and 6 H 2 molecules in the product box. The mole ratio is 2:6 or 1:3. So, 4.0 mol CH 4 can produce 12:0 mol H 2 .

Check .  Use proportions. 2 mol CH 4 /2 mol CO = 4 mol CH 4 /4 mol CO;

2 mol CH 4 /6 mol H 2 = 4 mol CH 4 /12 mol H 2 .

3.7 Analyze .  Given a box diagram and formulas of reactants, answer questions about the reaction mixture in the box.

Plan .  Write and balance the chemical equation. Determine combining ratios of elements and decide on limiting reactant. Decide the maximum number of NH 3 molecules that can be produced and the number of leftover reactant molecules.

Solve.

  1. N 2 + 3 H 2 → 2 NH 3
  2. H 2 is the limiting reactant. There are 4 N 2 molecules and 9 H 2 molecules in the box. According to the chemical reaction, each N 2 molecule requires 3 H 2 molecules for complete reaction. If all N 2 molecules were to react, 12 H 2 molecules would be required. There are only 9 H 2 molecules, so H 2 is the limiting reactant.

  1. 6 NH 3 molecules. Because H 2 is the limiting reactant, the amount of H 2 available determines the amount of NH 3 produced. Three H 2 molecules produce 2 NH 3 molecules, so 9 H 2 molecules produce 6 NH 3 molecules.
  2. One N 2 molecule is left over. The 9 H 2 molecules react with 3 N 2 molecules, leaving one N 2 molecule unreacted. All H 2 molecules are used up.

3.8

  1. 8 NO 2 molecules can be produced. The overall chemical reaction is

    2 NO + O 2 → 2 NO 2 .

    There are 8 NO molecules and 5 O 2 molecules in the box. Each NO molecule reacts with 1 O atom (1/2 of an O 2 molecule) to produce 1 NO 2 molecule. Eight NO molecules react with 8 O atoms (4 O 2 molecules) to produce 8 NO 2 molecules. One O 2 molecule doesn’t react (is in excess).

  2. NO is the limiting reactant, because it limits (determines) the amount of product that can be produced. It is completely used up if the reaction goes to completion.
  3. If the yield is 75%, 6 NO 2 molecules, 2 NO molecules and 2 O 2 molecules are present. A 100% yield is 8 NO 2 molecules. 75% of that is 6 NO 2 molecules. Formation of 6 NO 2 molecules requires 6 NO molecules and 6 O atoms or 3 O 2 molecules. This leaves 2 NO molecules and 2 O 2 molecules unreacted.

Chemical Equations and Simple Patterns of Chemical Reactivity (Sections 3.1 and 3.2)

3.9

  1. False. We balance chemical equations as we do because mass must be conserved.
  2. True. Mass is conserved.
  3. False. Subscripts in chemical formulas cannot be changed when balancing an equation. Changing a subscript changes the identity of a compound, which changes the overall reaction.

3.10

  1. CaO(s) + H 2 O(l) → Ca(OH) 2 (aq)
  2. The only way to write a balanced equation with CaOH(aq) as a product is to include OH−(aq) as a second product. Otherwise, the ratio of elements in the product is never the same as the ratio of elements in the reactants.

    CaO(s) + H 2 O(l) → CaOH(aq) + OH (aq)

    But, if CaOH is a neutral compound, this equation violates the principle of charge balance. The equation above cannot be the correct balanced equation for the reaction of calcium oxide with water.

3.11

  1. 2 CO(g) + O 2 (g) → 2 CO 2 (g)
  2. N 2 O 5 (g) + H 2 O(l) → 2 HNO 3 (aq)
  3. CH 4 (g) + 4 Cl 2 (g) → CCl 4 (l) + 4 HCl(g)
  4. Zn(OH) 2 (s) + 2 HNO 3 (aq) → Zn(NO 3 ) 2 (aq) + 2 H 2 O(l)

3.12

  1. 6 Li(s) + N 2 (g) → 2 Li 3 N(s)
  2. TiCl 4 (l) + 2 H 2 O(l) → TiO 2 (s) + 4 HCl(aq)
  3. 2 NH 4 NO 3 (s) → 2 N 2 (g) + O 2 (g) + 4 H 2 O(g)
  4. 2 AlCl 3 (s) + Ca 3 N 2 (s) → 2 AlN(s) + 3 CaCl 2 (s)

3.13

  1. Al 4 C 3 (s) + 12 H 2 O(l) → 4 Al(OH) 3 (s) + 3 CH 4 (g)
  2. 2 C 5 H 10 O 2 (l) + 13 O 2 (g) → 10 CO 2 (g) + 10 H 2 O(g)
  3. 2 Fe(OH) 3 (s) + 3 H 2 SO 4 (aq) → Fe 2 (SO 4 ) 3 (aq) + 6 H 2 O(l)
  4. Mg 3 N 2 (s) + 4 H 2 SO 4 (aq) → 3 MgSO 4 (aq) + (NH 4 ) 2 SO 4 (aq)

3.14

  1. Ca 3 P 2 (s) + 6 H 2 O(l) → 3 Ca(OH) 2 (aq) + 2 PH 3 (g)
  2. 2 Al(OH) 3 (s) + 3 H 2 SO 4 (aq) → Al 2 (SO 4 ) 3 (aq) + 6 H 2 O(l)
  3. 2 AgNO 3 (aq) + Na 2 CO 3 (aq) → Ag 2 CO 3 (s) + 2 NaNO 3 (aq)
  4. 4 C 2 H 5 NH 2 (g) + 15 O 2 (g) → 8 CO 2 (g) + 14 H 2 O(g) + 2 N 2 (g)

3.15

  1. CaC 2 (s) + 2 H 2 O(l) → Ca(OH) 2 (aq) + C 2 H 2 (g)
  2. 2 KCIO 3 ( s ) Δ 2 KCI(s ) + 3 O 2 ( g )
  3. Zn(s) + H 2 SO 4 (aq) → H 2 (g) + ZnSO 4 (aq)
  4. PCl 3 (l) + 3 H 2 O(l) → H 3 PO 3 (aq) + 3 HCl(aq)
  5. 3 H 2 S(g) + 2 Fe(OH) 3 (s) → Fe 2 S 3 (s) + 6 H 2 O(g)

3.16

  1. SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq)
  2. B 2 S 3 (s) + 6 H 2 O(l) → 2 H 3 BO 3 (aq) + 3 H 2 S(g)
  3. 4 PH 3 (g) + 8 O 2 (g) → P 4 O 10 (s) + 6 H 2 O(g)
  4. 2 Hg(NO 3 ) 2 ( s ) Δ 2 HgO(s) + 4 NO 2 ( g) + O 2 ( g)
  5. Cu(s) + 2 H 2 SO 4 (aq) → CuSO 4 (aq) + SO 2 (g) + 2 H 2 O(l)

Patterns of Chemical Reactivity (Section 3.2)

3.17

  1. NaBr. When a metal reacts with a nonmetal, an ionic compound forms. The combining ratio of the atoms is such that the total positive charge on the metal cation(s) is equal to the total negative charge on the nonmetal anion(s). Determine the formula by balancing the positive and negative charges in the ionic product.
  2. The product is a solid at room temperature. All ionic compounds are solids.
  3. In the balanced chemical equation, the coefficient in front of the product is two.

    2 Na(s) + Br 2 (l) → 2 NaBr(s)


3.18

  1. O 2 (g). Combustion is reaction with oxygen.
  2. The products are CO 2 (g) and H 2 O(l).
  3. The sum of the balanced coefficients is 11. (Remember that the coefficient in front of C 3 H 6 O(l) is one.)

    C 3 H 6 O(l) + 4 O 2 (g) → 3 CO 2 (g) + 3 H 2 O(l)

3.19

  1. Mg(s) + Cl 2 (g) → MgCl 2 (s)
  2. BaCO 3 ( s ) Δ BaO ( s ) + CO 2 ( g )
  3. C 8 H 8 (l) + 10 O 2 (g) → 8 CO 2 (g) + 4 H 2 O(l)
  4. CH 3 OCH 3 is C 2 H 6 O. C 2 H 6 O(g) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O(l)

3.20

  1. 2 Ti(s) + O 2 (g) → 2 TiO(s) [or Ti(s) + O 2 (g) → TiO 2 (s)]
  2. 2 Ag 2 O ( s ) Δ 4 Ag ( s ) + O 2 ( g )
  3. 2 C 3 H 7 OH(l) + 9 O 2 (g) → 6 CO 2 (g) + 8 H 2 O(l)
  4. 2 C 5 H 12 O(l) + 15 O 2 (g) → 10 CO 2 (g) + 12 H 2 O(l)

3.21

  1. 2 C 3 H 6 (g) + 9 O 2 (g) → 6 CO 2 (g) + 6 H 2 O(g) combustion
  2. NH 4 NO 3 (s) → N 2 O(g) + 2 H 2 O(g) decomposition
  3. C 5 H 6 O(l) + 6 O 2 (g) → 5 CO 2 (g) + 3 H 2 O(g) combustion
  4. N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) combination
  5. K 2 O(s) + H 2 O(l) → 2 KOH(aq) combination

3.22

  1. PbCO 3 (s) → PbO(s) + CO 2 (g) decomposition
  2. C 2 H 4 (g) + 3 O 2 (g) → 2 CO 2 (g) + 2 H 2 O(g) combustion
  3. 3 Mg(s) + N 2 (g) → Mg 3 N 2 (s) combination
  4. C 7 H 8 O 2 (l) + 8 O 2 (g) → 7 CO 2 (g) + 4 H 2 O(g) combustion
  5. 2 Al(s) + 3 Cl 2 (g) → 2 AlCl 3 (s)  combination

Formula Weights (Section 3.3)

3.23 Analyze .  Given molecular formula or name, calculate formula weight.

Plan .  If a name is given, write the correct molecular formula. Then, follow the method in Sample Exercise 3.5. Solve .

  1. HNO 3 : 1(1.0) + 1(14.0) + 3(16.0) = 63.0 amu
  2. KMnO 4 : 1(39.1) + 1(54.9) + 4(16.0) = 158.0 amu
  3. Ca 3 (PO 4 ) 2 : 3(40.1) + 2(31.0) + 8(16.0) = 310.3 amu
  4. SiO 2 : 1(28.1) + 2(16.0) = 60.1 amu
  5. Ga 2 S 3 : 2(69.7) + 3(32.1) = 235.7 amu
  6. Cr 2 (SO 4 ) 3 : 2(52.0) + 3(32.1) + 12(16.0) = 392.3 amu
  7. PCl 3 : 1(31.0) + 3(35.5) = 137.5 amu

3.24   Formula weight in amu to 1 decimal place.

  1. N 2 O: FW = 2(14.0) + 1(16.0) = 44.0 amu
  2. C 6 H 5 COOH: 7(12.0) + 6(1.0) + 2(16.0) = 122.0 amu
  3. Mg(OH) 2 : 1(24.3) + 2(16.0) + 2(1.0) = 58.3 amu
  4. (NH 2 ) 2 CO: 2(14.0) + 4(1.0) + 1(12.0) + 1(16.0) = 60.0 amu
  5. CH 3 CO 2 C 5 H 11 : 7(12.0) + 14(1.0) + 2(16.0) = 130.0 amu

3.25 Plan .  Calculate the formula weight (FW), then the mass % oxygen in the compound. Solve .

  1. C 17 H 19 NO 3 : FW = 17(12.0) + 19(1.0) + 1(14.0) + 3(16.0) = 285.0 amu

    % O = 3(16 .0) amu 285.0 amu × 100 = 16.842 = 16.8 %

  2. C 18 H 21 NO 3 : FW = 18(12.0) + 21(1.0) + 1(14.0) + 3(16.0) = 299.0 amu

    % O = 3(16 .0) amu 299.0 amu × 100 = 16.054 = 16.1 %

  3. C 17 H 21 NO 4 : FW = 17(12.0) + 21(1.0) + 1(14.0) + 4(16.0) = 303.0 amu

    % O = 4(16 .0) amu 303.0 amu × 100 = 21.122 = 21.1 %

  4. C 22 H 24 N 2 O 8 : FW = 22(12.0) + 24(1.0) + 2(14.0) + 8(16.0) = 444.0 amu

    % O = 8(16 .0) amu 4 44 .0 amu × 100 = 28.829 = 28.8 %

  5. C 41 H 64 O 13 : FW = 41(12.0) + 64(1.0) + 13(16.0) = 764.0 amu

    % O = 13(16 .0) amu 7 64 amu × 100 = 27.225 = 27.2 %

  6. C 66 H 75 Cl 2 N 9 O 24 : FW = 66(12.0) + 75(1.0) + 2(35.5) + 9(14.0) + 24(16.0) = 1448.0 amu

    % O = 24(16 .0) amu 1448 .0 amu × 100 = 26.519 = 26.5 %

3.26

  1. C 2 H 2 : FW = 2(12.0) + 2(1.0) = 26.0 amu

    % C = 2(12 .0) amu 26.0 amu × 100 = 92.3 %

  2. HC 6 H 7 O 6 : FW = 6(12.0) + 8(1.0) + 6(16.0) = 176.0 amu

    % H = 8(1 .0) amu 176.0 amu × 100 = 4.5 %

  3. (NH 4 ) 2 SO 4 : FW = 2(14.0) + 8(1.0) + 1(32.1) + 4(16.0) = 132.1 amu

    % H = 8(1 .0) amu 132.1 amu × 100 = 6.1 %

  4. PtCl 2 (NH 3 ) 2 : FW = 1(195.1) + 2(35.5) + 2(14.0) + 6(1.0) = 300.1 amu

    % Pt = 1(195 .1) amu 300.1 amu × 100 = 65.01 %


  1. C 18 H 24 O 2 : FW = 18(12.0) + 24(1.0) + 2(16.0) = 272.0 amu

    % O = 2(16 .0) amu 272.0 amu × 100 = 11.8 %

  1. C 18 H 27 NO 3 : FW = 18(12.0) + 27(1.0) + 1(14.0) + 3(16.0) = 305.0 amu

    % C = 18(12 .0) amu 305.0 amu × 100 = 70.8 %

3.27 Plan .  Follow the logic for calculating mass % C given in Sample Exercise 3.6. Solve .

  1. C 7 H 6 O: FW = 7(12.0) + 6(1.0) + 1(16.0) = 106.0 amu

    % C = 7(12 .0) amu 106.0 amu × 100 = 79.2 %

  2. C 8 H 8 O 3 : FW = 8(12.0) + 8(1.0) + 3(16.0) = 152.0 amu

    % C = 8(12 .0) amu 152.0 amu × 100 = 63.2 %

  3. C 7 H 14 O 2 : FW = 7(12.0) + 14(1.0) + 2(16.0) = 130.0 amu

    % C = 7(12 .0) amu 130.0 amu × 100 = 64.6 %

3.28

  1. CO 2 : FW = 1(12.0) + 2(16.0) = 44.0 amu

    % C = 12 .0 amu 44 .0 amu × 100 = 27.3 %

  2. CH 3 OH: FW = 1(12.0) + 4(1.0) + 1(16.0) = 32.0 amu

    % C = 12 .0 amu 32 .0 amu × 100 = 37.5 %

  3. C 2 H 6 : FW = 2(12.0) + 6(1.0) = 30.0 amu

    % C = 2(12 .0) amu 30.0 amu × 100 = 80.0 %

  4. CS(NH 2 ) 2 : FW = 1(12.0) + 1(32.1) + 2(14.0) + 4(1.0) = 76.1 amu

    % C = 12 .0 amu 76 .1 amu × 100 = 15.8 %

Avogadro’s Number and the Mole (Section 3.4)

3.29

  1. False. A mole of horses contains four moles of horse legs.
  2. True.
  3. False. The mass of one mole of water is 18.0 g.
  4. True. Electrically neutral NaCl is composed of Na + cations and Cl− anions.

3.30

  1. exactly 12 g
  2. 6.0221421 × 10 23 , Avogadro’s number

3.31 Plan .  Because the mole is a counting unit, use it as a basis of comparison; determine the total moles of atoms in each given quantity. Solve .

23 g Na contains 1 mol of atoms

0.5 mol H 2 O contains (3 atoms × 0.5 mol) = 1.5 mol atoms

6.0 × 10 23 N 2 molecules contains (2 atoms × 1 mol) = 2 mol atoms

3.32  42 g NaHCO 3 (molar mass = 84 g) contains (6 atoms × 0.5 mol) = 3 mol atoms

1.5 mol CO 2 contains (3 atoms × 1.5 mol) = 4.5 mol atoms

6.0 × 10 24 Ne atoms contains (1 atoms × 10 mol) = 10 mol atoms

3.33 Analyze .  Given: 160 lb/person; Avogadro’s number of people, 6.022 × 10 23 people. Find: mass in kg of Avogadro’s number of people; compare with mass of Earth.

Plan .  people → mass in lb → mass in kg; mass of people/mass of Earth

Solve 6.022 × 10 23 people × 1 6 0 lb person × 1 kg 2 . 2046 lb = 4 . 3 7 0 × 1 0 25 = 4.37 × 1 0 2 5 or 4 . 4 × 10 25 kg

4 . 370 × 1 0 2 5 kg of people 5 . 98 × 1 0 2 4 kg Earth = 7 . 31 or 7 . 3

One mole of people weighs 7.3 times as much as Earth.

Check .  This mass of people is reasonable because Avogadro’s number is large.

Estimate: 160 lb ≈ 70 kg; 6 × 10 23 × 70 = 420 × 10 23 = 4.2 × 10 25 kg

3.34  321 million = 321 × 10 6 = 3.21 × 10 8 people

6 .022 × 10 23 ¢ 3 .21 × 10 8 people × $1 100 ¢ = $6 .022 × 10 21 3 .21 × 10 8 people = 1 .876 × 10 13 = $1 .88 × 10 13 gift/person $17 .419 trillion = $1 .7419 × 10 13 = $1 .74 × 10 13 debt; $1 .876 × 10 13 gift 1 .7419 × 10 13 debt = 1 .1011 = 1 .10

Each person would receive an amount that is 1.10 times the dollar amount of the national debt.


3.35

  1. Analyze .  Given: 0.105 mol sucrose, C 12 H 22 O 11 . Find: mass in g.

    Plan .  Use molar mass (g/mol) of C 12 H 22 O 11 to find g C 12 H 22 O 11 .

    Solve .  molar mass = 12(12.0107) + 22(1.00794) + 11(15.9994) = 342.296 = 342.30

    0 .105 mol sucrose × 342 .30 g 1 mol = 35 .942 = 35 .9 g C 12 H 22 O 11

    Check .  0.1(342) = 34.2 g. The calculated result is reasonable.

  1. Analyze .  Given: mass. Find: moles. Plan .  Use molar mass of Zn(NO 3 ) 2 .

    Solve .  molar mass = 1(65.39) + 2(14.0067) + 6(15.9994) = 189.3998 = 189.40

    143 .50 g Zn(NO 3 ) 2 × 1 mol 189 .40 g Zn(NO 3 ) 2 = 0 .75766 mol Zn(NO 3 ) 2

    Check .  140/180 ≈ 7/9 = 0.78 mol

  1. Analyze .  Given: moles. Find: molecules. Plan .  Use Avogadro’s number.

    Solve. 1 .0 × 10 -6 mol CH 3 CH 2 OH × 6 .022 × 10 23 molecules 1 mol = 6 .022 × 10 17 6 .0×10 17 CH 3 CH 2 OH molecules

    Check .  (1.0 × 10 −6 )(6 × 10 23 ) = 6 × 10 17

  1. Analyze .  Given: mol NH 3 . Find: N atoms.

    Plan .  mol NH 3 → mol N atoms → N atoms

    S o l v e . 0 .410 mol NH 3 × 1 mol N atoms 1 mol NH 3 × 6.022 × 10 23 atoms 1 mol = 2.47 × 10 23 N atoms

    Check .  (0.4)(6 × 10 23 ) = 2.4 × 10 23 .

3.36

  1. molar mass = 1(112.41) + 1(32.07) = 144.48 g

    1.50 × 10 2 mol CdS × 144.48 g 1 mol = 2. 17 g CdS

  2. molar mass = 1(14.01) + 4(1.008) + 1(35.45) = 53.49 g/mol

    8 6 .6 g NH 4 Cl × 1 mol 53 .49 g = 1.6190 = 1 .62 mol NH 4 Cl

  3. 8.447 × 10 2 mol C 6 H 6 × 6.02214 × 10 23 molecules 1 mol = 5.087 × 10 22 C 6 H 6 molecules
  4. 6.25 × 10 3 mol Al(NO 3 ) 3 × 9 mol O 1 mol Al(NO 3 ) 3 × 6.022 × 10 23 O atoms 1 mol = 3.39 × 10 22 O atoms

3.37 Analyze/Plan .  See 3.35 for stepwise problem-solving approaches. Solve .

  1. (NH 4 ) 3 PO 4 molar mass = 3(14.007) + 12(1.008) + 1(30.974) + 4(16.00) = 149.091 = 149.1 g/mol

    2.50 × 10 3 mol (NH 4 ) 3 PO 4 × 149.1 g (NH 4 ) 3 PO 4 1 mol = 0.373 g (NH 4 ) 3 PO 4

  2. AlCl 3 molar mass = 26.982 + 3(35.453) = 133.341 = 133.34 g/mol

    0.2550 g AlCl 3 × 1 mol 133 .34 g AlCl 3 × 3 mol Cl 1 mol AlCl 3 = 5.737 × 10 3 mol  Cl


  1. C 8 H 10 N 4 O 2 molar mass = 8(12.01) + 10(1.008) + 4(14.01) + 2(16.00) = 194.20 = 194.2 g/mol

    7 . 7 0 × 1 0 2 0 molecules × 1 mol 6 . 0 2 2 × 1 0 2 3 molecules × 194 . 2 g C 8 H 1 0 N 4 O 2 1 mol caffeine = 0 . 248 g C 8 H 10 N 4 O 2

  1. 0.406 g cholesterol 0.00105 mol = 387 g cholesterol/mol

3.38

  1. Fe 2 (SO 4 ) 3 molar mass = 2(55.845) + 3(32.07) + 12(16.00) = 399.900 = 399.9 g/mol

    1 . 2 2 3 mol Fe 2 ( SO 4 ) 3 × 3 9 9 . 9 g Fe 2 ( SO 4 ) 3 1 mol = 489 . 077 = 489 . 1 g Fe 2 ( SO 4 ) 3

  2. (NH 4 ) 2 CO 3 molar mass = 2(14.007) + 8(1.008) + 12.011 + 3(15.9994) = 96.0872 = 96.087 g/mol

    6.955 g ( NH 4 ) 2 CO 3 × 1 mol 96.087 g (NH 4 ) 2 CO 3 × 2 mol NH 4 + 1 mol (NH 4 ) 2 CO 3 = 0.1448 mol NH 4 +

  3. C 9 H 8 O 4 molar mass = 9(12.01) + 8(1.008) + 4(16.00) = 180.154 = 180.2 g/mol

    1. 50 × 10 21 molecules × 1 mol 6 .022 × 10 23 molecules × 180.2 g C 9 H 8 O 4 1 mol aspirin = 0.449 g C 9 H 8 O 4

  4. 15.86 g diazepam 0.05570 mol = 284.7 g diazepam/mol

3.39

  1. C 6 H 10 OS 2 molar mass = 6(12.01) + 10(1.008) + 1(16.00) + 2(32.07) = 162.28 = 162.3 g/mol
  2. Plan .  mg → g → mol Solve .

    5.00 mg allicin × 1 × 10 3 g 1 mg × 1 mol 162 .3 g = 3.081 × 10 5 = 3.08 × 10 5 mol allicin

    Check .  5.00 mg is a small mass, so the small answer is reasonable.

    (5 × 10 −3 )/200 = 2.5 × 10 −5

  3. Plan .  Use mol from part (b) and Avogadro’s number to calculate molecules.

    S o l v e . 3 .081 × 10 5 mol allicin × 6 .022 × 10 23 molecules mol = 1.855 × 10 19 = 1.86 × 10 19 allicin molecules

    Check . (3 × 10 −5 )(6 × 10 23 ) = 18 × 10 18 = 1.8 × 10 19

  4. Plan .  Use molecules from part (c) and molecular formula to calculate S atoms.

    S o l v e . 1 .855 × 10 19 allicin molecules × 2 S atoms 1 allicin molecule = 3.71 × 10 19 S atoms

    Check .  Obvious.


3.40

  1. C 14 H 18 N 2 O 5 molar mass = 14(12.01) + 18(1.008) + 2(14.01) + 5(16.00) = 294.30 g/mol
  2. 1.00 mg aspartame × 1 × 10 3 g 1 mg × 1 mol 294 .3 g = 3.398 × 10 6 = 3.40 × 10 6 mol aspartame
  3. 3.398 × 10 6 mol aspartame × 6 .022 × 10 23 molecules 1 mol = 2.046 × 10 18 2 . 05 × 10 18 aspartame molecules
  4. 2.046 × 10 18 aspartame molecules × 18 H atoms 1 aspartame molecule = 3.68 × 10 19 H atoms

3.41

  1. Analyze .  Given: C 6 H 12 O 6 , 1.250 × 10 21 C atoms. Find: H atoms.

    Plan .  Use molecular formula to determine number of H atoms that are present with 1.250 × 10 21 C atoms. Solve .

    12 H atoms 6 C atoms = 2 H 1 C × 1.250 × 10 21 C atoms = 2 .500 × 10 21 H atoms

    Check .  (2 × 1 × 10 21 ) = 2 × 10 21

  2. Plan .  Use molecular formula to find the number of glucose molecules that contain 1.250 × 10 21 C atoms. Solve .

    1 C 6 H 12 O 6 molecule 6 C atoms × 1.250 × 10 21 C atoms = 2.0833 × 10 20 = 2.083 × 10 20 C 6 H 12 O 6 molecules

    Check .  (12 × 10 20 /6) = 2 × 10 20

  3. Plan .  Use Avogadro’s number to change molecules → mol. Solve .

    2.0833 × 10 20 C 6 H 12 O 6 molecules × 1 mol 6 .022 × 10 23 molecules = 3.4595 × 10 4 = 3.460 × 10 4 mol C 6 H 12 O 6

    Check .  (2 × 10 20 )/(6 × 10 23 ) = 0.33 × 10 −3 = 3.3 × 10 −4

  4. Plan .  Use molar mass to change mol → g. Solve .

    1 mole of C 6 H 12 O 6 weighs 180.0 g (Sample Exercise 3.9)

    3.4595 × 10 4 mol C 6 H 12 O 6 × 180.0 g C 6 H 12 O 6 1 mol = 0.06227 g C 6 H 12 O 6

    Check .  3.5 × 180 = 630; 630 × 10 −4 = 0.063

3.42

  1. 3. 88 × 10 21 H atoms × 19 C atoms 28 H atoms = 2. 63 × 10 21 C atoms
  2. 3 .88 × 1 0 2 1 H atoms × 1 C 19 H 28 O 2 molecule 2 8 H atoms = 1.3857 × 10 20 = 1 . 39 × 10 20 C 19 H 28 O 2 molecules

  1. 1.3857 × 10 20 C 19 H 28 O 2 molecules × 1 mol 6 .022 × 10 23 molecules = 2.301 × 10 4 = 2 .30 × 10 4 mol C 19 H 28 O 2
  1. C 19 H 28 O 2 molar mass = 19(12.01) + 28(1.008) + 2(16.00) = 288.41 = 288.4 g/mol

    2.301 × 10 4 mol C 19 H 28 O 2 × 288.4 g C 19 H 28 O 2 1 mol = 0.0664 g C 19 H 28 O 2

3.43 Analyze .  Given: g C 2 H 3 Cl/L. Find: mol/L, molecules/L.

Plan .  The /L is constant throughout the problem, so we can ignore it. Use molar mass for g → mol, Avogadro’s number for mol → molecules. Solve .

2.0 × 10 6 g C 2 H 3 Cl 1 L × 1 mol C 2 H 3 Cl 62.50 g C 2 H 3 Cl = 3.20 × 10 8 = 3 .2 × 10 8 mol C 2 H 3 Cl/L

3.20 × 10 8 mol C 2 H 3 Cl 1 L × 6.022 × 10 23 molecules 1 mol = 1.9 × 10 16 molecules/L

Check .  (200 × 10 −8 )/60 = 2.5 × 10 −8 mol

(2.5 × 10 −8 ) × (6 × 10 23 ) = 15 × 10 15 = 1.5 × 10 16

3.44 25 × 10 6 g C 21 H 30 O 2 × 1 mol C 21 H 30 O 2 314.5 g C 21 H 30 O 2 = 7.95 × 10 8 = 8.0 × 10 8 mol C 21 H 30 O 2

7.95 × 10 8 mol C 21 H 30 O 2 × 6.022 × 10 23 molecules 1 mol = 4.8 × 10 16 C 21 H 30 O 2 molecules

Empirical Formulas from Analyses (Section 3.5)

3.45

  1. Analyze .  Given: moles. Find: empirical formula.

    Plan .  Find the simplest ratio of moles by dividing by the smallest number of moles present.

    Solve .  0.0130 mol C / 0.0065 = 2

    0.0390 mol H / 0.0065 = 6

    0.0065 mol O / 0.0065 = 1

    The empirical formula is C 2 H 6 O.

    Check .  The subscripts are simple integers.

  2. Analyze .  Given: grams. Find: empirical formula.

    Plan .  Calculate the moles of each element present, then the simplest ratio of moles.

    S o l v e . 11 .66 g Fe × 1 mol Fe 55 .85 g Fe = 0.2088 mol Fe;  0 .2088/0 .2088 = 1 5 .01 g O × 1 mol O 16 .00 g O = 0.3131 mol O;  0 .3131/0 .2088 1 .5

    Multiplying by two, the integer ratio is 2 Fe : 3 O; the empirical formula is Fe 2 O 3 .

    Check .  The subscripts are simple integers.


  1. Analyze .  Given: mass %. Find: empirical formulas.

    Plan .  Assume 100 g sample, calculate moles of each element, find the simplest ratio of moles.

    S o l v e . 40 .0 g C × 1 mol C 12 .01 g C = 3.33 mol C;  3 .33/3 .33 = 1 6 .7 g H × 1 mol H 1 .008 mol H = 6.65 mol H;  6 .65/3 .33 2 53 .3 g O × 1 mol O 16 .00 mol O = 3.33 mol O;  3 .33/3 .33 = 1

    The empirical formula is CH 2 O.

    Check .  The subscripts are simple integers.

3.46

  1. Calculate the simplest ratio of moles.

    0.104 mol K / 0.052 = 2

    0.052 mol C / 0.052 = 1

    0.156 mol O / 0.052 = 3

    The empirical formula is K 2 CO 3 .

  2. Calculate moles of each element present, then the simplest ratio of moles.

    5.28 g Sn × 1 mol Sn 118.7 g Sn = 0.04448 mol Sn ; 0.04448 / 0.04448 = 1 3.37 g F × 1 mol F 19.00 g F = 0.1774 mol F ; 0.1774 / 0.04448 4 The integer ratio is 1 Sn : 4 F ; the empirical formula is SnF 4 .

  3. Assume 100 g sample, calculate moles of each element, find the simplest ratio of moles.

    87.5 % N = 87 .5 g N × 1 mol N 14 .01 g = 6.25 mol N;   6 .25/6 .25 = 1 12 .5% H = 12 .5 g H × 1 mol 1 .008 g = 12.4 mol H;   12 .4/6 .25 2

    The empirical formula is NH 2 .

3.47 Analyze/Plan .  The procedure in all these cases is to assume 100 g of sample, calculate the number of moles of each element present in that 100 g, then obtain the ratio of moles as smallest whole numbers. Solve .

  1. 10.4 g C × 1 mol C 12 .01 g C = 0.866 mol C;   0 .866/0 .866 = 1

    27.8 g S × 1 mol S 32 .07 g S = 0.867 mol S; 0 .867/0 .866 1

    61.7 g Cl × 1 mol Cl 35 .45 g Cl = 1.74 mol Cl;   1 .74/0 .866 2

    The empirical formula is CSCl 2 .


  1. 21 .7 g C× 1 mol C 12 .01 g C =1 .81 mol C;   1 .81/0 .600 3 9 .6 g O× 1 mol O 16 .00 g O =0 .600 mol O;   0 .600 / 0 .600 = 1 68 .7 g F× 1 mol F 19 .00 g F =3 .62 mol F;   3 .62 / 0 .600 6

    The empirical formula is C 3 OF 6 .

  1. The mass of F is [100 g total − (32.79 g Na + 13.02 Al)] = 54.19 g F

    32 .79 g Na× 1 mol Na 22 .99 g Na =1 .426 mol Na;   1 .426 / 0 .4826 3 13 .02 g Al× 1 mol Al 26 .98 g Al =0 .4826 mol Al;   0 .4826 / 0 .4826=1 54 .19 g F× 1 mol F 19 .00 g F =2 .852 mol F;   2 .852 / 0 .4826 6

    The empirical formula is Na 3 AlF 6 .

3.48  See Solution 3.47 for stepwise problem-solving approach.

  1. 55 .3 g K× 1 mol K 39 .10 g K =1 .414 mol K; 1 .414/0 .4714 3 14 .6 g P× 1 mol P 30 .97 g P =0 .4714 mol P;  0 .4714/0 .4714 = 1 30 .1 g O× 1 mol O 16 .00 g O =1 .881 mol O;  1 .881/0 .4714 4

    The empirical formula is K 3 PO 4 .

  2. 24 .5 g Na × 1 mol Na 22 .99 g Na = 1 .066 mol Na;  1 .066/0 .5304 2 14 .9 g Si × 1 mol Si 28 .09 Si = 0 .5304 mol si;  0 .5304/0 .5304 = 1 60 .6 g F × 1 mol F 19 .00 g F = 3 .189 mol F; 3 .189/0 .5304 6

    The empirical formula is Na 2 SiF 6 .

  3. The mass of O is [100 g total − (62.1 g C + 5.21 g H + 12.1 g N)] = 20.59 = 20.6 g O

    62 .1 g C× 1 mol C 12 .01 g C = 5 .17 mol C;   5 .17 / 0 .864 6 5 .21 g H× 1 mol H 1 .008 g H = 5 .17 mol O;   5 .17 / 0 .864 6 12 .1 g N× 1 mol N 14 .01 g N = 0 .864 mol N;   0 .864 / 0 .864 = 1 20 .6 g O× 1 mol O 16 .00 g O = 1 .29 mol O;   1 .29 / 0 .864 1 .5

    Multiplying by 2, the empirical formula is C 12 H 12 N 2 O 3 .


3.49 Analyze . Given: mass% F; empirical formula XF 3 implies 3:1 ratio of mol F to mol X.

Find: atomic mass (AM) of X.

Plan. Calculate mol F. This is 3 times mol X. mol X = 35 g X/AM X.

mol F/3 = 35 g X/AM X. Solve for AM X.

Solve. Mol F = 65/19.0 = 3.421 = 3.4; mol X = 3.421/3 = 1.14035 = 1.1

1.14035 mol X = 35 g X/AM X; AM X = 35 g X/1.14035 mol X = 30.69 = 31 g/mol

The element is likely phosphorus and the compound is then PF 3 .

3.50  Follow the logic in Solution 3.49. Match the calculated atomic mass to that of an element.

mol Cl = 75.0/35.453 = 2.1155 = 2.12; mol X = 2.1155/4 = 0.52887 = 0.529

0.52887 mol X = 25.0 g X/AM X; AM X = 25.0 g X/0.52887 mol X = 47.271 = 47.3 g/mol

The element with atomic mass closest to 47.3 is Ti , atomic mass = 47.867 g/mol.

3.51 Analyze .  Given: empirical formula, molar mass. Find: molecular formula.

Plan .  Calculate the empirical formula weight (FW); divide FW by molar mass (MM) to calculate the integer that relates the empirical and molecular formulas. Check. If FW/MM is an integer, the result is reasonable. Solve .

  1. FW CH 2 = 12.0 + 2 ( 1.01 ) = 14.0 MM FW = 84.0 14.0 = 6.00 = 6

    The subscripts in the empirical formula are multiplied by 6. The molecular formula is C 6 H 12 .

  2. FW NH 2 Cl = 14 .01 + 2(1 .008) + 35 .45 = 51 .48 . MM FW = 51.5 51.5 = 1

    The empirical and molecular formulas are NH 2 Cl.

3.52

  1. FW HCO 2 = 12 .01 + 1 .008 + 2(16 .00) = 45 .0 MM FW = 90.0 45.0 = 2

    The molecular formula is C 2 H 2 O 4 .

  2. FW C 2 H 4 O = 2( 12.0 ) + 4 ( 1.01 ) + 16.0 = 44.0 MM FW = 88.0 44.0 = 2.00 = 2

    The molecular formula is C 4 H 8 O 2 .

3.53 Analyze .  Given: mass %, molar mass. Find: molecular formula.

Plan .  Use the plan detailed in Solution 3.47 to find an empirical formula from mass % data. Then use the plan detailed in Solution 3.51 to find the molecular formula. Note that some indication of molar mass must be given, or the molecular formula cannot be determined. Check .  If there is an integer ratio of moles and MM/ FW is an integer, the result is reasonable. Solve .

  1. 92.3 g C × 1 mol C 12 .01 g C = 7.685 mol C;  7 .685/7 .639 = 1 .006 1

    7.7 g H × 1 mol H 1 .008 g H = 7.639 mol H;  7 .639/7 .639 = 1

    The empirical formula is CH, FW = 13.

    MM FW = 104 13 = 8 ; the molecular formula is C 8 H 8 .


  1. 49 .5 g C × 1 mol C 12 .01 g C = 4 .12 mol C;   4 .12/1 .03 4 5 .15 g H × 1 mol H 1 .008 g H = 5 .11 mol H;   5 .11/1 .03 5 28 .9 g N × 1 mol N 14 .01 g N = 2 .06 mol N;   2 .06/1 .03 2 16 .5 g O × 1 mol O 16 .00 g O = 1 .03 mol O;   1 .03/1 .03 = 1

    Thus, C 4 H 5 N 2 O, FW = 97. If the molar mass is about 195, a factor of 2 gives the molecular formula C 8 H 10 N 4 O 2 .

  1. 35 .51 g C × 1 mol C 12 .01 g C = 2 .96 mol C;  2 .96/0 .592 = 5 4 .77 g H × 1 mol H 1 .008 g H = 4 .73 mol H; 4 .73/0 .592 = 7 .99 8 37 .85 g O × 1 mol O 16 .00 g O = 2 .37 mol O; 2 .37/0 .592 = 4 8 .29 g N × 1 mol N 14 .01 g N = 0 .592 mol N;  0 .592/0 .592 = 1 13 .60 g Na × 1 mol Na 22 .99 g Na = 0 .592 mol Na;  0 .592/0 .592 = 1

    The empirical formula is C 5 H 8 O 4 NNa, FW = 169 g. Because the empirical formula weight and molar mass are approximately equal, the empirical and molecular formulas are both NaC 5 H 8 O 4 N.

3.54  Assume 100 g in the following problems.

  1. 75 .69 g C × 1 mol C 12 .01 g C =6 .30 mol C;  6 .30/0 .969 = 6 .5 8 .80 g H× 1 mol H 1 .008 g H =8 .73 mol H;  8 .73/0 .969 = 9 .0 15 .51 g O× 1 mol O 16 .00 g O =0 .969 mol O; 0 .969/0 .969 = 1

    Multiply by 2 to obtain the integer ratio 13:18:2. The empirical formula is C 13 H 18 O 2 , FW = 206 g. Because the empirical formula weight and the molar mass are equal (206 g), the empirical and molecular formulas are C 13 H 18 O 2 .

  1. 58 .55 g C× 1 mol C 12 .01 g C =4 .875 mol C;   4 .875/1 .956 2 .5 13 .81 g H× 1 mol H 1 .008 g H =13 .700 mol H;   13 .700/1 .956 7 .0 27 .40 g N× 1 mol N 14 .01 g N =1 .956 mol N;    1 .956/1 .956 = 1 .0

    Multiply by 2 to obtain the integer ratio 5:14:2. The empirical formula is C 5 H 14 N 2 ; FW = 102. Because the empirical formula weight and the molar mass are equal (102 g), the empirical and molecular formulas are C 5 H 14 N 2 .

  1. 59 .0 g C × 1 mol C 12 .01 g C =4 .91 mol C;   4 .91 / 0 .550 9 7 .1 g H × 1 mol H 1 .008 g H =7 .04 mol H;   7 .04 / 0 .550 13 26 .2 g O × 1 mol O 16 .00 g O =1 .64 mol O;   1 .64 / 0 .550 3 7 .7 g N × 1 mol N 14 .01 g N =0 .550 mol N;   0 .550 / 0 .550 = 1

    The empirical formula is C 9 H 13 O 3 N, FW = 183 amu (or g). Because the molar mass is approximately 180 amu, the empirical formula and molecular formula are the same, C 9 H 13 O 3 N.

3.55

  1. Analyze .  Given: mg CO 2 , mg H 2 O Find: empirical formula of hydrocarbon, C x H y

    Plan .  Upon combustion, all C → CO 2 , all H → H 2 O.

    mg CO 2 → g CO 2 → mol C; mg H 2 O → g H 2 O, mol H

    Find simplest ratio of moles and empirical formula. Solve .

    5 . 8 6 × 1 0 3 g CO 2 × 1 mol CO 2 4 4 . 0 1 g CO 2 × 1 mol C 1 mol CO 2 = 1 . 3 3 × 1 0 4 mol C 1 . 3 7 × 1 0 3 g H 2 O × 1 mol H 2 O 1 8 . 0 2 g H 2 O × 2 mol H 1 mol H 2 O = 1 . 5 2 × 1 0 4 mol H

    Dividing both values by 1.33 × 10 −4 gives C:H of 1:1.14. This is not “close enough” to be considered 1:1. No obvious multipliers (2, 3, 4) produce an integer ratio. Testing other multipliers (trial and error!), the correct factor seems to be 7. The empirical formula is C 7 H 8 .

    Check .  See discussion of C:H ratio above.

  2. Analyze .  Given: g of menthol, g CO 2 , g H 2 O, molar mass. Find: molecular formula.

    Plan/Solve .  Calculate mol C and mol H in the sample.

    0 . 2 8 2 9 g CO 2 × 1 mol CO 2 4 4 . 0 1 g CO 2 × 1 mol C 1 mol CO 2 = 0 . 0 0 6 4 2 8 1 = 0 . 0 0 6 4 2 8 mol C 0 . 1 1 5 9 g H 2 O × 1 mol H 2 O 1 8 . 0 2 g H 2 O × 2 mol H 1 mol H 2 O = 0 . 0 1 2 8 6 3 = 0 . 0 1 2 8 6 mol H

    Calculate g C, g H and get g O by subtraction.

    0 . 0 0 6 4 2 8 1 mol C × 1 2 . 0 1 g C 1 mol C = 0 . 0 7 7 2 0 g C


    0 . 0 1 2 8 6 3 mol H × 1 . 0 0 8 g H 1 mol H = 0 . 0 1 2 9 7 g H

    mass O = 0.1005 g sample − (0.07720 g C + 0.01297 g H) = 0.01033 g O

    Calculate mol O and find integer ratio of mol C: mol H: mol O.

    0 . 0 1 0 3 3 g O × 1 mol O 1 6 . 0 0 g O = 6 . 4 5 6 × 1 0 4 mol O

    Divide moles by 6.456 × 10 −4 .

    C : 0 . 0 0 6 4 2 8 6 . 4 5 6 × 1 0 4 1 0 ; H : 0 . 0 1 2 8 6 6 . 4 5 6 × 1 0 4 2 0 ; O : 6 . 4 5 6 × 1 0 4 6 . 4 5 6 × 1 0 4 = 1

    The empirical formula is C 10 H 20 O.

    FW = 1 0 ( 1 2 ) + 2 0 ( 1 ) + 1 6 = 1 5 6 ; M FW = 1 5 6 1 5 6 = 1

    The molecular formula is the same as the empirical formula, C 10 H 20 O.

    Check .  The mass of O wasn’t negative or greater than the sample mass; empirical and molecular formulas are reasonable.

3.56

  1. Plan .  Calculate mol C and mol H, then g C and g H; get g O by subtraction.

    Solve .

    6 . 3 2 × 1 0 3 g CO 2 × 1 mol CO 2 4 4 . 0 1 g CO 2 × 1 mol C 1 mol CO 2 = 1 . 4 3 6 × 1 0 4 = 1 . 4 4 × 1 0 4 mol C 2 . 5 8 × 1 0 3 g H 2 O × 1 mol H 2 O 1 8 . 0 2 g H 2 O × 2 mol H 1 mol H 2 O = 2 . 8 6 3 × 1 0 4 = 2 . 8 6 × 1 0 4 mol H 1 . 4 3 6 × 1 0 4 mol C × 1 2 . 0 1 g C 1 mol C = 1 . 7 2 5 × 1 0 3 g C = 1 . 7 3 mg C 2 . 8 6 3 × 1 0 4 mol H × 1 . 0 0 8 g H 1 mol H = 2 . 8 8 6 × 1 0 4 g H = 0 . 2 8 9 mg H

    mass of O = 2.78 mg sample − (1.725 mg C + 0.289 mg H) = 0.77 mg O

    0 . 7 7 × 1 0 3 g O × 1 mol O 1 6 . 0 0 g O = 4 . 8 1 × 1 0 5 mol O . Divide moles by 4 . 8 1 × 1 0 5 . C : 1 . 4 4 × 1 0 4 4 . 8 1 × 1 0 5 3 ; H : 2 . 8 6 × 1 0 4 4 . 8 1 × 1 0 5 6 ; O : 4 . 8 1 × 1 0 5 4 . 8 1 × 1 0 5 = 1

    The empirical formula is C 3 H 6 O.

  2. Plan .  Calculate mol C and mol H, then g C and g H. In this case, get N by subtraction. Solve .

    1 4 . 2 4 2 × 1 0 3 g CO 2 × 1 mol CO 2 4 4 . 0 1 g CO 2 × 1 mol C 1 mol CO 2 = 3 . 2 3 6 1 × 1 0 4 mol C 4 . 0 8 3 × 1 0 3 g H 2 O × 1 mol H 2 O 1 8 . 0 2 g H 2 O × 2 mol H 1 mol H 2 O = 4 . 5 3 1 6 × 1 0 4 = 4 . 5 3 2 × 1 0 4 mol H


    3 . 2 3 6 1 × 1 0 4 mol C × 1 2 . 0 1 g C 1 mol H = 3 . 8 8 6 6 × 1 0 3 g C = 3 . 8 8 6 6 mg C 4 . 5 3 2 × 1 0 4 mol H × 1 . 0 0 8 g H 1 mol H = 0 . 4 5 6 8 3 × 1 0 3 g H = 0 . 4 5 6 8 mg H

    mass of N = 5.250 mg sample − (3.8866 mg C + 0.4568 mg H) = 0.9066 = 0.907 mg N

    0 . 9 0 6 6 × 1 0 3 g N × 1 mol N 1 4 . 0 1 g N = 6 . 4 7 × 1 0 5 mol N . Divide moles by 6.47 × 10 −5 .

    C : 3 . 2 4 × 1 0 4 6 . 4 7 × 1 0 5 5 ; H : 4 . 5 3 × 1 0 4 6 . 4 7 × 1 0 5 7 ; N : 6 . 4 7 × 1 0 5 6 . 4 7 × 1 0 5 = 1

    The empirical formula is C 5 H 7 N, FW = 81. A molar mass of 160 ± 5 indicates a factor of 2 and a molecular formula of C 10 H 14 N 2 .

3.57 Analyze . Given: mass of H 2 O and mass of CO 2 from combustion of 0.165 g of valproic acid; molar mass of valproic acid. Find: empirical and molecular formulas of valproic acid.

Plan .  Calculate mol C and mol H, then g C and g H; get g O by subtraction.

Solve .

0. 403 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 1 mol C 1 mol CO 2 = 9.157 × 10 3 = 9.16 × 10 3 mol C 0. 166 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 2 mol H 1 mol H 2 O = 0.018424 = 0.0184 mol H 9 .157 × 10 3 mol C × 12 .01 g C 1 mol C = 0.10998 g C = 0 .110 g C 0. 018424 mol H × 1 .008 g H 1 mol H = 0.01857 g H = 0.0186 g H

mass of O = 0.165 g sample − (0.10998 g C + 0.01857 g H) = 0.03645 = 0.036 g O

0.03645 g O × 1 mol O 16 .00 g O = 2.278 × 10 3 = 2 .3 × 10 3 mol O .  Divide moles by 2.278 × 10 3 .

C: 9 .157 × 10 3 2.278 × 10 3 4 ; H: 0.018424 2.278 × 10 3 8 ; O: 2.278 × 10 3 2.278 × 10 3 = 1

The empirical formula is C 4 H 8 O, FW = 72. A molar mass of 144 g/mol indicates a factor of two and a molecular formula of C 8 H 16 O 2 .

3.58 Analyze/Plan. Calculate theoretical mass% C and H in propenoic acid, C 3 H 4 O 2 , and experimental mass% C and H in the combustion sample. Compare the theoretical and experimental mass percents to determine if the sample is propenic acid. (This is a common method in modern chemical analysis.)

Solve . The molar mass of C 3 H 4 O 2 is [3(12.0107) + 4(1.00794 + 2(15.9994)] = 72.0627 = 72.06 g/mol

% C(theo) = 3(12 .01) g 72.06 g × 100 = 50.0 %


% H(theo) = 4(1 .008) g 72.06 g × 100 = 5.60 %

0. 374 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 1 mol C 1 mol CO 2 = 8.498 × 10 3 = 8.50 × 10 3 mol C 0. 102 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 2 mol H 1 mol H 2 O = 0.01132 = 0.0113 mol H 8.498 × 10 3 mol C × 12 .01 g C 1 mol C = 0.10206 g C = 0 .102 g C 0. 01132 mol H × 1 .008 g H 1 mol H = 0.01141 g H = 0.0114 g H

% C(exptl) = 0 .10206 g C 0.275 g sample × 100 = 37.1 %

% H(exptl) = 0 .01141 g H 0.275 g sample × 100 = 4.15 %

Clearly theoretical and experimental mass percents don’t match. The unknown liquid is not propenoic acid.

3.59 Analyze .  Given 2.558 g Na 2 CO 3 • xH 2 O, 0.948 g Na 2 CO 3 . Find: x.

Plan .  The reaction involved is Na 2 CO 3 • xH 2 O(s) → Na 2 CO 3 (s) + xH 2 O(g).

Calculate the mass of H 2 O lost and then the mole ratio of Na 2 CO 3 and H 2 O.

Solve . g H 2 O lost = 2.558 g sample − 0.948 g Na 2 CO 3 = 1.610 g H 2 O

0.948 g Na 2 CO 3 × 1 mol Na 2 CO 3 106.0 g Na 2 CO 3 = 0.00894 mol Na 2 CO 3

1.610 g H 2 O × 1 mol H 2 O 18.02 g H 2 O = 0.08935 mol H 2 O

The formula is Na 2 CO 3 10 H 2 O.

Check .  x is an integer.

3.60  The reaction involved is MgSO 4 • xH 2 O(s) → MgSO 4 (s) + xH 2 O(g). First, calculate the number of moles of product MgSO 4 ; this is the same as the number of moles of starting hydrate.

2.472 g MgSO 4 × 1 mol MgSO 4 120.4 g MgSO 4 × 1 mol MgSO 4 x H 2 O 1 mol MgSO 4 = 0.02053 mol MgSO 4 x H 2 O

Thus, 5 .061 g MgSO 4 x H 2 O 0.02053 = 246.5 g/mol = FW of MgSO 4 x H 2 O

FW of MgSO 4 • xH 2 O = FW of MgSO 4 + x(FW of H 2 O).

246.5 = 120.4 + x(18.02). x = 6.998. The hydrate formula is MgSO 4 7 H 2 O.

Alternatively, we could calculate the number of moles of water represented by weight loss: (5.061 − 2.472) = 2.589 g H 2 O lost.

2.589 g H 2 O × 1 mol H 2 O 18.02 g H 2 O = 0.1437 mol H 2 O; mol H 2 O mol MgSO 4 = 0.1437 0.02053 = 7.000

Again the correct formula is MgSO 4 7 H 2 O.


Quantitative Information from Balanced Equations (Section 3.6)

3.61   Na 2 SiO 3 (s) + 8 HF(aq) → H 2 SiF 6 (aq) + 2 NaF(aq) + 3 H 2 O(l)

  1. Analyze .  Given: mol Na 2 SiO 3 . Find: mol HF. Plan .  Use the mole ratio 8 HF: 1 Na 2 SiO 3 from the balanced equation to relate moles of the two reactants.

    Solve . 0.300 mol Na 2 SiO 3 × 8 mol HF 1 mol Na 2 SiO 3 = 2.40 mol HF

    Check .  Mol HF should be greater than mol Na 2 SiO 3 .

  2. Analyze .  Given: mol HF. Find: g NaF. Plan .  Use the mole ratio 2 NaF : 8 HF to change mol HF to mol NaF, then molar mass to get NaF. Solve .

    0.500 mol HF × 2 mol NaF 8 mol HF × 41.99 g NaF 1 mol NaF = 5.25 g NaF

    Check .  (0.5/4) = 0.125; 0.13 × 42 > 4 g NaF

  3. Analyze .  Given: g HF Find: g Na 2 SiO 3 . P l a n . g HF mol HF ( mol ratio ) mol Na 2 SiO 3 g Na 2 SiO 3

    The mole ratio is at the heart of every stoichiometry problem. Molar mass is used to change to and from grams. Solve .

    0.800 g HF × 1 mol HF 20 .01g HF × 1 mol Na 2 SiO 3 8 mol HF × 122.1 g Na 2 SiO 3 1 mol Na 2 SiO 3 = 0.610 g Na 2 SiO 3

    Check .  0.8 (120/160) < 0.75 mol

3.62   4 KO 2 + 2 CO 2 → 2 K 2 CO 3 + 3 O 2

  1. 0.400 mol KO 2 × 3 mol O 2 4 mol KO 2 = 0.300 mol O 2
  2. 7.50 g O 2 × 1 mol O 2 3 2 .00 g O 2 × 4 mol KO 2 3 mol O 2 × 7 1 .10 g KO 2 1 mol KO 2 = 22.2 g KO 2
  3. 7.50 g O 2 × 1 mol O 2 3 2 .00 g O 2 × 2 mol CO 2 3 mol O 2 × 44.01 g CO 2 1 mol CO 2 = 6.88 g CO 2

3.63

  1. Al(OH) 3 (s) + 3 HCl(aq) → AlCl 3 (aq) + 3 H 2 O(l)
  2. Analyze .  Given mass of one reactant, find stoichiometric mass of other reactant and products.

    Plan .  Follow the logic in Sample Exercise 3.16. Calculate mol Al(OH) 3 in 0.500 g

    Al(OH 3 ) 3 separately, because it will be used several times.

    S o l v e . 0 .500 g Al(OH) 3 × 1 mol Al(OH) 3 78.00 g Al(OH) 3 = 6.410 × 10 3 = 6.41 × 10 3 mol Al(OH) 3

    6.410 × 10 3 mol Al(OH) 3 × 3 mol HCl 1 mol Al(OH) 3 × 36.46 g HCl 1 mol HCl = 0.7012 = 0.701 g HCl

    6 . 4 1 0 × 1 0 3 molAl ( OH ) 3 × 1 mol HCl 1 mol Al ( OH ) 3 × 1 3 3 . 3 4 g AlCl 3 1 mol AlCl 3 = 0 . 8 5 4 7 = 0 . 8 5 5 g AlCl 3


  1. 6.410 × 10 3 mol Al(OH) 3 × 3 mol H 2 O 1 mol Al(OH) 3 × 18.02 g H 2 O 1 mol H 2 O = 0.3465 = 0.347 g H 2 O
  1. Conservation of mass: mass of products = mass of reactants

    reactants: Al(OH) 3 + HCl, 0.500 g + 0.701 g = 1.201 g

    products: AlCl 3 + H 2 O, 0.855 g + 0.347 g = 1.202 g

    The 0.001 g difference is due to rounding (0.8547 + 0.3465 = 1.2012). This is an excellent check of results.

3.64

  1. Fe 2 O 3 (s) + 3 CO(g) → 2 Fe(s) + 3 CO 2 (g)
  2. 0.350 kg Fe 2 O 3 × 1000 g 1 kg × 1 mol Fe 2 O 3 159.688 g Fe 2 O 3 = 2.1918 = 2.19 mol Fe 2 O 3

    2. 1918 mol Fe 2 O 3 × 3 mol CO 1 mol Fe 2 O 3 × 28.01 g CO 1 mol CO = 184.17 = 184 g CO

  3. 2. 1918 mol Fe 2 O 3 × 2 mol Fe 1 mol Fe 2 O 3 × 55.845 g Fe 1 mol Fe = 244.80 = 245 g Fe

    2. 1918 mol Fe 2 O 3 × 3 mol CO 2 1 mol Fe 2 O 3 × 44.01 g CO 2 1 mol CO 2 = 289.38 = 289 g CO 2

  4. reactants: 350 g Fe 2 O 3 + 184.17 g CO = 534.17 = 534 g

    products: 244.80 g Fe + 289.38 g CO 2 = 534.18 = 534 g

    Mass is conserved.

3.65

  1. Al 2 S 3 (s) + 6 H 2 O(l) → 2 Al(OH) 3 (s) + 3 H 2 S(g)
  2. Plan .  g A → mol A → mol B → g B. See Solution 3.61 (c). Solve .

    14.2 g Al 2 S 3 × 1 mol Al 2 S 3 150.2 g Al 2 S 3 × 2 mol Al(OH) 3 1 mol Al 2 S 3 × 78.00 g Al(OH) 3 1 mol Al(OH) 3 = 14.7 g Al(OH) 3

    Check. 14 ( 2 × 78 150 ) 14(1) 14 g Al(OH) 3

3.66

  1. CaH 2 (s) + 2 H 2 O(l) → Ca(OH) 2 (aq) + 2 H 2 (g)
  2. 4.500 g H 2 × 1 mol H 2 2.016 g H 2 × 1 mol CaH 2 2 mol H 2 × 42.10 g CaH 2 1 mol CaH 2 = 46.99 g CaH 2

3.67

  1. Analyze .  Given: mol NaN 3 . Find: mol N 2 .

    Plan .  Use mole ratio from balanced equation. Solve .

    1.50 mol NaN 3 × 3 mol N 2 2 mol NaN 3 = 2.25 mol N 2

    Check .  The resulting mol N 2 should be greater than mol NaN 3 , (the N 2 :NaN 3 ratio is > 1), and it is.

  2. Analyze .  Given: g N 2 Find: g NaN 3 .

    Plan .  Use molar masses to get from and to grams, mol ratio to relate moles of the two substances. Solve .

    10.0 g N 2 × 1 mol N 2 28.01 g N 2 × 2 mol NaN 3 3 mol N 2 × 65.01 g NaN 3 1 mol NaN 3 = 15.5 g NaN 3

    Check .  Mass relations are less intuitive than mole relations. Estimating the ratio of molar masses is sometimes useful. In this case, 65 g NaN 3 /28 g N 2 ≈ 2.25. Then, (10 × 2/3 × 2.25) ≈ 15 g NaN 3 . The calculated result looks reasonable.

  1. Analyze .  Given: vol N 2 in ft 3 , density N 2 in g/L. Find: g NaN 3 .

    Plan .  First determine how many g N 2 are in 10.0 ft 3 , using the density of N 2 . Then proceed as in part (b).

    Solve .

    1 . 2 5 g 1 L × 1 L 1 0 0 0 cm 3 × ( 2 . 5 4 ) 3 cm 3 1 in 3 × ( 1 2 ) 3 in 3 1 ft 3 × 1 0 . 0 ft 3 = 3 5 4 . 0 = 3 5 4 g N 2 3 5 4 . 0 g N 2 × 1 mol N 2 2 8 . 0 1 g N 2 × 2 mol Na N 3 3 mol N 2 × 6 5 . 0 1 g Na N 3 1 mol Na N 3 = 5 4 8 g Na N 3

    Check .  1 ft 3 ~ 28 L; 10 ft 3 ~ 280 L; 280 L × 1.25 ~ 350 g N 2

    Using the ratio of molar masses from part (b), (350 × 2/3 × 2.25) ≈ 525 g NaN 3


3.68  2 C 8 H 18 (l) + 25 O 2 (g) → 16 CO 2 (g) + 18 H 2 O(l)

  1. 1.50 mol C 8 H 18 × 25 mol O 2 2 mol C 8 H 18 = 18.75 = 18.8 mol O 2
  2. 10.0 g C 8 H 18 × 1 mol C 8 H 18 114.2 g C 8 H 18 × 25 mol O 2 2 mol C 8 H 18 × 32.00 g O 2 1 mol O 2 = 35.0 g O 2
  3. 15.0 gal C 8 H 18 × 3.7854 L 1 gal × 1000 mL 1 L × 0.692 g 1 mL = 39 , 292 = 3 .93 × 10 4 g C 8 H 18

    3.9292 × 10 4 g C 8 H 18 × 1 mol C 8 H 18 114.2 g C 8 H 18 × 25 mol O 2 2 mol C 8 H 18 × 32.00 g O 2 1 mol O 2 = 137 , 627 g = 1 .38 × 10 5 g O 2

  4. 3.9292 × 10 4 g C 8 H 18 × 1 mol C 8 H 18 114.2 g C 8 H 18 × 1 6 mol CO 2 2 mol C 8 H 18 × 4 4 .01 g CO 2 1 mol O 2 = 121 , 139 g = 1 .21 × 10 5 g CO 2

3.69

  1. Analyze .  Given: dimensions of Al foil. Find: mol Al.

    Plan. Dimensions vol density mass molar mass mol Al

    Solve . 1 .00 cm × 1 .00 cm × 0 .550 mm × 1 cm 10 mm = 0.0550 cm 3 Al

    0.0550 cm 3 Al × 2 .699 g Al 1 cm 3 × 1 mol Al 26 .98 g Al = 5.502 × 10 3 = 5 .50 × 10 3 mol Al

    Check .  2.699/26.98 ≈ 0.1; (0.055 cm 3 × 0.1) = 5.5 × 10 −3 mol Al

  2. Plan .  Write the balanced equation to get a mole ratio; change mol Al → mol AlBr 3 → g AlBr 3 .

    Solve .  2 Al(s) + 3 Br 2 (l) → 2 AlBr 3 (s)

    5.502 × 10 3 mol Al × 2 mol AlBr 3 2 mol Al × 266.69 g AlBr 3 1 mol AlBr 3 = 1.467 = 1.47 g AlBr 3

    Check .  (0.006 × 1 × 270) ≈ 1.6 g AlBr 3


3.70

  1. Plan .  Calculate a “mole ratio” between nitroglycerine and total moles of gas produced. (12 + 6 + 1 + 10) = 29 mol gas; 4 mol nitro: 29 total mol gas. Solve .

    2.00 mL nitro × 1 .592 g mL × 1 mol nitro 227.1 g nitro × 29 mol gas 4 mol nitro = 0.10165 = 0.102 mol gas

  2. 0.10165 mol gas × 55 L mol = 5.5906 = 5.6 L
  3. 2.00 mL nitro × 1 .592 g mL × 1 mol nitro 227.1 g nitro × 6 mol N 2 4 mol nitro × 28.01 g N 2 1 mol N 2 = 0.589 g N 2

3.71 Analyze/Plan. We are given heat produced by combustion of one mole of ethanol and a mass of ethanol to burn. Change mass ethanol to moles ethanol and multiply by kJ/mol.

Solve. The molar mass of CH 3 CH 2 OH is [2(12.0107) + 6(1.00794 + 15.9994] = 46.06844 = 46.07 g/mol

235 .0 g CH 3 CH 2 OH × 1 mol CH 3 CH 2 OH 46.068 g CH 3 CH 2 OH × 1367 kJ 1 mol CH 3 CH 2 OH = 6973.3 = 6.973 × 10 3 kJ

3.72  The molar mass of CH 3 (CH 2 ) 6 CH 3 or C 8 H 18 is [8(12.0107) + 18(1.00794)] = 114.22852 = 114.2 g/mol

1.000 gal C 8 H 18 × 3.7854 L 1 gal × 1000 mL 1 L × 0.692 g 1 mL = 2619.5 = 2.62 × 10 3 g C 8 H 18

2 .6195 × 10 3 g C 8 H 18 × 1 mol C 8 H 18 114.23 g C 8 H 18 × 5470 kJ 1 mol C 2 H 5 OH = 125 , 437 = 1.25 × 10 5 kJ

[Three sig figs in the density dictate 3 sig figs in the result.]

Limiting Reactants (Section 3.7)

3.73

  1. The limiting reactant determines the maximum number of product moles resulting from a chemical reaction; any other reactant is an excess reactant .
  2. The limiting reactant regulates the amount of products, because it is completely used up during the reaction; no more product can be made when one of the reactants is unavailable.
  3. Combining ratios are molecule and mole ratios. Since different molecules have different masses, equal masses of different reactants will not have equal numbers of molecules. By comparing initial moles, we compare numbers of available reactant molecules, the fundamental combining units in a chemical reaction.

3.74

  1. Theoretical yield is the maximum amount of product possible, as predicted by stoichiometry, assuming that the limiting reactant is converted entirely to product.

    Actual yield is the amount of product actually obtained, less than or equal to the theoretical yield. Percent yield is the ratio of (actual yield to theoretical yield) × 100.


  1. No reaction is perfect. Not all reactant molecules come together effectively to form products; alternative reaction pathways may produce secondary products and reduce the amount of desired product actually obtained, or it might not be possible to completely isolate the desired product from the reaction mixture. In any case, these factors reduce the actual yield of a reaction.
  1. No, 110% actual yield is not possible. Theoretical yield is the maximum possible amount of pure product, assuming all available limiting reactant is converted to product, and that all product is isolated. If an actual yield of 110% is obtained, the product must contain impurities which increase the experimental mass.

3.75

  1. 2 C 2 H 5 OH + 6 O 2 → 4 CO 2 + 6 H 2 O

    The equation above corresponds to the contents of the diagram shown in Exercise 3.75, but does not have the simplest ratio of coefficients. Divide all integer coefficients by 2 to obtain C 2 H 5 OH + 3 O 2 → 2 CO 2 + 3 H 2 O

  2. C 2 H 5 OH is the limiting reactant. According to the balanced equation, two molecules of C 2 H 5 OH require six molecules of O 2 for reaction. In the box there are two molecules of C 2 H 5 OH and seven molecules of O 2 . O 2 is present in excess and C 2 H 5 OH limits.
  3. If the reaction goes to completion, there will be four molecules of CO 2 , six molecules of H 2 O, zero molecules of C 2 H 5 OH (the limiting reactant is completely consumed), and one molecule of O 2 (excess reactant).

3.76

  1. C 3 H 8 + 5 O 2 → 3 CO 2 (g) + 4 H 2 O
  2. O 2 is the limiting reactant. According to the balanced equation, one molecule of C 3 H 8 requires five molecules of O 2 for reaction. The three C 3 H 8 molecules in the box would require fifteen molecules of O 2 for complete reaction, but only ten O 2 molecules are present. C 3 H 8 is present in excess and O 2 limits.
  3. If the reaction goes to completion, there will be six molecules of CO 2 , eight molecules of H 2 O, one molecule of C 3 H 8 (excess reactant), and zero molecules of O 2 (the limiting reactant is completely consumed).

3.77 Analyze .  Given: 1.85 mol NaOH, 1.00 mol CO 2 . Find: mol Na 2 CO 3 .

Plan .  Amounts of more than one reactant are given, so we must determine which reactant regulates (limits) product. Then apply the appropriate mole ratio from the balanced equation.

Solve .  The mole ratio is 2 NaOH: 1 CO 2 , so 1.00 mol CO 2 requires 2.00 mol NaOH for complete reaction. Less than 2.00 mol NaOH are present, so NaOH is the limiting reactant.

1.85 mol NaOH × 1 mol Na 2 CO 3 2 mol NaOH = 0.925 mol Na 2 CO 3 can be produced

The Na 2 CO 3 :CO 2 ratio is 1:1, so 0.925 mol Na 2 CO 3 produced requires 0.925 mol CO 2 consumed. (Alternately, 1.85 mol NaOH × 1 mol CO 2 /2 mol NaOH = 0.925 mol CO 2 reacted). 1.00 mol CO 2 initial − 0.925 mol CO 2 reacted = 0.075 mol CO 2 remain.


Check . 2 NaOH(s) + CO 2 (g) Na 2 CO 3 (s) + H 2 O(l)
initial 1.85 mol 1.00 mol 0 mol
change (reaction) −1.85 mol −0.925 mol +0.925 mol
final 0 mol 0.075 mol 0.925 mol

Note that the “change” line (but not necessarily the “final” line) reflects the mole ratios from the balanced equation.

3.78 0.500 mol Al(OH) 3 × 3 mol H 2 SO 4 2 mol Al(OH) 3 = 0.750 mol H 2 SO 4 needed for complete reaction

Only 0.500 mol H 2 SO 4 available, so H 2 SO 4 limits.

0.500 mol H 2 SO 4 × 1 mol Al 2 ( SO 4 ) 3 3 mol H 2 SO 4 = 0.1667 = 0.167 mol Al 2 ( SO 4 ) 3 can form

0.500 mol H 2 SO 4 × 2 mol Al(OH) 3 3 mol H 2 SO 4 = 0.3333 = 0.333 mol Al(OH) 3 react

0.500 mol Al(OH) 3 initial − 0.333 mol react = 0.167 mol Al(OH) 3 remain

3.79   3 NaHCO 3 (aq) + H 3 C 6 H 5 O 7 (aq) → 3 CO 2 (g) + 3 H 2 O(l) + Na 3 C 6 H 5 O 7 (aq)

  1. Analyze/Plan .  Abbreviate citric acid as H 3 Cit. Follow the approach in Sample Exercise 3.19. Solve .

    1 . 0 0 g Na H CO 3 × 1 mol Na H CO 3 8 4 . 0 1 g Na H CO 3 = 1 . 1 9 0 × 1 0 2 = 1 . 1 9 × 1 0 2 mol Na H CO 3

    1.00 g H 3 C 6 H 5 O 7 × 1 mol H 3 Cit 192.1 g H 3 Cit = 5.206 × 10 3 = 5.21 × 10 3 mol H 3 Cit

    But NaHCO 3 and H 3 Cit react in a 3:1 ratio, so 5.21 × 10 −3 mol H 3 Cit require 3(5.21 × 10 −3 ) = 1.56 × 10 −2 mol NaHCO 3 . We have only 1.19 × 10 −2 mol NaHCO 3 , so NaHCO 3 is the limiting reactant.

  2. 1.190 × 10 2 mol NaHCO 3 × 3 mol CO 2 3 mol NaHCO 3 × 44.01 g CO 2 1 mol CO 2 = 0.524 g CO 2
  3. 1 . 1 9 0 × 1 0 2 mol Na H CO 3 × 1 mol H 3 Cit 3 mol Na H CO 3 = 3 . 9 6 8 × 1 0 3 = 3 . 9 7 × 1 0 3 mol H 3 Cit react

    5.206 × 10 3 mol H 3 Cit 3.968 × 10 3 mol react = 1 .238 × 10 3 = 1.24 × 10 3 mol H 3 Cit remain

    1.238 × 10 3 mol H 3 Cit × 192 .1 g H 3 Cit mol H 3 Cit = 0.238 g H 3 Cit remain

3.80   4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 O(g)

  1. Follow the approach in Sample Exercise 3.19.

    2. 00 g NH 3 × 1 mol NH 3 17.03 g NH 3 = 0.11744 = 0.117 mol NH 3


    2.50 g O 2 × 1 mol O 2 32.00 g O 2 = 0.07813 = 0 .0781 mol O 2

    0.07813 mol O 2 × 4 mol NH 3 5 mol O 2 = 0.06250 = 0.0625 mol NH 3 required

    More than 0.0625 mol NH 3 is available, so O 2 is the limiting reactant.

  1. 0.07813 mol O 2 × 4 mol NO 5 mol O 2 × 30.01 g NO 1 mol NO = 1.8756 = 1.88 g NO produced

    0 .07813 mol O 2 × 6 mol H 2 O 5 mol O 2 × 18 .02 g H 2 O 1 mol H 2 O = 1 .6894 = 1 .69 g H 2 O produced

  1. 0.11744 mol NH 3 − 0.0625 mol NH 3 reacted = 0.05494 = 0.0549 mol NH 3 remain

    0.05494 mol NH 3 × 17.03 g NH 3 1 mol NH 3 = 0.93563 = 0.936 g NH 3 remain

  1. mass products = 1.8756 g NO + 1.6894 g H 2 O + 0.9356 g NH 3 remaining = 4.50g products

    mass reactants = 2.00 g NH 3 + 2.50 g O 2 = 4.50 g reactants

    (For comparison purposes, the mass of excess reactant can be either added to the products, as above, or subtracted from reactants.)

3.81 Analyze .  Given: initial g Na 2 CO 3 , g AgNO 3 . Find: final g Na 2 CO 3 , AgNO 3 , Ag 2 CO 3 , NaNO 3

Plan .  Write balanced equation; determine limiting reactant; calculate amounts of excess reactant remaining and products, based on limiting reactant.

Solve .  2 AgNO 3 (aq) + Na 2 CO 3 (aq) → Ag 2 CO 3 (s) + 2 NaNO 3 (aq)

3.50 g Na 2 CO 3 × 1 mol Na 2 CO 3 106.0 g Na 2 CO 3 = 0.03302 = 0.0330 mol Na 2 CO 3

5 . 0 0 g AgNO 3 × 1 mol AgNO 3 1 6 9 . 9 g AgNO 3 = 0 . 0 2 9 4 3 = 0 . 0 2 9 4 mol AgNO 3

0 . 0 2 9 4 3 mol AgNO 3 × 1 mol Na 2 CO 3 2 mol AgNO 3 = 0 . 0 1 4 7 1 = 0 . 0 1 4 7 mol Na 2 CO 3 required

AgNO 3 is the limiting reactant and Na 2 CO 3 is present in excess.

2 AgNO 3 (aq) + Na 2 CO 3 (aq) Ag 2 CO 3 (s) + 2 NaNO 3 (aq)
initial 0.0294 mol 0.0330 mol 0 mol 0 mol
reaction −0.0294 mol −0.0147 mol +0.0147 mol +0.0294 mol
final 0 mol 0.0183 mol 0.0147 mol 0.0294 mol

0.01830 mol Na 2 CO 3 × 106.0 g/mol = 1.940 = 1.94 g Na 2 CO 3

0.01471 mol Ag 2 CO 3 × 275.8 g/mol = 4.057 = 4.06 g Ag 2 CO 3

0.02943 mol NaNO 3 × 85.00 g/mol = 2.502 = 2.50 g NaNO 3

Check .  The initial mass of reactants was 8.50 g, and the final mass of excess reactant and products is 8.50 g; mass is conserved.


3.82 Plan .  Write balanced equation; determine limiting reactant; calculate amounts of excess reactant remaining and products, based on limiting reactant.

Solve .  H 2 SO 4 (aq) + Pb(C 2 H 3 O 2 ) 2 (aq) → PbSO 4 (s) + 2 HC 2 H 3 O 2 (aq)

5.00 g H 2 SO 4 × 1 mol H 2 SO 4 98.09 g H 2 SO 4 = 0.05097 = 0.0510 mol H 2 SO 4

5.00 g Pb(C 2 H 3 O 2 ) 2 × 1 mol Pb(C 2 H 3 O 2 ) 2 325.3 g Pb(C 2 H 3 O 2 ) 2 = 0.015370 = 0.0154 mol Pb(C 2 H 3 O 2 ) 2

1 mol H 2 SO 4 :1 mol Pb(C 2 H 3 O 2 ) 2 , so Pb(C 2 H 3 O 2 ) 2 is the limiting reactant.

0 mol Pb(C 2 H 3 O 2 ) 2 , (0.05097 − 0.01537) = 0.0356 mol H 2 SO 4 , 0.0154 mol PbSO 4 ,

(0.01537 × 2) = 0.0307 mol HC 2 H 3 O 2 are present after reaction

0.03560 mol H 2 SO 4 × 98.09 g/mol = 3.4920 = 3.49 g H 2 SO 4

0.01537 mol PbSO 4 × 303.3 g/mol = 4.6619 = 4.66 g PbSO 4

0.03074 mol HC 2 H 3 O 2 × 60.05 g/mol = 1.8460 = 1.85 g HC 2 H 3 O 2

Check .  The initial mass of reactants was 10.00 g; and the final mass of excess reactant and products is 10.00 g; mass is conserved.

3.83 Analyze .  Given: amounts of two reactants. Find: theoretical yield.

Plan .  Determine the limiting reactant and the maximum amount of product it could produce. Then calculate % yield. Solve .

  1. 30.0 g C 6 H 6 × 1 mol C 6 H 6 78.11 g C 6 H 6 = 0.3841 = 0.384 mol C 6 H 6

    65.0 g Br 2 × 1 mol Br 2 159.8 g Br 2 = 0.4068 = 0.407 mol Br 2

    Because C 6 H 6 and Br 2 react in a 1:1 mole ratio, C 6 H 6 is the limiting reactant and determines the theoretical yield.

    0.3841 mol C 6 H 6 × 1 mol C 6 H 5 Br 1 mol C 6 H 6 × 157.0 g C 6 H 5 Br 1 mol C 6 H 5 Br = 60 .30 = 60 .3 g C 6 H 5 Br

    Check .  30/78 ~ 3/8 mol C 6 H 6 . 65/160 ~ 3/8 mol Br 2 . Because moles of the two reactants are similar, a precise calculation is needed to determine the limiting reactant. 3/8 × 160 ≈ 60 g product

  2. % yield = 42 .3 g C 6 H 5 Br actual 60.3 g C 6 H 5 Br theoretical × 100 = 70.149 = 70.10 %

3.84

  1. C 2 H 6 + Cl 2 → C 2 H 5 Cl + HCl

    125 g C 2 H 6 × 1 mol C 2 H 6 30.07 g C 2 H 6 = 4.157 = 4.16 mol C 2 H 6

    255 g Cl 2 × 1 mol Cl 2 70.91 g Cl 2 = 3.596 = 3 .60 mol Cl 2


    Because the reactants combine in a 1:1 mole ratio, Cl 2 is the limiting reactant. The theoretical yield is:

    3.596 mol Cl 2 × 1 mol C 2 H 5 Cl 1 mol Cl 2 × 64.51 g C 2 H 5 Cl 1 mol C 2 H 5 Cl = 231.98 = 232 g C 2 H 5 Cl

  1. % yield = 206 g C 2 H 5 Cl actual 232 g C 2 H 5 Cl theoretical × 100 = 88.8 %

3.85 Analyze .  Given: g of two reactants, % yield. Find: g S 8 .

Plan .  Determine limiting reactant and theoretical yield. Use definition of % yield to calculate actual yield. Solve .

30.0 g H 2 S × 1 mol H 2 S 34 .08 g H 2 S = 0.8803 = 0.880 mol H 2 S

50.0 g O 2 × 1 mol O 2 32.00 g H 2 S = 1.5625 = 1.56 mol O 2

0.8803 mol H 2 S × 4 mol O 2 8 mol H 2 S = 0.4401 = 0 .440 mol O 2 required

Because there is more than enough O 2 to react exactly with 0.880 mol H 2 S, O 2 is present in excess and H 2 S is the limiting reactant.

0 .8803 mol H 2 S × 1 mol S 8 8 mol H 2 S × 256 .56 g S 8 1 mol S 8 = 28 .231 = 28 .2 g S 8 theoretical yield

Check . 30/34 ≈ 1 mol H 2 S; 50/32 ≈ 1.5 mol O 2 . Twice as many mol H 2 S as mol O 2 are required, so H 2 S limits. 1 × (260/8) ≈ 30 g S 8 theoretical.

% yield = actual theoretical × 100 ; % yield × theoretical 100 = actual yield

98 % 100 × 28.231 g S 8 = 27 .666 = 28 g S 8 actual

3.86  H 2 S(g) + 2 NaOH(aq) → Na 2 S(aq) + 2 H 2 O(l)

1.25 g H 2 S × 1 mol H 2 S 34.08 g H 2 S = 0.03668 = 0.0367 mol H 2 S 2 .00 g NaOH × 1 mol NaOH 40 .00 g NaOH = 0.0500 mol NaOH

By inspection, twice as many mol NaOH as H 2 S are needed for exact reaction, but mol NaOH given is less than twice mol H 2 S, so NaOH limits.

0 . 0 5 0 0 mol NaOH × 1 mol Na 2 S 2 mol NaOH × 7 8 . 0 5 g Na 2 S 1 mol Na 2 S = 1 . 9 5 1 2 5 = 1 . 9 5 g Na 2 S theoretical

92.0 % 100 × 1.95125 g Na 2 S theoretical = 1 .7951 = 1 .80 g Na 2 S actual


Additional Exercises

3.87

  1. CH 3 COOH = C 2 H 4 O 2 . At room temperature and pressure, pure acetic acid is a liquid.  C 2 H 4 O 2 (l) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)
  2. Ca(OH) 2 (s) → CaO(s) + H 2 O(g)
  3. Ni(s) + Cl 2 (g) → NiCl 2 (s)

3.88  C 2 H 5 OH(l) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O(g)

C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(g)

CH 3 CH 2 COCH 3 (l) + 11/2 O 2 (g) → 4 CO 2 (g) + 4 H 2 O(l)

In a combustion reaction, all H in the fuel is transformed to H 2 O in the products. The reactant with most mol H/mol fuel will produce the most H 2 O. C 3 H 8 and CH 3 CH 2 COCH 3 (C 4 H 8 O) both have 8 mol H/mol fuel, so 1.5 mol of either fuel will produce the same amount of H 2 O. 1.5 mol C 2 H 5 OH will produce less H 2 O.

3.89  The formulas of the fertilizers are NH 3 , NH 4 NO 3 , (NH 4 ) 2 SO 4 and (NH 2 ) 2 CO. Qualitatively, the more heavy, non-nitrogen atoms in a molecule, the smaller the mass % of N. By inspection, the mass of NH 3 is dominated by N, so it will have the greatest % N, (NH 4 ) 2 SO 4 will have the least. In order of increasing % N:

(NH 4 ) 2 SO 4 < NH 4 NO 3 > (NH 2 ) 2 CO < NH 3 .

Check by calculation:

(NH 4 ) 2 SO 4 : FW = 2(14.0) + 8(1.0) + 1(32.1) + 4(16.0) = 132.1 amu

% N = [2(14.0)/132.1] × 100 = 21.2%

NH 4 NO 3 : FW = 2(14.0) + 4(1.0) + 3(16.0) = 80.0 amu

% N = [2(14.0)/80.0] × 100 = 35.0%

(NH 2 ) 2 CO: FW = 2(14.0) + 4(1.0) = 1(12.0) + 1(16.0) = 60.0 amu

% N = [2(14.0)/60.0] × 100 = 46.7% N

NH 3 : FW = 1(14.0) + 3(1.0) = 17.0 amu

% N = [14.0/17.0] × 100 = 82.4 % N

3.90

  1. 0.500 g C 9 H 8 O 4 × 1 mol C 9 H 8 O 4 180.2 g C 9 H 8 O 4 = 2.7747 × 10 3 = 2.77 × 10 3 mol C 9 H 8 O 4
  2. 0.0027747 mol C 9 H 8 O 4 × 6.022 × 10 23 molecules 1 mol = 1.67 × 10 21 C 9 H 8 O 4 molecules
  3. 1.67 × 10 21 C 9 H 8 O 4 molecules × 9 C atoms 1 C 9 H 8 O 4 molecule = 1.50 × 10 22 C atoms

3.91

  1. Analyze . Given: diameter of CdSe sphere (dot), density of CdSe. Find: mass of dot.

    Plan . Calculate volume of sphere in cm 3 , use density to calculate mass of the sphere (dot).

    Solve. V = 4/3π r 3 ; r = d/2

    radius of dot = 2.5 nm 2 × 1 × 10 9 m 1 nm × 1 cm 1 × 10 2 m = 1.25 × 10 7 cm

    volume of dot = (4/3) × π × (1.25 × 10 −7 ) 3 = 8.1812 × 10 −21 = 8.2 × 10 −21 cm 3

    8 .1812 × 10 21 cm 3 × 5 .82 g CdSe cm 3 = 4 .7615 × 10 20 = 4 .8 × 10 20 g/dot

  2. Plan . Change g CdSe to mol Cd using molar mass, then mol Cd to atoms Cd using Avogadro’s number. Solve .

    4 .7615 × 10 20 g CdSe × 1 mol CdSe 191.385 g CdSe × 1 mol Cd 1 mol CdSe × 6.0221 × 10 23 Cd atoms mol Cd = 149 . 82 = 150 Cd atoms

  3. volume of dot = (4/3) × π × (3.25 × 10 −7 ) 3 = 1.4379 × 10 −19 = 1.4 × 10 −19 cm 3

    1 .4379 × 10 19 cm 3 × 5 .82 g CdSe cm 3 = 8 .3688 × 10 19 = 8 .4 × 10 19 g/dot

  4. 8 .3688 × 10 19 g CdSe × 1 mol CdSe 191.385 g CdSe × 1 mol Cd 1 mol CdSe × 6.022 × 10 23 Cd atoms mol Cd = 2633 . 3 = 2 . 6 × 10 3 Cd atoms
  5. We can calculate the number of 2.5-nm dots required to make a 6.5-nm dot using either the ratio of the volumes of the dots, or the ratio of the number of CdSe units in the two dots.

    1 .4379 × 10 19 cm 3 8 .1812 × 10 21 cm 3 = 17.5758 2.5 -nm dots; 2633.3 CdSe units 149.82 CdSe units = 17 .5764 2 .5-nm dots

    The number of 2.5-nm dots required to make a 6.5-nm dot and the number of CdSe units left over are integer numbers. We have included extra figures in the ratio calculations to show that the results are amazingly close. Both ratios indicate that 18 of the smaller dots are needed to produce one 6.5 nm dot. There will be a few CdSe units left over, slightly less than half of one dot. Taking the average of the “leftovers,”

    0.424 small dot x 150 CdSe units/small dot = 63.6 = 64 CdSe units left over

3.92

  1. 5.342 × 10 21 g 1 molecule penicillin G × 6.0221 × 10 23 molecules 1 mol = 3217 g/mol penicillin G
  2. 1.00 g hemoglobin (hem) contains 3.40 × 10 −3 g Fe. 1.00 g hem 3.40 × 10 3 g Fe × 55.85 g Fe 1 mol Fe × 4 mol Fe 1 mol hem = 6.57 × 10 4 g/mol hemoglobin

3.93 Plan .  Assume 100 g, calculate mole ratios, empirical formula, then molecular formula from molar mass. Solve .

68.2 g C × 1 mol C 12 .01 g C = 5.68 mol C;  5 .68/0 .568 10

6.86 g H × 1 mol H 1 .008 g H = 6.81 mol H;  6 .81/0 .568 12

15.9 g N × 1 mol N 14 .01 g N = 1.13 mol N;  1 .13/0 .568 2

9.08 g O × 1 mol O 16 .00 g O = 0.568 mol O;  0 .568/0 .568 = 1

The empirical formula is C 10 H 12 N 2 O, FW = 176 amu (or g). Because the molar mass is 176, the empirical and molecular formula are the same, C 10 H 12 N 2 O.

3.94 Plan .  Assume 1.000 g and get mass O by subtraction. Solve .

  1. 0.7787 g C × 1 mol C 12 .01 g C = 0.06484 mol C

    0.1176 g H × 1 mol H 1 .008 g H = 0.1167 mol H

    0.1037 g O × 1 mol C 16 .00 g O = 0.006481 mol O

    Dividing through by the smallest of these values we obtain C 10 H 18 O.

  2. The formula weight of C 10 H 18 O is 154. Thus, the empirical formula is also the molecular formula.

3.95  Because all the C in the vanillin must be present in the CO 2 produced, get g C from g CO 2 .

2.43 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 12.01 g C 1 mol C = 0.6631 = 0.663 g C

Because all the H in vanillin must be present in the H 2 O produced, get g H from g H 2 O.

0.50 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 2 mol H 1 mol H 2 O × 1. 008 g H 1 mol H = 0.0559 = 0.056 g H

Get g O by subtraction. (Because the analysis was performed by combustion, an unspecified amount of O 2 was a reactant, and thus not all the O in the CO 2 and H 2 O produced came from vanillin.) 1.05 g vanillin − 0.663 g C − 0.056 g H = 0.331 g O

0 . 6 6 3 1 g C × 1 mol C 1 2 . 0 1 g C = 0 . 0 5 5 2 mol C ; 0 . 0 5 5 2 / 0 . 0 2 0 7 = 2 . 6 7

0.0559 g H × 1 mol H 1 .008 g H = 0.0555 mol C;  0 .0555/0 .0207 = 2 .68

0 . 3 3 1 g O × 1 mol O 1 6 . 0 0 g O = 0 . 0 2 0 7 mol O ; 0 . 0 2 0 7 / 0 . 0 2 0 7 = 1 . 0 0

Multiplying the numbers above by 3 to obtain an integer ratio of moles, the empirical formula of vanillin is C 8 H 8 O 3 .


3.96 Plan .  Because different sample sizes were used to analyze the different elements, calculate mass % of each element in the sample.

  1. Calculate mass % C from g CO 2 .
  2. Calculate mass % Cl from AgCl.
  3. Get mass % H by subtraction.
  4. Calculate mole ratios and the empirical formulas.

Solve .

  1. 3.52 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 1 mol C 1 mol CO 2 × 12.01 g C 1 mol C = 0.9606 = 0.961 g C

    0.9606 g C 1.50 g sample × 100 = 64.04 = 64.0 % C

  2. 1.27 g AgCl × 1 mol AgCl 143.3 g AgCl × 1 mol Cl 1 mol AgCl × 35.45 g Cl 1 mol Cl = 0.3142 = 0.314 g Cl

    0.3142 g Cl 1.00 g sample × 100 = 31.42 = 31.4 % Cl

  3. % H = 100.0 − (64.04% C + 31.42% Cl) = 4.54 = 4.5% H
  4. Assume 100 g sample.

64.04 g C × 1 mol C 12 .01 g C = 5.33 mol C; 5 .33/0 .886 = 6 .02

31.42 g Cl × 1 mol Cl 35 .45 g Cl = 0.886 mol Cl;  0 .886/0 .886 = 1 .00

4.54 g H × 1 mol H 1 .008 g H = 4.50 mol H;  4 .50/0 .886 = 5 .08

The empirical formula is probably C 6 H 5 Cl.

The subscript for H, 5.08, is relatively far from 5.00, but C 6 H 5 Cl makes chemical sense. More significant figures in the mass data are required for a more accurate mole ratio.

3.97  The mass percentage is determined by the relative number of atoms of the element times the atomic weight, divided by the total formula mass. Thus, the mass percent of bromine in KBrO x is given by 0.5292 = 79.91 39.10 + 79.91 + x ( 16.00 ) .

Solving for x, we obtain x = 2.00. Thus, the formula is KBrO 2 .

3.98

  1. Let AW = the atomic weight of X.

    According to the chemical reaction, moles XI 3 reacted = moles XCl 3 produced

    0.5000 g XI 3 × 1 mol XI 3 /(AW + 380 .71) g XI 3 = 0.2360 g XCl 3 × 1 mol XCl 3 ( AW + 106 .36) g XCl 3

    0.5000 (AW + 106.36) = 0.2360 (AW + 380.71)

    0.5000 AW + 53.180 = 0.2360 AW + 89.848

    0.2640 AW = 36.67; AW = 138.9 g

  2. X is lanthanum, La, atomic number 57.

3.99  O 3 (g) + 2 NaI(aq) + H 2 O(l) → O 2 (g) + I 2 (s) + 2 NaOH(aq)

  1. 5.95 × 10 6 mol O 3 × 2 mol NaI 1 mol O 3 = 1.19 × 10 5 mol NaI
  2. 1 .3 mg O 3 × 1 × 10 3 g 1 mg × 1 mol O 3 48.00 g O 3 × 2 mol NaI 1 mol O 3 × 149.9 g NaI 1 mol NaI = 8 . 120 × 10 −3 = 8 . 1 × 10 −3 g NaI = 8 . 1 mg NaI

3.100  2 NaCl(aq) + 2 H 2 O(l) → 2 NaOH(aq) + H 2 (g) + Cl 2 (g)

Calculate mol Cl 2 and relate to mol H 2 , mol NaOH.

1.5 × 10 6 kg × 1000 g 1 kg × 1 mol Cl 2 70. 91 g Cl 2 = 2.115 × 10 7 = 2.1 × 10 7 mol Cl 2

2.115 × 10 7 mol Cl 2 × 1 mol H 2 1 mol Cl 2 × 2.016 g H 2 1 mol H 2 = 4.26 × 10 7 g H 2 = 4.3 × 10 4 kg H 2

4.3 × 10 7 g × 1 metric ton 1 × 10 6 g (1 Mg) = 43 metric tons H 2

2.115 × 10 7 mol Cl 2 × 2 mol NaOH 1 mol Cl 2 × 40.0 g NaOH 1 mol NaOH = 1.69 × 10 9 = 1.7 × 10 9 g NaOH

1.7 × 10 9 g NaOH = 1.7 × 10 6 kg NaOH = 1.7 × 10 3 metric tons NaOH

3.101  2 C 57 H 110 O 6 + 163 O 2 → 114 CO 2 + 110 H 2 O

molar mass of fat = 57(12.01) + 110(1.008) + 6(16.00) = 891.5

1.0 kg fat × 1000 g 1 kg × 1 mol fat 891 .5 g fat × 110 mol H 2 O 2 mol fat × 18.02 g H 2 O 1 mol H 2 O × 1 kg 1000 g = 1.1 kg H 2 O

3.102

  1. Plan .  Calculate the total mass of C from g CO and g CO 2 . Calculate the mass of H from g H 2 O. Calculate mole ratios and the empirical formula. Solve .

    0.467 g CO × 1 mol CO 28 .01 g CO × 1 mol C 1 mol CO × 12.01 g C = 0 .200 g C

    0.733 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 1 mol C 1 mol CO 2 × 12.01 g C = 0 .200 g C

    Total mass C is 0.200 g + 0.200 g = 0.400 g C.

    0.450 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 2 mol H 1 mol H 2 O × 1.008 g H 1 mol H = 0.0503 g H

    (Because hydrocarbons contain only the elements C and H, g H can also be obtained by subtraction: 0.450 g sample − 0.400 g C = 0.050 g H.)

    0.400 g C × 1 mol C 12 .01 g C = 0.0333 mol C;  0 .0333/0 .0333 = 1 .0

    0.0503 g H × 1 mol H 1 .008 g H = 0.0499 mol H;   0 .0499/0 .0333 = 1 .5

    Multiplying by a factor of 2, the empirical formula is C 2 H 3 .


  1. Mass is conserved. Total mass products − mass sample = mass O 2 consumed.

    0.467 g CO + 0.733 g CO 2 + 0.450 g H 2 O − 0.450 g sample = 1.200 g O 2 consumed

  1. For complete combustion, 0.467 g CO must be converted to CO 2 .

    2 CO(g) + O 2 (g) → 2 CO 2 (g)

    0.467 g CO × 1 mol CO 28 .01 g CO × 1 mol O 2 2 mol CO × 32.00 g O 2 1 mol O 2 = 0.267 g O 2

    The total mass of O 2 required for complete combustion is

    1.200 g + 0.267 g = 1.467 g O 2 .

3.103  N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

Determine the moles of N 2 and H 2 required to form the 3.0 moles of NH 3 present after the reaction has stopped.

3.0 mol NH 3 × 3 mol H 2 2 mol NH 3 = 4.5 mol H 2 reacted

3.0 mol NH 3 × 1 mol N 2 2 mol NH 3 = 1.5 mol N 2 reacted

mol H 2 initial = 3.0 mol H 2 remain + 4.5 mol H 2 reacted = 7.5 mol H 2

mol N 2 initial = 3.0 mol N 2 remain + 1.5 mol N 2 reacted = 4.5 mol N 2

In tabular form: N 2 (g) + 3 H 2 (g) 2 NH 3 (g)
initial 4.5 mol 7.5 mol 0 mol
reaction −1.5 mol −4.5 mol +3.0 mol
final 3.0 mol 3.0 mol 3.0 mol

(Tables like this will be extremely useful for solving chemical equilibrium problems in Chapter 15 .)

3.104  All of the O 2 is produced from KClO 3 ; get g KClO 3 from g O 2 . All of the H 2 O is produced from KHCO 3 ; get g KHCO 3 from g H 2 O. The g H 2 O produced also reveals the g CO 2 from the decomposition of KHCO 3 . The remaining CO 2 (13.2 g CO 2− g CO 2 from KHCO 3 ) is due to K 2 CO 3 and g K 2 CO 3 can be derived from it.

4.00 g O 2 × 1 mol O 2 32.00 g O 2 × 2 mol KClO 3 3 mol O 2 × 122.6 g KClO 3 1 mol KClO 3 = 10.22 = 10.2 g KClO 3

1.80 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 2 mol KHCO 3 1 mol H 2 O × 100.1 g KHCO 3 1 mol KHCO 3 = 20.00 = 20.0 g KHCO 3

1.80 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 2 mol CO 2 1 mol H 2 O × 44.01 g CO 2 1 mol CO 2 = 8.792 = 8 .79 g CO 2 from KHCO 3

13.20 g CO 2 total − 8.792 CO 2 from KHCO 3 = 4.408 = 4.41 g CO 2 from K 2 CO 3

4.408 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 1 mol K 2 CO 3 1 mol CO 2 × 138.2 g K 2 CO 3 1 mol K 2 CO 3 = 13.84 = 13.8 g K 2 CO 3

100.0 g mixture − 10.22 g KClO 3 − 20.00 g KHCO 3 − 13.84 g K 2 CO 3 = 55.9 g KCl


3.105

  1. 2 C 2 H 2 (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O(g)
  2. Following the approach in Sample Exercise 3.18,

    10.0 g C 2 H 2 × 1 mol C 2 H 2 26.04 g C 2 H 2 × 5 mol O 2 2 mol C 2 H 2 × 32.00 g O 2 1 mol O 2 = 30.7 g O 2 required

    Only 10.0 g O 2 are available, so O 2 limits.

  3. Because O 2 limits, 0.0 g O 2 remain.

    Next, calculate the g C 2 H 2 consumed and the amounts of CO 2 and H 2 O produced by reaction of 10.0 g O 2 .

    10.0 g O 2 × 1 mol O 2 32.00 g O 2 × 2 mol C 2 H 2 5 mol O 2 × 26.04 g C 2 H 2 1 mol C 2 H 2 = 3.26 g C 2 H 2 consumed

    10.0 g C 2 H 2 initial − 3.26 g consumed = 6.74 = 6.7 g C 2 H 2 remain

    10.0 g O 2 × 1 mol O 2 32.00 g O 2 × 4 mol CO 2 5 mol O 2 × 44.01 g CO 2 1 mol CO 2 = 11.0 g CO 2 produced

    10.0 g O 2 × 1 mol O 2 32.00 g O 2 × 2 mol H 2 O 5 mol O 2 × 18.02 g H 2 O 1 mol H 2 O = 2.25 g H 2 O produced

Integrative Exercises

3.106 Plan . Write a balanced chemical equation for the synthesis of GaAs, assuming the reaction is homogeneous and occurs in the gas phase. Use given masses of each reactant, along with molar masses, to find moles of each reactant. Use mole ratios from the equation determine limiting reactant and mass of GaAs produced. Use density to change mass from part (a) to volume of GaAs, then calculate area of the thin film produced in part (c). S olve .

  1. Ga(CH 3 ) 3 (g) + AsH 3 (g) → GaAs(s) + 3 CH 4 (g)

    Determine the moles of Ga(CH 3 ) 3 (g) and AsH 3 (g) available for reaction. The molar masses are 114.827 g/mol and 77.9454 g/mol, respectively.

    450.0 g Ga(CH 3 ) 3 × 1 mol Ga(CH 3 ) 3 114.827 g Ga(CH 3 ) 3 = 3.91894 = 3.919 mol Ga(CH 3 ) 3

    300.0 g AsH 3 × 1 mol AsH 3 77.9454 g AsH 3 = 3.84885 = 3 .849 mol AsH 3

    According to the balanced equation, Ga(CH 3 ) 3 and AsH 3 react in a 1: 1 mole ratio. We have nearly equal moles of the two reactants, but there are fewer moles of AsH 3 . The amount of AsH 3 determines the amount of GaAs that can be made.

    3.84885 mol AsH 3 × 1 mol GaAs 1 mol AsH 3 × 144.645 g GaAs mol GaAs = 556.717 = 556 .7 g GaAs

  2. Ga(CH 3 ) 3 is present in excess. Because of the 1: 1 reacting ratio, the amount of Ga(CH 3 ) 3 left over is

    (3.91894 mol available − 3.84885 mol reacted) = 0.07009 mol Ga(CH 3 ) 3 left over

  3. 556 .717 g GaAs × 1 cm 3 5.32 g × 1 40 nm × 1 nm 1 × 10 7 cm = 26 , 161 , 513 = 2.62 × 10 7 cm 2

    The area of the 40-nm thick GaAs film is 2.62 × 10 7 cm 2 .


3.107  (average yield) 11 = 0.05; average yield = (0.05) 1/11 = 0.7616 = 0.8; 80% average yield

Think of two reactions in a sequence that each has a yield of 50%. Relative to the amount of starting material, the first product is 50% of the maximum possible product. The second product is 50% of the maximum relative to the product of the first reaction. Relative to the amount of starting material, the quantity of final product is 25% of the maximum possible product. That is,

(0.5)(0.5) = 0.25 or (average yield) 2 = 0.25

Extending this example to eleven reactions in a sequence, the net yield, relative to the initial amount of starting material, is 5%.

(average yield) 11 = 0.05; average yield = 80%

3.108 P l a n . Volume cube density mass CaCO 3 moles CaCO 3 moles O O atoms

Solve. (2 .005) 3 in 3 × ( 2.54 ) 3 cm 3 1 in 3 × 2.71 g CaCO 3 1 cm 3 × 1 mol CaCO 3 100.1 g CaCO 3 × 3 mol O 1 mol CaCO 3 × 6.022 × 10 23 O atoms 1 mol O = 6.46 × 10 24 O atoms

3.109

  1. P l a n . volume of Ag cube density mass of Ag mol Ag Ag atoms

    Solve. (1 .000) 3 cm 3 Ag × 10 .5 g Ag 1 cm 3 Ag × 1 mol Ag 107.87 g Ag × 6.022 × 10 23 atoms 1 mol = 5.8618 × 10 22 = 5.86 × 10 22 Ag atoms

  2. 1.000 cm 3 cube volume, 74% is occupied by Ag atoms

    0.74 cm 3 = volume of 5.86 × 10 22 Ag atoms

    0.7400 cm 3 5.8618 × 10 22 Ag atoms = 1.2624 × 10 23 = 1.3 × 10 23 cm 3 / Ag atom

    Because atomic dimensions are usually given in Å, we will show this conversion.

    1.2624 × 10 23 cm 3 × ( 1 × 10 2 ) 3 m 3 1 cm 3 × 1 Å 3 ( 1 × 10 10 ) 3 m 3 = 12.62 = 13 Å 3 / Ag atom

  3. V = 4/3πr 3 ; r 3 = 3V/4π; r = (3V/4π) 1/3

    r A = (3 × 12.62 Å 3 / 4π) 1/3 = 1.444 = 1.4 Å

3.110

  1. Analyze .  Given: gasoline = C 8 H 18 , density = 0.692 g/mL, 20.5 mi/gal, 225 mi.

    Find: kg CO 2 .

    Plan .  Write and balance the equation for the combustion of octane. Change mi → gal octane → mL → g octane. Use stoichiometry to calculate g and kg CO 2 from g octane.


    Solve .  2 C 8 H 18 (l) + 25 O 2 (g) → 16 CO 2 (g) + 18 H 2 O(l)

    2 2 5 mi × 1 gal 2 0 . 5 mi × 3 . 7 8 5 4 L 1 gal × 1 mL 1 × 1 0 3 L × 0 . 6 9 2 g octane 1 mL = 2 . 8 7 5 1 × 1 0 4 g = 2 8 . 8 kg octane

    2 . 8 7 5 1 × 1 0 4 g C 8 H 1 8 × 1 mol C 8 H 1 8 1 1 4 . 2 g C 8 H 1 8 × 1 6 mol CO 2 2 mol C 8 H 1 8 × 4 4 . 0 1 g CO 2 1 mol CO 2 = 8 . 8 6 3 8 × 1 0 4 g = 8 8 . 6 kg CO 2

    Check. ( 225 × 4 × 0 .7 20 ) × 10 3 = ( 45 × 0.7 ) × 10 3 = 30 × 10 3 g = 30 kg octane

    44 114 1 3 ; 30 kg × 8 3 80 kg CO 2

  1. Plan . Use the same strategy as part (a). Solve.

    225 mi × 1 gal 5 mi × 3.7854 L 1 gal × 1 mL 1 × 10 3 L × 0.69 g octane 1 mL = 1.1754 × 10 5 = 1 × 10 2 kg octane

    1.1754 × 10 5 g C 8 H 18 × 1 mol C 8 H 18 114.2 g C 8 H 18 × 16 mol CO 2 2 mol C 8 H 18 × 44.01 g CO 2 1 mol CO 2 = 3.624 × 10 5 g = 4 × 10 2 kg CO 2

Check. Mileage of 5 mi/gal requires ~4 times as much gasoline as mileage of 20.5 mi/gal, so it should produce ~4 times as much CO 2 . 90 kg CO 2 [from (a)] × 4 = 360 = 4 × 10 2 kg CO 2 [from (b)].

3.111  Structural isomers, like 1-propanol and 2-propanol, have the same number and kinds of atoms, but different arrangements of these atoms. Because molecular weight is the sum of atomic weights, and number and kinds of atoms are the same, the molecular weights of structural isomers are the same. Again, because number and kinds of atoms are the same, percent composition and therefore combustion analysis results will be the same. Physical properties, like boiling point and density, are influenced by structure as well as molecular weight, and are different for structural isomers.

The properties (a) boiling point and (d) density will distinguish between 1-propanol and 2-propanol. This is confirmed by comparing these properties from either Wolfram Alpha (WA) or the CRC Handbook of Chemistry and Physics (CRC).

Compound Boiling Point
(WA)
Boiling Point
(CRC)
Density
(WA)
Density
(CRC)
1-propanol 97 o C 97.4 o C 0.804 g/cm 3 0.8035 g/cm 3
2-propanol 82 o C 82.4 o C 0.785 g/cm 3 0.7855 g/cm 3

3.112

  1. S(s) + O 2 (g) → SO 2 (g); SO 2 (g) + CaO(s) → CaSO 3 (s)
  2. If the coal contains 2.5% S, then 1 g coal contains 0.025 g S.

    2 0 0 0 . 0 tons Coal day × 2 0 0 0 lb 1 ton × 1 kg 2 . 2 0 4 6 lb × 1 0 0 0 g 1 kg × 0 . 0 2 5 g S 1 g Coal × 1 mol S 3 2 . 0 6 5 g S × 1 mol SO 2 1 mol S × 1 mol CaO 1 mol SO 2 × 5 6 . 0 7 7 g CaO 1 mol CaO × 1 kg CaO 1 0 0 0 g CaO = 7 9 , 3 2 7 . 4 9 = 7 . 9 × 1 0 4 kg CaO or 7 . 9 × 1 0 7 g CaO

  3. 1 mol CaO = 1 mol CaSO 3

    7.9327 × 10 7 g CaO × 1 mol CaO 5 6 . 0 7 7 g CaO × 1 mol CaSO 3 1 mol CaO × 1 2 0 . 1 4 g CaSO 3 1 mol CaSO 3

    = 1 . 6 9 9 5 × 1 0 8 = 1 . 7 × 1 0 8 g CaSO 3

    This corresponds to about 190 tons of CaSO 3 per day as a waste product.

3.113

  1. Plan .  Calculate the kg of air in the room and then the mass of HCN required to produce a dose of 300 mg HCN/kg air. Solve .

    12 ft × 15 ft × 8.0 ft = 1440 = 1.4 × 10 3 ft 3 of air in the room

    1 4 4 0 ft 3 air × ( 1 2 in ) 3 1 ft 3 × ( 2 . 5 4 cm ) 3 1 in 3 × 0 . 0 0 1 1 8 g air 1 cm 3 air × 1 kg 1 0 0 0 g = 4 8 . 1 2 = 4 8 kg air

    4 8 . 1 2 kg air × 3 0 0 mg HCN 1 kg air × 1 g 1 0 0 0 mg = 1 4 . 4 3 = 1 4 g HCN

  2. 2 NaCN(s) + H 2 SO 4 (aq) → Na 2 SO 4 (aq) + 2 HCN(g)

    The question can be restated as: What mass of NaCN is required to produce 14 g of HCN according to the above reaction?

    1 4 . 4 3 g HCN × 1 mol HCN 2 7 . 0 3 g HCN × 2 mol NaCN 2 mol HCN × 4 9 . 0 1 g NaCN 1 mol NaCN = 2 6 . 2 = 2 6 g NaCN
  3. 1 2 ft × 1 5 ft × 1 yd 2 9 ft 2 × 3 0 oz 1 yd 2 × 1 lb 1 6 oz × 4 5 4 g 1 lb = 1 7 , 0 2 5 = 1 . 7 × 1 0 4 g CH 2 CHCN in the room

    50% of the carpet burns, so the starting amount of CH 2 CHCN is

    0.50(17,025) = 8513 = 8.5 × 10 3 g

    8 5 1 3 g C H 2 CHCN × 5 0 . 9 g HCN 1 0 0 g C H 2 CHCN = 4 3 3 3 = 4 . 3 × 1 0 3 g HCN possible

    If the actual yield of combustion is 20%, actual g HCN = 4333(0.20) = 866.6 = 8.7 × 10 2 g HCN produced. From part (a), 14 g of HCN is a lethal dose. The fire produces much more than a lethal dose of HCN.


3.114

  1. N 2 (g) + O 2 (g) → 2 NO(g); 2 NO(g) + O 2 → 2 NO 2 (g)
  2. 1 million = 1 × 10 6

    1 9 × 1 0 6 tons  N O 2 × 2000 Lb 1 ton × 4 5 3 . 6 g 1 Lb = 1 . 7 2 4 × 1 0 1 3 = 1 . 7 × 1 0 1 3 g N O 2

  3. Plan . Calculate g O 2 needed to burn 500 g octane. This is 85% of total O 2 in the engine. 15% of total O 2 is used to produce NO 2 , according to the second equation in part (a).

    Solve . 2 C 8 H 18 (l) + 25 O 2 (g) → 16 CO 2 (g) + 18 H 2 O(l)

    5 0 0 g C 8 H 1 8 × 1 mol C 8 H 1 8 1 1 4 . 2 g C 8 H 1 8 × 2 5 mol O 2 2 mol C 8 H 1 8 × 3 2 . 0 0 g O 2 mol O 2 = 1 7 5 1 = 1 . 7 5 × 1 0 3 g O 2

    1 7 5 1 g O 2 total g O 2 = 0 . 8 5 ; 2 0 6 0 = 2 . 1 × 1 0 3 g O 2 total in engine

    2060 g O 2 total × 0.15 = 309.1 = 3.1 × 10 2 g O 2 used to produce NO 2 . One mol O 2 produces 2 mol NO. Then 2 mol NO react with a second mol O 2 to produce 2 mol NO 2 . Two mol O 2 are required to produce 2 mol NO 2 ; one mol O 2 per mol NO 2 .

    3 0 9 . 1 g O 2 × 1 mol O 2 3 2 . 0 0 g × 1 mol NO 2 1 mol O 2 × 4 6 . 0 1 g NO 2 1 mol O 2 = 4 4 4 . 4 = 4 . 4 × 1 0 2 g NO 2

3.115

  1. Fe 2 O 3 (s) + 2 Al(s) → Al 2 O 3 (s) + 2 Fe(l)
  2. 5 0 0 . 0 g Fe 2 O 3 × 1 mol Fe 2 O 3 1 5 9 . 6 8 8 g Fe 2 O 3 × 2 mol Al 1 mol Fe 2 O 3 × 2 6 . 9 8 1 5 g Al 1 mol Al = 1 6 8 . 9 6 4 = 1 6 9 . 0 g Al
  3. 1 . 0 0 × 1 0 4 kJ × 1 mol Fe 2 O 3 8 5 2 kJ × 1 5 9 . 6 8 8 g Fe 2 O 3 1 mol Fe 2 O 3 = 1 8 7 4 . 2 7 = 1 . 8 7 × 1 0 3 g Fe 2 O 3
  4. Heat is a reactant in the reverse reaction.

3.116

  1. Represent the hexyl group, CH 3 CH 2 CH 2 CH 2 CH 2 CH 2− , as C 6 H 13 . The Ugi product is quite complex. Moving from left to right, we can write

    C 6 H 13 CON(C 6 H 13 )C(C 6 H 13 ) 2 CON(H)C 6 H 13 . The parentheses indicate hexyl groups that are not in the backbone of the molecule. The molecular formula of the product is (C 6 H 13 ) 5 C 3 HN 2 O 2 or C 33 H 66 N 2 O 2 . The molecular formula of the product is convenient for calculating molar mass in part (b).

    (C 6 H 13 ) 2 CO + C 6 H 13 NH 2 + C 6 H 13 COOH + C 6 H 13 NC → C 6 H 13 CON(C 6 H 13 )C(C 6 H 13 ) 2 CON(H)C 6 H 13

  2. 435.0 g C 6 H 13 NH 2 × 1 mol C 6 H 13 NH 2 101.190 g C 6 H 13 NH 2 × 1 mol C 33 H 66 N 2 O 2 1 mol C 6 H 13 NH 2 × 5 2 2 . 8 8 9 g C 3 3 H 6 6 N 2 O 2 1 mol C 3 3 H 6 6 N 2 O 2 = 2 2 4 7 . 8 2 = 2 2 4 8 g C 3 3 H 6 6 N 2 O 2

 

4 Reactions in Aqueous Solution

Visualizing Concepts

4.1 Analyze .  Correlate the formula of the solute with the charged spheres in the diagrams.

Plan .  Determine the electrolyte properties of the solute and the relative number of cations, anions, or neutral molecules produced when the solute dissolves.

Solve .  Li 2 SO 4 is a strong electrolyte, a soluble ionic solid that dissociates into separate Li + and SO 4 2– when it dissolves in water. There are twice as many Li + cations as SO 4 2– anions. Diagram (c) represents the aqueous solution of a 2:1 electrolyte.

4.2 Analyze/Plan .  Correlate the neutral molecules, cations, and anions in the diagrams with the definitions of strong, weak, and nonelectrolytes. Solve .

  1. AX is a nonelectrolyte because no ions form when the molecules dissolve.
  2. AY is a weak electrolyte because a few molecules ionize when they dissolve, but most do not.
  3. AZ is a strong electrolyte because all molecules break up into ions when they dissolve.

4.3 Analyze/Plan . From the molecular representations, write molecular formulas for the compounds. Using Table 4.2 and molecular formulas (there are no ionic compounds in this exercise), classify the compounds as strong acid, strong base, weak acid, weak base (NH 3 ), or nonelectrolyte. Strong acids and bases are strong electrolytes, weak acids and bases are weak electrolytes. Solve.

  1. HCOOH. The molecule has a COOH group; it is a weak acid and weak electrolyte (it is not one of the strong acids listed in Table 4.2).
  2. HNO 3 . The molecule is a strong acid (Table 4.2) and a strong electrolyte.
  3. CH 3 CH 2 OH. The molecule is neither an acid nor a base; it is a nonelectrolyte.

4.4   Statement (b) is most correct. Statement (a) is incorrect because, at equilibrium, the chemical reactions are ongoing. Statement (c) is incorrect because the concentration of the product (or reactant) is changing until equilibrium is reached, but not at equilibrium. Equilibrium is a state of dynamic constancy.

4.5 Analyze/Plan .  From the names and/or formulas of the three possible solids, determine which exhibits the described solubility properties. Use Table 4.1.

Solve .  The three possible compounds are BaCl 2 , PbCl 2 , and ZnCl 2 . PbCl 2 does not dissolve in water to give a clear solution, so it can be eliminated. Of the remaining possibilities, Ba 2+ has a sulfate precipitate, but Zn 2+ does not. The compound is indeed BaCl 2 .


4.6 Analyze .  Given the formulas of some ions, determine whether these ions ever form precipitates in aqueous solution. Plan . Use Table 4.1 to determine if the given ions can form precipitates. If not, they will always be spectator ions. Solve .

  1. Cl– can form precipitates with Ag + , Hg 2 2+ , and Pb 2+ .
  2. NO 3 never forms precipitates, so it is always a spectator.
  3. NH 4 + never forms precipitates, so it is always a spectator.
  4. S 2– usually forms precipitates.
  5. SO 4 2– can form precipitates with Sr 2+ , Ba 2+ , Hg 2 2+ , and Pb 2+ .

Check .  NH 4 + is a soluble exception for sulfides, phosphates, and carbonates which usually form precipitates, so all rules indicate that it is a perpetual spectator.

4.7 Analyze/Plan. Given three metal powders and three 1 M solutions, use Table 4.5, the activity series of metals, to find a scheme to distinguish the metals.

Solve. In the activity series, any metal on the list can be oxidized by the ions of elements below it. The nitric acid solution contains H + (aq). This solution will oxidize and thus dissolve Zn(s) and Pb(s), which appear above H 2 (g) on the list. Platinum, Pt(s), is distinguished by its lack of reaction with nitric acid.

To distinguish between Zn and Pb, use a metal ion that occurs between them on the list. We have such an ion, Ni 2+ (aq) in the nickel nitrate solution. Ni 2+ (aq) will oxidize and thus dissolve Zn(s), which is above it on the list. Ni 2+ (aq) will not oxidize or dissolve Pb(s), which is below it on the list.

To summarize, Pt(s) will neither be oxidized by nor dissolve in any of the three available solutions. Pb(s) is oxidized by and will dissolve in the nitric acid solution, but not the nickel nitrate solution. Zn(s) is oxidized by and will dissolve in both nitric acid and nickel nitrate solutions.

4.8   In a redox reaction, one reactant loses electrons and a different reactant gains electrons; electrons are transferred. Acids ionize in aqueous solution to produce (donate) hydrogen ions (H + , protons). Bases are substances that react with or accept protons (H + ). In an neutralization reaction, protons are transferred from an acid to a base. We characterize redox reactions by tracking electron transfer using oxidation numbers. We characterize neutralization reactions by tracking H + (proton) transfer via molecular formulas of reactants and products.

4.9   The answer is: (c) a redox reaction. The “water-splitting” reaction is

2 H 2 O(l) → 2 H 2 (g) + O 2 (g)

In the reaction, the oxidation number of hydrogen decreases from +1 to 0, and the oxidation number of oxygen increases from –2 to 0. Hydrogen is reduced and oxygen is oxidized.

4.10 Analyze/Plan . We are given a total ion concentration for two salt solutions and the formulas of the ionic solutes. Decide the total moles of ions produced when one mole of the salt dissolves, and the ratio of the moles of the ion in question to the total moles of ions. Use this ratio to calculate the concentration of the ion in question.

  1. When one mole of NaCl dissolves, it produces 1 mole of Cl ions and 2 total moles of ions. The concentration of Cl is: ½(1.2 m M ) = 0.60 m M Cl (aq).

  1. When one mole of FeCl 3 dissolves, it produces 3 moles of Cl ions and 4 total moles of ions. The concentration of Cl is: ¾ (1.2 m M ) = 0.90 m M Cl (aq).

4.11 Analyze/Plan . The plot shows indicator color versus volume of standard solution added. Consider how the indicator changes color in a titration like the one in Figure 4.18, and relate this behavior to the shapes shown in the graph.

Solve . The “green” data set is expected from a titration like the one in Figure 4.18. The indicator color remains constant until the reaction is very near the equivalence point. Within a very small volume of standard solution added, the indicator color changes rapidly. This behavior is shown in green. The red graph shows a constantly changing indicator color.

4.12 Analyze/Plan . The purpose of every titration is to determine the equivalence point. Based on the indicator behavior described in the exercise, decide how the amount of reactants and products in the titration beaker relate to the equivalence point. Given this reaction mixture, design an experiment to reach the equivalence point.

Solve . This indicator is colorless in acid and blue in base. When it is added to the beaker, the solution is dark blue, so the solution is quite basic. This means that the amount of base added is already greater than the amount of acid initially present. To reach the equivalence point, more acid must be added to the titration beaker.

First, record the volume of base added. Then, find a standard acid solution (an acid of very well-known concentration) or standardize an acid solution (probably HCl). Rinse and fill a clean buret with the standard acid. Carefully titrate the mixture in the beaker until the blue color fades and finally disappears. Record the volume of standard acid added. Subtract the amount of added acid from the total amount of base added. The remaining amount of base is the volume required to reach equivalence with the original acid sample. This procedure is called “back titration.”

General Properties of Aqueous Solutions (Section 4.1)

4.13

  1. False. Electrolyte solutions conduct electricity because ions are moving through the solution.
  2. True. The conductivity is unchanged as long as the concentration of electrolytes is unchanged. Because ions are mobile in solution, the added presence of uncharged molecules does not inhibit conductivity.

4.14

  1. False. Methanol is an organic alcohol, and the –OH group is not ionizable. The neutral methanol molecules in solution do not support the movement of charge. The solution does not conduct electricity.
  2. True. CH 3 COOH is a weak electrolyte. When it dissolves in water, a small percentage of the molecules ionize to form H + and CH 3 COO– ions. The presence of a small concentration of ions produces a weakly conducting solution, and the specific presence of H + makes the solution acidic.

4.15  Statement (b) is most correct. Statements (a) and (c) are incorrect because water is not a strong acid and the hydrogen and oxygen bonds of water are not broken by ionic solids.


4.16  Anions are negatively charged. They will be attracted to and thus physically closer to the partially positive portion of the water molecule, which is the hydrogens.

4.17 Analyze/Plan .  Given the solute formula, determine the separate ions formed upon dissociation. Solve .

  1. FeCl 2 (aq) → Fe 2+ (aq) + 2 Cl (aq)
  2. HNO 3 (aq) → H + (aq) + NO 3 (aq)
  3. (NH 4 ) 2 SO 4 (aq) → 2 NH 4 + (aq) + SO 4 2– (aq)
  4. Ca(OH) 2 (aq) → Ca 2+ (aq) + 2 OH (aq)

4.18

  1. MgI 2 (aq) → Mg 2+ (aq) + 2 I (aq)
  2. K 2 CO 3 (aq) → 2 K + (aq) + CO 3 2– (aq)
  3. HClO 4 (aq) → H + (aq) + ClO 4 (aq)
  4. NaCH 3 COO(aq) → Na + (aq) + CH 3 COO (aq)

4.19 Analyze/Plan . Apply the definition of a weak electrolyte to HCOOH.

Solve .  When HCOOH dissolves in water, neutral HCOOH molecules, H + ions and HCOO ions are all present in the solution. HCOOH(aq) ⇌ H + (aq) + HCOO (aq)

4.20

  1. acetone (nonelectrolyte): CH 3 COCH 3 (aq) molecules only; hypochlorous acid (weak electrolyte): HClO(aq) molecules, H + (aq), ClO– (aq); ammonium chloride (strong electrolyte): NH 4 + (aq), Cl (aq)
  2. NH 4 Cl, 0.2 mol solute particles; HClO, between 0.1 and 0.2 mol particles; CH 3 COCH 3 , 0.1 mol of solute particles

Precipitation Reactions (Section 4.2)

4.21 Analyze .  Given: formula of compound. Find: solubility.

Plan .  Follow the guidelines in Table 4.1, in light of the anion present in the compound and notable exceptions to the “rules.” Solve .

  1. Mg Br 2 : soluble
  2. Pb I 2 : insoluble, Pb 2+ is an exception to soluble iodides
  3. (NH 4 ) 2 CO 3 : soluble, NH 4 + is an exception to insoluble carbonates
  4. Sr (OH) 2 : soluble, Sr 2+ is an exception to insoluble hydroxides
  5. Zn SO 4 : soluble

4.22  According to Table 4.1:

  1. Ag I : insoluble (an exception to the generally soluble iodides)
  2. Na 2 CO 3 : soluble, Na + is an exception to insoluble carbonates
  3. Ba Cl 2 : soluble
  4. Al (OH) 3 : insoluble
  5. Zn( CH 3 COO) 2 : soluble

4.23 Analyze .  Given: formulas of reactants. Find: balanced equation including precipitates.

Plan .  Follow the logic in Sample Exercise 4.3.

Solve .  In each reaction, the precipitate is in bold type.

  1. Na 2 CO 3 (aq) + 2 AgNO 3 (aq) → Ag 2 CO 3 (s) + 2 NaNO 3 (aq)
  2. No precipitate (all nitrates and most sulfates are soluble).
  3. FeSO 4 (aq) + Pb(NO 3 ) 2 (aq) → PbSO 4 (s) + Fe(NO 3 ) 2 (aq)

4.24  In each reaction, the precipitate is in bold type.

  1. No precipitate. [CH 3 COO gains H + to form CH 3 COOH(aq), which is soluble.]
  2. Cu(NO 3 ) 2 (aq) + 2 KOH(aq) → Cu(OH) 2 (s) + 2 KNO 3 (aq)
  3. Na 2 S(aq) + CdSO 4 (aq) → CdS(s) + Na 2 SO 4 (aq)

4.25 Analyze/Plan .  Follow the logic in Sample Exercise 4.4. From the complete ionic equation, identify the ions that don’t change during the reaction; these are the spectator ions. Solve .

  1. 2 K + (aq) + CO 3 2– (aq) + Mg 2+ (aq) + SO 4 2– (aq) → MgCO 3 (s) + 2 K + (aq) + SO 4 2– (aq) Spectators: K + , SO 4 2–
  2. Pb 2+ (aq) + 2 NO 3 (aq) + 2 Li + (aq) + S 2– (aq) → PbS(s) + 2 Li + (aq) + 2 NO 3 (aq) Spectators: Li + , NO 3
  3. 6 NH 4 + (aq) + 2 PO 4 3– (aq) + 3 Ca 2+ (aq) + 6 Cl (aq) → Ca 3 (PO 4 ) 2 (s) + 6 NH 4 + (aq) + 6 Cl (aq) Spectators: NH 4 + , Cl

4.26  Spectator ions are those that do not change during reaction.

  1. 2 Cr 3+ (aq) + 3 CO 3 2– (aq) → Cr 2 (CO 3 ) 3 (s); spectators: NH 4 + , SO 4 2–
  2. Ba 2+ (aq) + SO 4 2– (aq) → BaSO 4 (s); spectators: K + , NO 3
  3. Fe 2+ (aq) + 2 OH (aq) → Fe(OH) 2 (s); spectators: K + , NO 3

4.27 Analyze .  Given: reactions of unknown salt with HBr, H 2 SO 4 , NaOH. Find: Does the unknown salt contain K + or Pb 2+ or Ba 2+ ?

Plan .  Analyze solubility guidelines for Br , SO 4 2– , and OH and select the cation that produces a precipitate with each of the anions.

Solve .  K + forms no precipitates with any of the anions. BaSO 4 is insoluble, but BaCl 2 and Ba(OH) 2 are soluble. Because the unknown salt forms precipitates with all three anions, it must contain Pb 2+ .

Check .  PbBr 2 , PbSO 4 , and Pb(OH) 2 are all insoluble according to Table 4.1, so our process of elimination is confirmed by the insolubility of the Pb 2+ compounds.

4.28  Br and NO 3 can be ruled out because the BaBr 2 is soluble and all NO 3 salts are

soluble. CO 3 2– forms insoluble salts with the three cations given; it must be the anion in question.


4.29 Analyze / Plan .  Using Table 4.1, determine the precipitates that could form when each of the unknowns is reacted with Ba(NO 3 ) 2 and NaCl. Solve .

  1. True. If the unknown is Al 2 (SO 4 ) 3 , BaSO 4 could precipitate.
  2. True. If the unknown is AgNO 3 , AgCl could precipitate.
  3. False (two ways). Ag 2 SO 4 is a soluble ionic compound, and no combination of the possible unknowns and the two reagents could produce Ag 2 SO 4 .
  4. True. This is the overall correct answer to the question.
  5. False, because (c) is false.

4.30  Consider all the possible combinations of Pb 2+ (aq), Na + (aq), Ca 2+ (aq), CH 3 COO (aq), S 2– (aq), and Cl (aq). Which of these compounds are insoluble ionic compounds that will precipitate when the solutions are combined?

PbS(s) and PbCl 2 (s) will precipitate.

Acids, Bases, and Neutralization Reactions (Section 4.3)

4.31 Analyze .  Given: solute and concentration of three solutions. Find: the solution that is most acidic.

Plan :  See Sample Exercise 4.6 and Table 4.2. Determine whether solutes are strong or weak acids or bases or nonelectrolytes. For solutions of equal concentration, strong acids will have greatest concentration of solvated protons. Take varying concentration into consideration when evaluating the same class of solutions.

Solve .  (a) LiOH is a strong base, (b) HI is a strong acid, (c) CH 3 OH is a molecular compound and nonelectrolyte. Solution (b), 0.2 M HI, is the most acidic solution.

Check .  The solution concentrations weren’t needed to answer the question.

4.32  NH 3 (aq) is a weak base, whereas KOH and Ba(OH) 2 are strong bases. NH 3 (aq) is only slightly ionized, so even (a) 0.6 M NH 3 is less basic than (b) 0.150 M KOH. Ba(OH) 2 has twice as many OH per mole as KOH, so (c) 0.100 M Ba(OH) 2 is more basic than (b) 0.150 M KOH. The most basic solution is (c) 0.100 M Ba(OH) 2 .

4.33

  1. False. Sulfuric acid, H 2 SO 4 , is a diprotic acid; it has two ionizable hydrogen atoms.
  2. False. According to Table 4.2, HCl is a strong acid.
  3. False. Methanol, CH 3 OH, is a molecular nonelectrolyte.

4.34

  1. True. NH 3 produces OH in aqueous solution by reacting with H 2 O (hydrolysis): NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH (aq). The OH causes the solution to be basic. NH 3 (aq) attracts an H + from water, leaving OH (aq) in the solution.
  2. False. According to Table 4.2, HF is not one of the strong acids.
  3. True. H 2 SO 4 is a diprotic acid; it has two ionizable hydrogens. The first hydrogen completely ionizes to form H + and HSO 4– , but HSO 4– only partially ionizes into H + and SO 4 2– (HSO 4– is a weak electrolyte). Thus, an aqueous solution of H 2 SO 4 contains a mixture of H + , HSO 4– , and SO 4 2– , with the concentration of HSO 4– greater than the concentration of SO 4 2– .

4.35 Analyze .  Given: chemical formulas. Find: classify as acid, base, salt; strong, weak, or nonelectrolyte.

Plan .  See Tables 4.2 and 4.3. Ionic or molecular? Ionic, soluble: OH , strong base and strong electrolyte; otherwise, salt, strong electrolyte. Molecular: NH 3 , weak base and weak electrolyte; H-first, acid; strong acid (Table 4.2), strong electrolyte; otherwise weak acid and weak electrolyte. Solve .

  1. HF: acid, mixture of ions and molecules (weak electrolyte)
  2. CH 3 CN: none of the above, entirely molecules (nonelectrolyte)
  3. NaClO 4 : salt, entirely ions (strong electrolyte)
  4. Ba(OH) 2 : base, entirely ions (strong electrolyte)

4.36  Because the solution does conduct some electricity, but less than an equimolar NaCl solution (a strong electrolyte), the unknown solute must be a weak electrolyte. The weak electrolytes in the list of choices are NH 3 and H 3 PO 3 ; because the solution is acidic, the unknown must be H 3 PO 3 .

4.37 Analyze .  Given: chemical formulas. Find: electrolyte properties.

Plan .  To classify as electrolytes, formulas must be identified as acids, bases, or salts as in Solution 4.35. Solve .

  1. H 2 SO 3 : H first, so acid; not in Table 4.2, so weak acid; therefore, weak electrolyte
  2. CH 3 CH 2 OH: not acid, not ionic (no metal cation), contains OH group, but not as anion so not a base; therefore, nonelectrolyte
  3. NH 3 : common weak base; therefore, weak electrolyte
  4. KClO 3 : ionic compound, so strong electrolyte
  5. Cu(NO 3 ) 2 : ionic compound, so strong electrolyte

4.38

  1. LiClO 4 : strong
  2. HClO: weak
  3. CH 3 CH 2 CH 2 OH: non
  4. HClO 3 : strong
  5. CuSO 4 : strong
  6. C 12 H 22 O 11 : non

4.39 Plan .  Follow Sample Exercise 4.7. Solve .

  1. 2 HBr(aq) + Ca(OH) 2 (aq) → CaBr 2 (aq) + 2 H 2 O(l)

    H + (aq) + OH (aq) → H 2 O(l)

  2. Cu(OH) 2 (s) + 2 HClO 4 (aq) → Cu(ClO 4 ) 2 (aq) + 2 H 2 O(l)

    Cu(OH) 2 (s) + 2 H + (aq) → 2 H 2 O(l) + Cu 2+ (aq)

  3. Al(OH) 3 (s) + 3 HNO 3 (aq) → Al(NO 3 ) 3 (aq) + 3 H 2 O(l)

    Al(OH) 3 (s) + 3 H + (aq) → 3 H 2 O(l) + Al 3+ (aq)

4.40

  1. 2 CH 3 COOH(aq) + Ba(OH) 2 (aq) → Ba(CH 3 COO) 2 (aq) + 2 H 2 O(l)

    CH 3 COOH(aq) + OH (aq) → CH 3 COO (aq) + H 2 O(l)

  2. Cr(OH) 3 (s) + 3 HNO 2 (aq) → Cr(NO 2 ) 3 (aq) + 3 H 2 O(l)

    Cr(OH) 3 (s) + 3 HNO 2 (aq) → 3 H 2 O(l) + Cr 3+ (aq) + 3 NO 2– (aq)


  1. HNO 3 (aq) + NH 3 (aq) → NH 4 NO 3 (aq)

    H + (aq) + NH 3 (aq) → NH 4 + (aq)

4.41 Analyze .  Given: names of reactants. Find: gaseous products.

Plan .  Write correct chemical formulas for the reactants, complete and balance the metathesis reaction, and identify either H 2 S or CO 2 products as gases. Solve .

  1. CdS(s) + H 2 SO 4 (aq) → CdSO 4 (aq) + H 2 S(g)

    CdS(s) + 2 H + (aq) → H 2 S(g) + Cd 2+ (aq)

  2. MgCO 3 (s) + 2 HClO 4 (aq) → Mg(ClO 4 ) 2 (aq) + H 2 O(l) + CO 2 (g)

    MgCO 3 (s) + 2 H + (aq) → H 2 O(l) + CO 2 (g) + Mg 2+ (aq)

4.42

  1. FeO(s) + 2 H + (aq) → H 2 O(l) + Fe 2+ (aq)
  2. NiO(s) + 2 H + (aq) → H 2 O(l) + Ni 2+ (aq)

4.43 Analyze .  Given the formulas or names of reactants, write balanced molecular and net ionic equations for the reactions.

Plan .  Write correct chemical formulas for all reactants. Predict products of the neutralization reactions by exchanging ion partners. Balance the complete molecular equation, identify spectator ions by recognizing strong electrolytes, write the corresponding net ionic equation (omitting spectators). Solve .

  1. MgCO 3 (s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 O(l) + CO 2 (g)

    MgCO 3 (s) + 2 H + (aq) → Mg 2+ (aq) + H 2 O(l) + CO 2 (g)

    MgO(s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 O(l)

    MgO(s) + 2 H + (aq) → Mg 2+ (aq) + H 2 O(l)

    Mg(OH) 2 (s) + 2 HCl(aq) → MgCl 2 (aq) + 2 H 2 O(l)

    Mg(OH) 2 (s) + 2 H + (aq) → Mg 2+ (aq) + 2 H 2 O(l)

  2. We can distinguish magnesium carbonate, MgCO 3 (s), because its reaction with acid produces CO 2 (g), which appears as bubbles. The other two compounds are indistinguishable because the products of the two reactions are exactly the same.

4.44

  1. K 2 O(aq) + H 2 O(l) → 2 KOH(aq), molecular; O 2– (aq) + H 2 O(l) → 2 OH (aq), net ionic
  2. base (H + ion acceptor): O 2– (aq);
  3. acid (H + ion donor): H 2 O(aq);
  4. spectator: K +

Oxidation–Reduction Reactions (Section 4.4)

4.45

  1. False. O xidation is loss of electrons; reduction is gain of electrons. (LEO says GER.)
  2. True. When a substance is oxidized, its oxidation number increases. When a substance is reduced, its oxidation number decreases.

4.46

  1. True. Oxidation is loss of electrons; it can occur in the presence of any electron acceptor, not just oxygen.
  2. False. Oxidation and reduction can only occur together, not separately. When a substance is oxidized, it loses electrons, but free electrons do not exist under normal conditions. If electrons are lost by one substance they must be gained by another, and vice versa.

4.47 Analyze .  Given the labeled periodic chart, determine which regions are most and least easily oxidized.

Plan . Review the definition of oxidation and apply it to the properties of elements in the indicated regions of the chart. Solve .

  1. Oxidation is loss of electrons. Elements easily oxidized form positive ions; these are metals. Elements in regions A and B are metals, and their ease of oxidation is shown in Table 4.5.
  2. Elements not readily oxidized tend to gain electrons and form negative ions; these are nonmetals. Elements in region D are nonmetals and are least easily oxidized.

4.48

  1. Ba S O 4 ; +6
  2. H 2 S O 3 ; +4
  3. Sr S ; –2
  4. H 2 S ; –2
  5. Sulfur is the third row of group 6A, the third column from the right on the periodic table. That is in region D on the designated chart.
  6. Based on these compounds, the range of oxidation numbers for sulfur is +6 to –2. Sulfur and other nonmetals in region D can adopt both positive and negative oxidation numbers. This is also true for the metalloids in region C. These elements have properties of both metals and nonmetals and can thus adopt both positive and negative oxidation numbers.

4.49 Analyze .  Given the chemical formula of a substance, determine the oxidation number of a particular element in the substance.

Plan .  Follow the logic in Sample Exercise 4.8. Solve.

  1. +4
  2. +4
  3. +7
  4. +1
  5. +3
  6. –1 (O 2 2– is peroxide ion)

4.50

  1. +3
  2. +3
  3. –2
  4. –3
  5. +3
  6. +6

4.51 Analyze .  Given: chemical reaction. Find: element oxidized or reduced. Plan . Assign oxidation numbers to all species. The element whose oxidation number increases (becomes more positive) is oxidized; the one whose oxidation number decreases (becomes more negative) is reduced. Solve .

  1. N 2 (g) [N, 0] → 2 NH 3 (g) [N, ‒3], N is reduced; 3 H 2 (g) [H, 0] → 2 NH 3 (g) [H, +1], H is oxidized.
  2. Fe 2+ → Fe, Fe is reduced; Al → Al 3+ , Al is oxidized
  3. Cl 2 → 2 Cl , Cl is reduced; 2 I → I 2 , I is oxidized
  4. S 2– → SO 4 2– (S, +6), S is oxidized; H 2 O 2 (O, –1) → H 2 O (O, –2); O is reduced

4.52

  1. oxidation–reduction reaction; P is oxidized, Cl is reduced
  2. oxidation–reduction reaction; K is oxidized, Br is reduced

  1. oxidation–reduction reaction; C is oxidized, O is reduced
  1. precipitation reaction

4.53 Analyze .  Given: reactants. Find: balanced molecular and net ionic equations.

Plan .  Metals oxidized by H + form cations. Predict products by exchanging cations and balance. The anions are the spectator ions and do not appear in the net ionic equations.

Solve .

  1. Mn(s) + H 2 SO 4 (aq) → MnSO 4 (aq) + H 2 (g);

    Mn(s) + 2 H + (aq) → Mn 2+ (aq) + H 2 (g)

    Products with the metal in a higher oxidation state are possible, depending on reaction conditions and acid concentration.

  2. 2 Cr(s) + 6 HBr(aq) → 2 CrBr 3 (aq) + 3 H 2 (g);

    2 Cr(s) + 6 H + (aq) → 2 Cr 3+ (aq) + 3 H 2 (g)

  3. Sn(s) + 2 HCl(aq) → SnCl 2 (aq) + H 2 (g); Sn(s) + 2 H + (aq) → Sn 2+ (aq) + H 2 (g)
  4. 2 Al(s) + 6 HCOOH(aq) → 2 Al(HCOO) 3 (aq) + 3 H 2 (g);

    2 Al(s) + 6 HCOOH(aq) → 2 Al 3+ (aq) + 6 HCOO (aq) + 3 H 2 (g)

4.54

  1. 2 HCl(aq) + Ni(s) → NiCl 2 (aq) + H 2 (g); Ni(s) + 2 H + (aq) → Ni 2+ (aq) + H 2 (g)
  2. H 2 SO 4 (aq) + Fe(s) → FeSO 4 (aq) + H 2 (g); Fe(s) + 2 H + (aq) → Fe 2+ (aq) + H 2 (g)

    Products with the metal in a higher oxidation state are possible, depending on reaction conditions and acid concentration.

  3. 2 HBr(aq) + Mg(s) → MgBr 2 (aq) + H 2 (g); Mg(s) + 2 H + (aq) → Mg 2+ (aq) + H 2 (g)
  4. 2 CH 3 COOH(aq) + Zn(s) → Zn(CH 3 COO) 2 (aq) + H 2 (g);

    Zn(s) + 2 CH 3 COOH(aq) → Zn 2+ (aq) + 2 CH 3 COO (aq) + H 2 (g)

4.55 Analyze .  Given: a metal and an aqueous solution. Find: balanced equation.

Plan .  Use Table 4.5. If the metal is above the aqueous solution, reaction will occur; if the aqueous solution is higher, NR. If reaction occurs, predict products by exchanging cations (a metal ion or H + ), then balance the equation. Solve.

  1. Fe(s) + Cu(NO 3 ) 2 (aq) → Fe(NO 3 ) 2 (aq) + Cu(s)
  2. Zn(s) + MgSO 4 (aq) → NR
  3. Sn(s) + 2 HBr(aq) → SnBr 2 (aq) + H 2 (g)
  4. H 2 (g) + NiCl 2 (aq) → NR
  5. 2 Al(s) + 3 CoSO 4 (aq) → Al 2 (SO 4 ) 3 (aq) + 3 Co(s)

4.56

  1. Ni(s) + Cu(NO 3 ) 2 (aq) → Ni(NO 3 ) 2 (aq) + Cu(s)
  2. Zn(NO 3 ) 2 (aq) + MgSO 4 (aq) → NR
  3. Au(s) + HCl(aq) → NR
  4. 2 Cr(s) + 3 CoCl 2 (aq) → 2 CrCl 3 (aq) + 3 Co(s)
  5. H 2 (g) + 2 AgNO 3 (aq) → 2 Ag(s) + 2 HNO 3 (aq)

4.57

    1. Zn(s) + Cd 2+ (aq) → Cd(s) + Zn 2+ (aq)
    2. Cd(s) + Ni 2+ (aq) → Ni(s) + Cd 2+ (aq)

    Observation (i) indicates that Cd is less active than Zn; observation (ii) indicates that Cd is more active than Ni. Cd is between Zn and Ni on the activity series.

  1. Chromium, iron, and cobalt, the three elements between Zn and Ni in Table 4.5, more closely define the position of Cd in the activity series.
  2. Place an iron strip in CdCl 2 (aq). If Cd(s) is deposited, Cd is less active than Fe; if there is no reaction, Cd is more active than Fe. Do the same test with Co if Cd is less active than Fe or with Cr if Cd is more active than Fe.

4.58  Br 2 + 2 NaI → 2 NaBr + I 2 indicates that Br 2 is more easily reduced than I 2 .

Cl 2 + 2 NaBr → 2 NaCl + Br 2 shows that Cl 2 is more easily reduced than Br 2 .

The order for ease of reduction is Cl 2 > Br 2 > I 2 . Conversely, the order for ease of oxidation is I > Br > Cl .

  1. From the information above, the halogen I 2 is most stable (less likely to react) when mixed with other halides, X–.
  2. Cl 2 + 2 KI → 2 KCl + I 2
  3. Br 2 + LiCl → no reaction

Concentrations of Solutions (Section 4.5)

4.59

  1. Concentration is an intensive property; it is the ratio of the amount of solute present in a certain quantity of solvent or solution. This ratio remains constant regardless of how much solution is present.
  2. The term 0.50 mol HCl defines an amount (~18 g) of the pure substance HCl. The term 0.50 M HCl is a ratio; it indicates that there are 0.50 mol of HCl solute in 1.0 liter of solution. This same ratio of moles solute to solution volume is present regardless of the volume of solution under consideration.

4.60 Analyze/Plan . Follow the logic in Sample Exercise 4.11. Solve .

  1. M = mol solute L solution ; 35 .0 g C 12 H 22 O 11 1 .000 L × 1 mol C 12 H 22 O 11 342 .3 g C 12 H 22 O 11 = 0.102 M C 12 H 22 O 11
  2. Add 1.000 L of water to reduce the molarity by a factor of 2. Adding water does not change to amount of solute, but it does change the total volume of solution. A total solution volume of 2.000 L will reduce the molarity by a factor of 2.

4.61 Analyze/Plan .  Follow the logic in Sample Exercises 4.11 and 4.12. Solve .

  1. M = mol solute L solution ; 0 .175 mol ZnCl 2 150 mL × 1000 mL 1 L = 1.17 M ZnCl 2

    Check .  (0.175 / 0.150) > 1.0 M


  1. mol = M × L ; 4 .50 mol HNO 3 1 L × 1 mol H + mol HNO 3 × 0.0350 L = 0.158 mol H +

    Check .  (4.5 × .04) ≈ 0.16 mol

  1. L = mol M ; 0 .350 mol NaOH 6 .00 mol NaOH/L = 0.0583 L or 58 .3 mL of 6.00 M NaOH

    Check .  (0.325/6.0) > 0.50 L.

4.62

  1. M = mol solute L solution ; 12.5 g Na 2 CrO 4 0.750 L × 1 mol Na 2 CrO 4 161.97 g Na 2 CrO 4 = 0.103 M Na 2 CrO 4
  2. mol = M × L ; 0.112 mol KBr 1 L × 0.150 L = 1.68 × 10 2 mol KBr
  3. L = mol M ; 0.150 mol HCl 6.1 mol HCl/L = 2.5 × 10 2 L or 25 mL

4.63 Analyze .  Given molarity, M , and volume, L, find mass of Na + (aq) in the blood.

Plan .  Calculate moles Na + (aq) using the definition of molarity: M = mol L ; mol = M × L.

Calculate mass Na + (aq) using the definition moles: mol = g/MM; g = mol × MM. (MM is the symbol for molar mass in this manual.)

Solve . 0.135 mol L × 5.0 L × 23.0 g Na + mol Na + = 15.525 = 16 g Na + ( aq )

Check .  Because there are more than 0.1 mol/L and we have 5.0 L, there should be more than half a mol (11.5 g) of Na + . The calculation agrees with this estimate.

4.64  Calculate the mol of Na + at the two concentrations; the difference is the mol NaCl required to increase the Na + concentration to the desired level.

0.118 mol L × 4.6 L = 0.5428 = 0.54 mol Na +

0 .138 mol L × 4.6 L = 0.6348 = 0.63 mol Na +

(0.6348 – 0.5428) = 0.092 = 0.09 mol NaCl (2 decimal places and 1 sig fig)

0.092 mol NaCl × 58.5 g NaCl mol = 5.38 = 5 g NaCl

4.65 Analyze . Given: g alcohol/100 mL blood; molecular formula of alcohol. Find: molarity (mol/L) of alcohol. Plan . Use the molar mass (MM) of alcohol to change (g/100) mL to (mol/100 mL) then mL to L.

Solve . MM of alcohol = 2(12.01) + 6(1.1008) + 1(16.00) = 46.07 g alcohol/mol

BAC = 0 .08 g alcohol 100 mL blood × 1 mol alcohol 46 .07 g alcohol × 1000 mL 1 L = 0 . 0174 = 0 . 02 M alcohol


4.66 Analyze . Given: BAC (definition from Exercise 4.65), vol of blood. Find: mass alcohol in bloodstream.

Plan . Change BAC (g/100 mL) to (g/L), then times vol of blood in L.

Solve . BAC = 0.10 g/100 mL

0 .10 g alcohol 100 mL blood × 1000 mL 1 L × 5 .0 L blood = 5 .0 g alcohol

4.67 Plan .  Proceed as in Sample Exercises 4.13.

M = mol L ; mol = g MM (MM  is the symbol for molar mass in this manual .)

Solve .

  1. 6.86 mol CH 3 CH 2 OH 1 L × 1.00 L × 46.07 g CH 3 CH 2 OH 1 mol CH 3 CH 2 OH = 316.04 = 316 g CH 3 CH 2 OH

    Check .  (7 × 50) ≈ 350 g ethanol (this is an upper limit)

  2. 316.04 g CH 3 CH 2 OH × 1 mL 0.789 g CH 3 CH 2 OH = 400.56 = 401 mL = 0.401 L CH 3 CH 2 OH

    Check .  (320/0.8) ≈ 400 mL ethanol

4.68 M = mol L ; mol = g MM (MM is the symbol for molar mass in this manual .)

M = mol solute L solution ; 124 mg C 6 H 8 O 6 0 .2366 L × 1 g 1000 mg × 1 mol C 6 H 8 O 6 176 .12 g C 6 H 8 O 6 = 2.98 × 10 3 M C 6 H 8 O 6 6

4.69 Analyze .  Given: formula and concentration of each solute. Find: concentration of K + in each solution. Plan . Note mol K + /mol solute and compare concentrations or total moles.

Solve .

  1. KCl → K + + Cl ; 0.20 M KCl = 0.20 M K +

    K 2 CrO 4 2 K + + CrO 4 2– ; 0.15 M K 2 CrO 4 = 0.30 M K +

    K 3 PO 4 3 K + + PO 4 3– ; 0.080 M K 3 PO 4 = 0.24 M K +

    0.15 M K 2 CrO 4 has the highest K + concentration.

  2. K 2 CrO 4 : 0.30 M K + × 0.0300 L = 0.0090 mol K +

    K 3 PO 4 : 0.24 M K + × 0.0250 L = 0.0060 mol K +

    30.0 mL of 0.15 M K 2 CrO 4 has more K + ions.

4.70

  1. 0.10 M BaI 2 = 0.2 M I ; 0.25 M KI = 0.25 M I

    0.25 M KI has the higher I concentration.

  2. 0.10 M KI = 0.1 M I ; 0.040 M ZnI 2 = 0.080 M I ; 0.10 M KI has a higher I concentration than 0.040 M ZnI 2 . Total volume does not affect concentration.

  1. 3.2 M HI = 3.2 M I

    145 g NaI × 1 mol NaI 149.9 g NaI × 1 0.150 L = 6.45 M NaI = 6.45 M I

    The NaI solution has the higher I concentration.

4.71 Analyze .  Given: molecular formula and solution molarity. Find: concentration ( M ) of each ion.

Plan . Follow the logic in Sample Exercise 4.12.

Solve .

  1. NaNO 3 → Na + , NO 3 ; 0.25 M NaNO 3 = 0.25 M Na + , 0.25 M NO 3
  2. MgSO 4 → Mg 2+ , SO 4 2– ; 1.3 × 10 –2 M MgSO 4 = 1.3 × 10 –2 M Mg 2+ , 1.3 × 10 –2 M SO 4 2–
  3. C 6 H 12 O 6 → C 6 H 12 O 6 (molecular solute); 0.0150 M C 6 H 12 O 6 = 0.0150 M C 6 H 12 O 6
  4. Plan . There is no reaction between NaCl and (NH 4 ) 2 CO 3 , so this is just a dilution problem, M 1 V 1 = M 2 V 2 . Then account for ion stoichiometry.

    Solve . 45.0 mL + 65.0 mL = 110.0 mL total volume

    0.272 M NaCl × 45.0 mL 110.0 mL = 0 . 111 M NaCl ; 0 . 111 M Na + , 0 . 111 M Cl

    0.0247 M (NH 4 ) 2 CO 3 × 65.0 mL 110.0 mL = 0 . 0146 M ( NH 4 ) 2 CO 3

    2 × (0.0146 M ) = 0.0292 M NH 4 + , 0.0146 M CO 3 2–

    Check . By adding the two solutions (with no common ions or chemical reaction), we have approximately doubled the solution volume, and reduced the concentration of each ion by approximately a factor of two.

4.72

  1. Plan . These two solutions have common ions. Find the ion concentration resulting from each solution, then add.

    Solve . total volume = 42.0 mL + 37.6 mL = 79.6 mL

    0 . 1 7 0 M NaOH × 4 2 . 0 mL 7 9 . 6 mL = 0 . 08970 = 0 . 0897 M NaOH;

    0.0897 M Na + , 0.0897 M OH

    0.400 M NaOH × 37.6 mL 79.6 mL = 0 . 18894 = 0 . 189 M NaOH ;

    0.189 M Na + , 0.189 M OH

    M Na + = 0.08970 M + 0.18894 M = 0.27864 = 0.279 M Na +

    M OH = M Na + = 0.279 M OH


  1. Plan . No common ions; just dilution.

    Solve . 44.0 mL + 25.0 mL = 69.0 mL

    0.100 M Na 2 SO 4 × 44.0 mL 69.0 mL = 0 . 06377 = 0 . 0638 M Na 2 SO 4

    2 × (0.06377 M ) = 0.1275 = 0.128 M Na + ; 0.0638 M SO 4 2–

    0.150 M KCl × 25.0 mL 69.0 mL = 0 . 054348 = 0 . 0543 M KCl

    0.0543 M K + , 0.0543 M Cl

  1. Plan . Calculate concentration of K + and Cl due to the added solid. Then sum to get total concentration of Cl .

    Solve . 3 .60 g KCl 75 .0 mL soln × 1 mol KCl 74 .55 g KCl × 1000 mL 1 L = 0 . 6439 = 0 . 644 M KCl

    0.250 M CaCl 2 ; 2(0.250 M) = 0.500 M Cl

    total Cl = 0.644 M + 0.500 M = 1.144 M Cl , 0.644 M K + , 0.250 M Ca 2+

4.73  A nalyze/Plan .  Follow the logic of Sample Exercise 4.14.

Solve .

  1. V 1 = M 2 V 2 / M 1 ; 0.250 M NH 3 × 1000.0 mL 14.8 M NH 3 = 16.89 = 16.9 mL 14.8 M NH 3

    Check .  250/15 ≈ 15 M

  2. M 2 = M 1 V 1 /V 2 ; 14.8 M NH 3 × 10.0 mL 500 mL = 0.296 M NH 3

    Check .  150/500 ≈ 0.30 M

4.74

  1. V 1 = M 2 V 2 / M 1 ; 0.500 M HNO 3 × 0.110 L 6.0 M HNO 3 = 0.00917 L = 9.2 mL 6.0 M HNO 3
  2. M 2 = M 1 V 1 / V 2 ; 6.0 M HNO 3 × 10.0 mL 250 mL = 0.240 M HNO 3

4.75 Analyze/Plan . Calculate the number of drug molecules in 1.00 mL of the stock solution, using M × L = moles and Avogadro’s number. Then calculate the desired ratio.

Solve . 1.00 mL = 0.00100 L

1.5 × 10 9 mol L × 0.0010 L × 6.022 × 10 23 molecules mole = 9.033 × 10 11 = 9.0 × 10 11 molecules

9.033 × 10 11 drug molecules 2.0 × 10 5 cancer cells = 4.517 × 10 6 = 4.5 × 10 6


4.76 Analyze/Plan . The 25.00 mL of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 –9 M . Then,

L stock solution = mol drug/ M solution; L = mol/5.00 × 10 –9 M .

1.0 × 10 8 molecules × 1 mol 6.022 × 10 23 molecules × 1 L 5.00 × 10 9 mol × 1000 mL 1 L = 3.3 × 10 5 mL

A volume of 3.3 × 10 –5 mL corresponds to 33 × 10 –9 L, or 33 nL (nanoliters).

Two points are of note. First, desired results can be achieved with a very small amount of the drug, which reduces the cost. And, delivering such a small volume of stock solution may be a challenge. A dilution scheme for the stock solution could be employed. If 1.00 mL of the 5.00 × 10 –9 M antibiotic stock solution is diluted to 1.00 L, and 1.00 mL of this solution is diluted to 0.500 L, the resulting concentration of the diluted stock solution is then 1.00 × 10 –14 M . Using the relationship L = mol/ M , we find that 0.017 L or 17 mL of the 1.00 × 10 –14 M stock solution diluted to 25.00 mL would kill the desired amount of bacteria.

4.77 Analyze .  Given: density of pure acetic acid, volume pure acetic acid, volume new solution. Find: molarity of new solution. Plan . Calculate the mass of acetic acid, CH 3 COOH, present in 20.0 mL of the pure liquid. Solve .

20.00 mL acetic acid × 1 .049 g acetic acid 1 mL acetic acid = 20.98 g acetic acid

20.98 g CH 3 COOH × 1 mol CH 3 COOH 60.05 g CH 3 COOH = 0.349375 = 0.3494 mol CH 3 COOH

M = mol/L = 0 .349375 mol CH 3 COOH 0.2500 L solution = 1.39750 = 1.398 M CH 3 COOH

Check .  (20 × 1) ≈ 20 g acid; (20/60) ≈ 0.33 mol acid; (0.33/0.25 = 0.33 × 4) ≈ 1.33 M

4.78 50.000 mL glycerol × 1 .2656 g glycerol 1 mL glycerol = 63.280 g glycerol

63.280 g C 3 H 8 O 3 × 1 mol C 3 H 8 O 3 92.094 g C 3 H 8 O 3 = 0.687124 = 0.68712 mol C 3 H 8 O 3

M = 0.687124 mol C 3 H 8 O 3 0.25000 L solution = 2.7485 M C 3 H 8 O 3

Solution Stoichiometry and Chemcial Analysis (Section 4.6)

4.79 Analyze .  Given: volume and molarity AgNO 3 , molarity HCl. Find: volume HCl or mass of KCl.

  1. Plan . M × L = mol AgNO 3 = mol Ag + ; balanced equation gives ratio mol HCl/mol AgNO 3 ; mol HCl → vol HCl. Solve .

    0.200 mol AgNO 3 1 L × 0.0150 L = 3 .00 × 10 3 mol AgNO 3 ( aq )


  1. AgNO 3 (aq) + HCl(aq) → AgCl(s) + HNO 3 (aq)

    mol HCl = mol AgNO 3 = 3.00 × 10 –3 mol KCl

    3.00 × 10 3 mol HCl × 1 L 0.150 mol HCl = 0.0200 L = 20 .0 mL 0 .15 M HCl

    Check .  (0.2 × 0.015) = 0.003 mol; (0.003/0.15) ≈ 0.02 L HCl

  1. Plan . M × L = mol AgNO 3 = mol Ag + ; balanced equation gives ratio mol KCl/mol AgNO 3 ; mol KCl → vol KCl. Solve .

    0.200 mol AgNO 3 1 L × 0.0150 L = 3 .00 × 10 3 mol AgNO 3 ( aq )

    AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq)

    mol KCl = mol AgNO 3 = 3.00 × 10 –3 mol KCl

    3.00 × 10 3 mol KCl × 74 .55 g KCl 1 mol KCl = 0.224 g KCl

    Check .  (0.2 × 0.015) = 0.003 mol; (0.003 × 75) ≈ 0.225 g KCl

  1. Clearly, the KCl reagent is virtually free relative to the HCl solution. The KCl analysis is more cost-effective.

4.80 Plan . M × L = mol Cd(NO 3 ) 2 ; balanced equation → mol ratio → mol NaOH → g NaOH

Solve . 0.500 mol Cd(NO 3 ) 2 1 L × 0.0350 L = 0.0175 mol Cd(NO 3 ) 2

Cd(NO 3 ) 2 (aq) + 2 NaOH(aq) → Cd(OH) 2 (s) + 2 NaNO 3 (aq)

0.0175 mol Cd ( NO 3 ) 2 × 2 mol NaOH 1 mol Cd(NO 3 ) 2 × 40.00 g NaOH 1 mol NaOH = 1.40 g NaOH

4.81

  1. Analyze .  Given: M and vol base, M acid. Find: vol acid

    Plan/Solve .  Write the balanced equation for the reaction in question:

    HClO 4 (aq) + NaOH(aq) → NaClO 4 (aq) + H 2 O(l)

    Calculate the moles of the known substance, in this case NaOH.

    moles NaOH = M × L = 0.0875 mol NaOH 1 L × 0.0500 L = 0.004375 = 0.00438 mol NaOH

    Apply the mole ratio (mol unknown/mol known) from the chemical equation.

    0.004375 mol NaOH × 1 mol HClO 4 1 mol NaOH = 0.004375 mol HClO 4

    Calculate the desired quantity of unknown, in this case the volume of 0.115 M HClO 4 solution.

    L = mol/ M ; L = 0.004375 mol HClO 4 × 1 L 0.115 mol HClO 4 = 0.0380 L = 38.0 mL

    Check .  (0.09 × 0.05) = 0.0045 mol; (0.0045/0.11) ≈ 0.040 L ≈ 40 mL


  1. Following the logic outlined in part (a):

    2 HCl(aq) + Mg(OH) 2 (s) → MgCl 2 (aq) + 2 H 2 O(l)

    2.87 g Mg(OH) 2 × 1 mol Mg(OH) 2 58.32 g Mg(OH) 2 = 0.049211 = 0.0492 mol Mg(OH) 2

    0.0492 mol Mg(OH) 2 × 2 mol HCl 1 mol Mg(OH) 2 = 0.0984 mol HCl

    L = mol/ M = 0.09840 mol HCl × 1 L HCl 0 .128 mol HCl = 0.769 L = 769 mL

  1. AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq)

    785 mg KCl × 1 × 10 3 g 1 mg × 1 mol KCl 74.55 g KCl × 1 mol AgNO 3 1 mol KCl = 0.01053 = 0.0105 mol AgNO 3

    M = mol/L = 0 .01053 mol AgNO 3 0.0258 L = 0.408 M AgNO 3

  1. HCl(aq) + KOH(aq) → KCl(aq) + H 2 O(l)

    0.108 mol HCl 1 L × 0.0453 L × 1 mol KOH 1 mol HCl × 56.11 g KOH 1 mol KOH = 0.275 g KOH

4.82

  1. 2 HCl(aq) + Ba(OH) 2 (aq) → BaCl 2 (aq) + 2 H 2 O(l)

    0.101 mol Ba(OH) 2 1 L Ba(OH) 2 × 0.0500 L Ba(OH) 2 × 2 mol HCl 1 mol Ba(OH) 2 × 1 L HCl 0 .120 mol HCl = 0.0842 L or 84 .2 mL HCl soln

  2. H 2 SO 4 (aq) + 2 NaOH(aq) → Na 2 SO 4 (aq) + 2 H 2 O(l)

    0.200 g NaOH × 1 mol NaOH 40.00 g NaOH × 1 mol H 2 SO 4 2 mol NaOH × 1 L H 2 SO 4 0.125 mol H 2 SO 4 = 0.0200 L or 20 .0 mL H 2 SO 4 soln

  3. BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq)

    752 mg = 0 .752 g Na 2 SO 4 × 1 mol Na 2 SO 4 142.1 g Na 2 SO 4 × 1 mol BaCl 2 1 mol Na 2 SO 4 × 1 0.0558 L = 0.0948 M BaCl 2

  4. 2 HCl(aq) + Ca(OH) 2 (aq) → CaCl 2 (aq) + 2 H 2 O(l)

    0.0427 L HCl × 0 .208 mol HCl 1 L HCl × 1 mol Ca(OH) 2 2 mol HCl × 74.10 g Ca(OH) 2 1 mol Ca(OH) 2 = 0.329 g Ca(OH) 2

4.83 Analyze/Plan .  See Exercise 4.81(a) for a more detailed approach. Solve .

6.0 mol H 2 SO 4 1 L × 0.027 L × 2 mol NaHCO 3 1 mol H 2 SO 4 × 84.01 g NaHCO 3 1 mol NaHCO 3 = 27 g NaHCO 3


4.84   See Exercise 4.81(a) for a more detailed approach.

0.115 mol NaOH 1 L × 0.0425 L × 1 mol CH 3 COOH 1 mol NaOH × 60.05 g CH 3 COOH 1 mol CH 3 COOH = 0.29349 = 0.293 g CH 3 COOH in 3 .45 mL

1.00 qt vinegar × 1 L 1 .057 qt × 1000 mL 1 L × 0.29349 g CH 3 COOH 3.45 mL vinegar = 80.5 g CH 3 COOH/qt

4.85 Analyze .  Given: M and vol HCl. Find: MM of base, an alkali metal hydroxide.

Plan .  Alkali metal ions have a 1+ charge, so the general formula of an alkali metal hydroxide is MOH. One mol of MOH requires one mol of HCl for neutralization.

  1. M × L = mol HCl = mol MOH. MM = g MOH 1 mol MOH . Solve.

    2. 50 mol HCl 1 L × 0.0170 L = 0.0425 mol HCl = 0 .0425 mol MOH

    MM of MOH = 4.36 g MOH 0.0425 mol MOH = 102.59 = 103 g/mol

  2. MM of alkali metal = MM of MOH – (17.01 g) Solve.

    MM of alkali metal = (102.59 g/mol – 17.01 g/mol) = 85.58 = 86 g/mol

    The experimental molar mass most closely fits that of Rb + , 85.47 g/mol

    Check .  The experimental molar mass matches one of the alkali metals.

4.86 Analyze/Plan. Follow the logic in Exercise 4.85. The unknown is a group 2A metal hydroxide, general formula M(OH) 2 . Two mol HCL are required to neutralize 1 mol M(OH) 2 . Solve.

  1. 2. 50 mol HCl 1 L × 0.0569 L × 1 mol M(OH) 2 2 mol HCl = 0.071125 = 0 .0711 mol M(OH) 2

    MM of M ( OH ) 2 = 8 . 6 5 g M ( OH ) 2 0 . 0 7 1 1 2 5 mol M ( OH ) 2 = 1 2 1 . 6 2 = 1 2 2 g/mol

  2. MM of group 2A metal = MM of M(OH) 2 – 2(17.01 g)

    MM of group 2A metal = (121.62 g/mol – 34.02 g/mol) = 87.60 = 88 g/mol

    The experimental molar mass most closely fits that of Sr 2+ , 87.62 g/mol

    Check .  The experimental molar mass matches one of the group 2A metals.

4.87

  1. NiSO 4 (aq) + 2 KOH(aq) → Ni(OH) 2 (s) + K 2 SO 4 (aq)
  2. The precipitate is Ni(OH) 2 .
  3. Plan .  Compare mol of each reactant; mol = M × L

    Solve .  0.200 M KOH × 0.1000 L KOH = 0.0200 mol KOH

    0.150 M NiSO 4 × 0.2000 L NiSO 4 = 0.0300 mol NiSO 4

    1 mol NiSO 4 requires 2 mol KOH, so 0.0300 mol NiSO 4 requires 0.0600 mol KOH. Because only 0.0200 mol KOH is available, KOH is the limiting reactant.


  1. Plan .  The amount of the limiting reactant (KOH) determines amount of product, in this case Ni(OH) 2 .

    Solve . 0.0200 mol KOH × 1 mol Ni(OH) 2 2 mol KOH × 92.71 g Ni(OH) 2 1 mol Ni(OH) 2 = 0.927 g Ni(OH) 2

  1. Plan/Solve .  Limiting reactant: OH : no excess OH remains in solution.

    Excess reactant: Ni 2+ : M Ni 2+ remaining = mol Ni 2+ remaining/L solution

    0.0300 mol Ni 2+ initial – 0.0100 mol Ni 2+ reacted = 0.0200 mol Ni 2+ remaining

    0.0200 mol Ni 2+ /0.3000 L = 0.0667 M Ni 2+ (aq)

    Spectators: SO 4 2– , K + . These ions do not react, so the only change in their concentration is dilution. The final volume of the solution is 0.3000 L.

    M 2 = M 1 V 1 /V 2 : 0.200 M K + × 0.1000 L/0.3000 L = 0.0667 M K + (aq)

    0.150 M SO 4 2– × 0.2000 L/0.3000 L = 0.100 M SO 4 2– (aq)

4.88

  1. 2 HNO 3 (aq) + Sr(OH) 2 (s) → Sr(NO 3 ) 2 (aq) + 2 H 2 O(l)
  2. Determine the limiting reactant, then the identity and concentration of ions remaining in solution. Assume that the H 2 O(l) produced by the reaction does not increase the total solution volume.

    15.0 g Sr(OH) 2 × 1 mol Sr(OH) 2 1 21 .64 g Sr(OH) 2 = 0.1233 = 0.123 mol Sr(OH) 2

    mol OH = 2(0.1233) mol Sr(OH) 2 = 0.2466 = 0.247 mol OH

    0.200 M HNO 3 × 0.0550 L HNO 3 = 0.0110 mol HNO 3 .

    Two mol HNO 3 react with one mol Sr(OH) 2 , so HNO 3 is the limiting reactant. No excess H + remains in solution. The remaining ions are OH (excess reactant), Sr 2+ , and NO 3 (spectators).

    OH : 0.2466 mol OH initial – 0.0110 mol OH react =

    0.2356 = 0.236 mol OH remain

    0.2356 mol OH /0.0550 L soln = 4.28 M OH (aq)

    Sr 2+ :  0.123 mol Sr 2+ /0.0550 L soln = 2.24 M Sr 2+ (aq)

    NO 3 : 0.0110 mol NO 3 /0.0550 L = 0.200 M NO 3 (aq)

  3. The resulting solution is basic because of the large excess of OH (aq).

4.89 Analyze .  Given: mass impure Mg(OH) 2 ; M and vol excess HCl; M and vol NaOH.

Find: mass % Mg(OH) 2 in sample. Plan/Solve . Write balanced equations.

Mg(OH) 2 (s) + 2 HCl(aq) → MgCl 2 (aq) + 2 H 2 O(l)

HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

Calculate total moles HCl = M HCl × L HCl

0.2050 mol HCl 1 L soln × 0.1000 L = 0 .02050 mol HCl total


mol excess HCl = mol NaOH used = M NaOH × L NaOH

0.1020 mol NaOH 1 L soln × 0.01985 L = 0 .0020247 = 0 .002025 mol NaOH

mol HCl reacted with Mg(OH) 2 = total mol HCl – excess mol HCl

0.02050 mol total – 0.0020247 mol excess = 0.0184753 = 0.01848 mol HCl reacted

(The result has 5 decimal places and 4 sig. figs.)

Use mol ratio to get mol Mg(OH) 2 in sample, then molar mass of Mg(OH) 2 to get g pure Mg(OH) 2 .

0.0184753 mol HCl × 1 mol Mg(OH) 2 2 mol HCl × 58.32 g Mg(OH) 2 1 mol Mg(OH) 2 = 0.53874 = 0.5387 g Mg(OH) 2

mass % Mg(OH) 2 = g Mg(OH) 2 g sample × 100 = 0.53874 g Mg(OH) 2 0.5895 g sample × 100 = 91.39 % Mg(OH) 2

4.90 Plan .  CaCO 3 (s) + 2 HCl(aq) → CaCl 2 (aq) + H 2 O(l) + CO 2 (g)

HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

total mol HCl – excess mol HCl = mol HCl reacted; mol CaCO 3 = (mol HCl)/2;

g CaCO 3 = mol CaCO 3 × molar mass; mass % = (g CaCO 3 /g sample) × 100

Solve :

1.035 mol HCl 1 L soln × 0.03000 L = 0 .031050 = 0 .03105 mol HCl total

1 . 0 1 0 m o l N a O H 1 L s o l n × 0 . 0 1 1 5 6 L = 0 . 0 1 1 6 7 6 = 0 . 0 1 1 6 8 m o l H C l e x c e s s

0.031050 total – 0.011676 excess = 0.019374 = 0.01937 mol HCl reacted

0.019374 mol HCl × 1 mol CaCO 3 2 mol HCl × 100.09 g CaCO 3 1 mol CaCO 3 = 0.96959 = 0.9696 g CaCO 3

mass % CaCO 3 = g CaCO 3 g rock × 100 = 0 .96959 1 .248 × 100 = 77.69 % CaCO 3

Additional Exercises

4.91

  1. U(s) + 2 ClF 3 (g) → UF 6 (g) + Cl 2 (g)
  2. This is not a metathesis reaction. The compounds involved are molecular rather than ionic, so the reaction is more complex than ions changing “partners.”
  3. It is a redox reaction. U is oxidized, Cl is reduced.

4.92

  1. The precipitate is CdS(s).
  2. Na + (aq) and NO 3 (aq) are spectator ions and remain in solution. Any excess reactant ions also remain in solution.
  3. Cd 2+ (aq) + S 2– (aq) → CdS(s).
  4. This is not a redox reaction. (It is a metathesis reaction.)

4.93  The two precipitates formed are AgCl(s) and SrSO 4 (s). Because no precipitate forms on addition of hydroxide ion to the remaining solution, the other two possibilities, Ni 2+ and Mn 2+ , are absent.

4.94  (a,b) Expt. 1 No reaction

Expt. 2 2 Ag + (aq) + CrO 4 2– (aq) → Ag 2 CrO 4 (s) red precipitate

Expt. 3 Ca 2+ (aq) + CrO 4 2– (aq) → CaCrO 4 (s) yellow precipitate

Expt. 4 2 Ag + (aq) + C 2 O 4 2– (aq) → Ag 2 C 2 O 4 (s) white precipitate

Expt. 5 Ca 2+ (aq) + C 2 O 4 2– (aq) → CaC 2 O 4 (s) white precipitate

Expt. 6 Ag + (aq) + Cl (aq) → AgCl(s) white precipitate

4.95

  1. Al(OH) 3 (s) + 3 H + (aq) → Al 3+ (aq) + 3 H 2 O(l)
  2. Mg(OH) 2 (s) + 2 H + (aq) → Mg 2+ (aq) + 2 H 2 O(l)
  3. MgCO 3 (s) + 2 H + (aq) → Mg 2+ (aq) + H 2 O(l) + CO 2 (g)
  4. NaAl(CO 3 )(OH) 2 (s) + 4 H + (aq) → Na + (aq) + Al 3+ (aq) + 3 H 2 O(l) + CO 2 (g)
  5. CaCO 3 (s) + 2 H + (aq) → Ca 2+ (aq) + H 2 O(l) + CO 2 (g)

[In (c), (d), and (e), one could also write the equation for formation of bicarbonate, e.g., MgCO 3 (s) + H + (aq) → Mg 2+ + HCO 3– (aq).]

4.96 4 NH 3 ( g ) + 5 O 2 ( g ) 4 NO(g ) + 6 H 2 O(g ) .

N = 7–3 O = 0 N = +2 O = –2

  1. redox reaction
  2. N is oxidized, O is reduced

2 NO ( g ) + O 2 ( g ) 2 NO 2 ( g ) .

N = +2 O = 0 N = +4, O = –2

  1. redox reaction
  2. N is oxidized, O is reduced

3 NO 2 ( g ) + H 2 O ( l ) 2 HNO 3 ( aq ) + NO ( g ) .

N = +4 N = +5 N = +2

  1. redox reaction
  2. N is oxidized (NO 2 → HNO 3 ), N is reduced (NO 2 → NO). A reaction where the same element is both oxidized and reduced is called disproportionation.
  3. 0.150 M HNO 3 × 1000.0 L = 150.00 = 150 mol HNO 3 required

    150 mol HNO 3 × 3 mol NO 2 2 mol HNO 3 × 2 mol NO 2 mol NO 2 × 4 mol NH 3 4 mol NO = 225 mol NH 3

    225 mol HNO 3 × 17.0305 g NH 3 1 mol NH 3 = 3831.86 = 3.83 × 10 3 g NH 3


4.97  A metal on Table 4.5 can be oxidized by ions of the elements below it. Or, a metal on the table is able to displace the (metal) cations below it from their compounds.

  1. Zn(s) + 2 HNO 3 (aq) → Zn(NO 3 ) 2 (aq) + H 2 (g)

    The substance that inflates the balloon is H 2 (g). Of Zn, Cu, and Hg, only Zn is above H on Table 4.5, so only Zn can displace H from HCl.

  2. 35.0 g Zn × 1 mol Zn 6 5 .39 g Zn = 0.53525 = 0.535 mol Zn

    3 .00 mol HNO 3 L × 0 .150 L = 0.450 mol HNO 3

    One mol Zn reacts with 2 mol HNO 3 , so HNO 3 is the limiting reactant.

    0.450 mol HNO 3 × 1 mol H 2 2 mol HNO 3 × 2. 016 g H 2 1 mol H 2 = 0.4536 = 0.454 g H 2

  3. Both Zn and Cu are above Ag on Table 4.5, so both Zn and Cu can displace Ag from AgNO 3 . Note that H 2 (g) would also displace Ag, but H + (aq) will not.

    Zn(s) + 2 AgNO 3 (aq) → 2 Ag(s) + Zn(NO 3 ) 2 (aq); Zn 2+ (aq) and NO 3 (aq) remain

    Cu(s) + 2 AgNO 3 (aq) → 2 Ag(s) + Cu(NO 3 ) 2 (aq); Cu 2+ (aq) and NO 3 (aq) remain

  4. 0.535 mol Zn [from part (b)]; 42.0 g Zn × 1 mol Zn 6 3 .546 g Cu = 0.6609 = 0.661 mol Cu

    0 .750 mol AgNO 3 L × 0 .150 L = 0.1125 = 0.113 mol AgNO 3

    One mol metal reacts with 2 mol AgNO 3 , so AgNO 3 is the limiting reactant for both Zn and Cu.

    0.1125 mol AgNO 3 × 1 07 .87 g Ag 1 mol Ag = 12.135 = 12.1 g Ag in both reactions

4.98

  1. Copper is the solvent and tin is the solute. In a solution, the solvent is present in greater amount and the solute is present in lesser amount. There can be more than one solute, but only one solvent.
  2. The molarity of Sn in the bronze is mol Sn/L bronze. If the bronze contains 10.0% Sn, 100.0 g of bronze contains 10.0 g Sn. Use AW of Sn to calculate mol Sn. Use the density of the bronze, 7.9 g/cm 3 , to calculate the volume of 100.0 g of this bronze.

    10.0 g Sn × 1 mol Sn 118 .71 g Sn = 0.08424 = 0 .0842 mol Sn

    100.0 g bronze × 1 cm 3 7 .9 g = 12.658 = 13 cm 3 = 13 mL

    M = mol Sn L bronze = 0.08424 mol Sn 12 .658 cm 3 bronze × 1000 cm 3 1 L = 6.655 = 6.7 M


  1. Sn(s) + 2 HCl(aq) → SnCl 2 (aq) + H 2 (g)

    According to the Activity Series of Metals, Table 4.5, Sn is above H and Cu is below it. That is, Sn can be oxidized to Sn 2+ (aq) by HCl(aq), but Cu(s) cannot. We can treat bronze with HCl(aq) to remove Sn as Sn 2+ (aq) and leave a pure Cu sample.

    [A closer look shows that this method has its problems. In fact, HCl causes some corrosion of the Cu, regardless of its position in the activity series, so we would not be left with a pure Cu sample. And, it would probably remove tin only on the surface of the bronze. A better method, which relies on principles presented in later chapters of the text, follows. Dissolve all metal with nitric acid, precipitate the tin, isolate the remaining solution and electroplate the copper from it.]

4.99 Plan .  Calculate moles KBr from the two quantities of solution (mol = M × L). Moles AgNO 3 required equals total moles KBr present. Change grams AgNO 3 to moles AgNO 3 . Solve .

1.00 M KBr × 0.0350 L = 0.0350 mol KBr; 0.600 M KBr × 0.060 L = 0.0360 mol KBr

0.0350 mol KBr + 0.0360 mol KBr = 0.0710 mol KBr total requires 0.0710 mol AgNO 3

0.0710 mol AgNO 3 × 169.87 g AgNO 3 mol AgNO 3 = 12.0607 = 12.1 g AgNO 3

4.100

  1. Acid-base. Because of the –NH 2 group, dopamine is an organic base similar to NH 3 . (The –OH groups are not basic because they are not hydroxide ions. In fact, when they are bonded to carbon atoms in this kind of six-membered ring, they are slightly acidic.)
  2. Determine the molecular formula of dopamine and calculate molar mass. Then, moles dopamine = g/molar mass; M = mol/L.

    C 8 H 11 NO 2 , molar mass = 153.18; 400.0 mg dopamine = 0.4000 g dopamine;

    250.0 mL solution = 0.2500 L solution

    0.4000 g dopamine × 1 mol 153.18 g × 1 0.2500 L = 0.01045 M dopamine

  3. Find the number of dopamine molecules in a 5.00 mm 3 brain that has an increased dopamine concentration of 0.75 μM . M = mol/L; 1 L = 1 dm 3

    0.75 μM dopamine solution = 0.75 × 10 –6 M dopamine solution

    5.00 mm 3 × 1 L 1 dm 3 × 1 dm 3 ( 100 ) 3 mm 3 = 5.00 × 10 6 L

    0.75 × 10 6 mol dopamine 1 L soln × 5.00 × 10 6 L × 6.022 × 10 23 molecules mol = 2.258 × 10 12 = 2.3 × 10 12 dopamine molecules


4.101

  1. Na + must replace the total positive (+) charge due to Ca 2+ and Mg 2+ . Think of this as moles of charge rather than moles of particles.

    0.020 mol Ca 2 + 1 L water × 1.5 × 10 3 L × 2 mol ( + ) charge 1 mol Ca 2 + = 60 mol of ( + ) charge

    0.0040 mol Mg 2 + 1 L water × 1.5 × 10 3 L × 2 mol ( + ) charge 1 mol Mg 2 + = 12 mol of ( + ) charge

    72 moles of (+) charge must be replaced; 72 mol Na + are needed.

  2. 7 2 mol Na + × 1 mol Na + 1 mol NaCl × 5 8 .44 g NaCl 1 mol NaCl = 4208 g = 4.2 × 10 3 g NaCl

4.102  H 2 C 4 H 4 O 6 + 2 OH (aq) → C 4 H 4 O 6 2– (aq) + 2 H 2 O(l)

0.02465 L NaOH soln × 0 .2500 mol NaOH 1 L × 1 mol H 2 C 4 H 4 O 6 2 mol NaOH × 1 0.0500 L H 2 C 4 H 4 O 6 = 0 . 06163 M H 2 C 4 H 4 O 6 soln

4.103

  1. 1 2 .50 g Sr(OH) 2 × 1 mol Sr(OH) 2 1 21 .64 g Sr(OH) 2 × 1 0. 05000 L = 2.0552 = 2.055 M Sr(OH) 2
  2. 2 HNO 3 (aq) + Sr(OH) 2 (aq) → Sr(NO 3 ) 2 (aq) + 2 H 2 O(l)
  3. Plan .  mol Sr(OH) 2 = M × L → mol ratio → mol HNO 3 M HNO 3 . Solve .

    2.0552 mol Sr(OH) 2 L × 0.0239 L Sr(OH) 2 × 2 mol HNO 3 1 mol Sr(OH) 2 × 1 0.0375 L HNO 3 = 2 . 6197 = 2 . 62 M HNO 3

4.104  mol OH from NaOH(aq) + mol OH from Zn(OH) 2 (s) = mol H + from HBr

mol H + = M HBr × L HBr = 0.500 M HBr × 0.350 L HBr = 0.175 mol H +

mol OH from NaOH = M NaOH × L NaOH = 0.500 M NaOH × 0.0885 L NaOH = 0.04425 = 0.0443 mol OH

mol OH from Zn(OH) 2 (s) = 0.175 mol H + – 0.04425 mol OH from NaOH = 0.13075 = 0.131 mol OH from Zn(OH) 2

0.13075 mol OH × 1 mol Zn(OH) 2 2 mol OH × 99.41 g Zn(OH) 2 1 mol Zn(OH) 2 = 6.50 g Zn(OH) 2

Integrative Exercises

4.105

  1. A metal can be oxidized by ions of the elements below it on Table 4.5. Of the three substances given, K + (aq) is above Mg(s), but Ag + (aq) is below it, so AgNO 3 (aq) will react with Mg(s).
  2. Mg(s) + 2 Ag + (aq) → Mg 2+ (aq) + 2 Ag(s)

  1. g Mg → mol Mg → [via mole ratio] mol Ag + → via (mol/M)] vol AgNO 3 (aq)

    5.00 g Mg × 1 mol Mg 2 4.305 g Mg = 0.2057 = 0.206 mol Mg

    0.2057 mol Mg × 2 mol AgNO 3 1 mol Mg × 1.00 L 2. 00 mol AgNO 3 = 0.2057 = 0.206 L AgNO 3 ( aq )

  1. 0 .2057 mol Mg 2 + 0.2057 L so ln = 1.00 M Mg 2 + ( aq )

4.106

  1. At the equivalence point of a titration, mol NaOH added = mol H + present

    M NaOH × L NaOH = g acid MM acid ( for an acid with 1 acidic hydrogen)

    MM acid = g acid M NaOH × L NaOH = 0.2053 g 0 .1008 M × 0.0150 L = 136 g/mol

  2. Assume 100 g of acid.

    70.6 g C × 1 mol C 12 .01 g C = 5.88 mol C; 5 .88/1 .47 4

    5.89 g H × 1 mol H 1 .008 g H = 5.84 mol H; 5 .84/1 .47 4

    23.5 g O × 1 mol O 16 .00 g O = 1.47 mol O; 1 .47/1 .47 = 1

    The empirical formula is C 4 H 4 O.

    MM FW = 136 68.1 = 2 ; the molecular formula is 2 × the empirical formula .

    The molecular formula is C 8 H 8 O 2 .

4.107

  1. Gold atoms from Au(s) are oxidized. Oxygen atoms from O 2 (g) are reduced. The oxidation number of Au changes from 0 to +1 {in Na[Au(CN) 2 ]}. The oxidation number of O changes from 0 to –2 (in NaOH).
  2. 2 [Au(CN) 2 ] (aq) + Zn(s) → 2 Au(s) + Zn 2+ (aq) + 4 CN–(aq)

    The oxidation of one Zn atom produces 2 electrons, so 2 Au atoms {in the form of [Au(CN) 2 ] (aq)} are reduced.

  3. Use mass % Au in rocks to get mass Au. Calculate moles Au, apply mole ratio of NaCN(aq) to Au(s) from the balanced equation given in the exercise, calculate volume of 0.200 M NaCN(aq) required. 2.00% Au = 2.00 kg Au/100 kg rocks

    40 .0 kg rocks × 2.00 kg Au 100 kg rocks × 1000 g kg = 8.00 × 10 2 g Au

    8 .00 × 10 2 g Au × 1 mol Au 196.97 g Au × 8 mol NaCN 4 mol Au × 1 L soln 0.200 mol NaCN = 40.615 = 40 .6 L


4.108 Plan .  Abbreviate the commercial aqueous ammonia solution as NH 3 soln. Abbreviate citric acid as H 3 Cit. Write the balanced equation.

vol NH 3 soln density mass NH 3 soln mass % mass NH 3

mass NH 3 mol NH 3 mol H 3 Cit mass H 3 Cit

Solve . H 3 Cit(aq) + 3 NH 3 ( aq) 3 NH 4 + ( aq ) + Cit 3 (aq)

3.43 × 10 4 gal NH 3 soln × 3.785 L gal × 1000 mL L × 0.88 g soln mL soln = 1.14246 × 10 8 = 1.1 × 10 8 g NH 3 soln

1.14246 × 10 8 g NH 3 soln × 30 g NH 3 100 g NH 3 soln = 3.4274 × 10 7 = 3.4 × 10 7 g NH 3

3.4274 × 10 7 g NH 3 × 1 mol NH 3 17.0305 g NH 3 × 1 mol H 3 Cit 3 mol NH 3 × 192.124 g H 3 Cit 1 mol H 3 Cit = 1 .3 × 10 8 g H 3 Cit

4.109

  1. Mg(OH) 2 (s) + 2 HNO 3 (aq) → Mg(NO 3 ) 2 (aq) + 2 H 2 O(l)
  2. 7.75 g Mg(OH) 2 × 1 mol Mg(OH) 2 58.32 g Mg(OH) 2 = 0.13289 = 0.133 mol Mg(OH) 2

    0.200 M HNO 3 × 0.0250 L = 0.00500 mol HNO 3

    The 0.00500 mol HNO 3 would neutralize 0.00250 mol Mg(OH) 2 and much more Mg(OH) 2 is present, so HNO 3 is the limiting reactant.

  3. Because HNO 3 limits, 0 mol HNO 3 is present after reaction.

    0.00250 mol Mg(NO 3 ) 2 is produced.

    0.13289 mol Mg(OH) 2 initial – 0.00250 mol Mg(OH) 2 react = 0.130 mol Mg(OH) 2 remain

4.110

  1. Calculate the volume of the first part of the river. Get mass of MCHM using density. Then, mass MCHM → mol MCHM → M MCHM

    7.00 ft × 1 yd 3 ft × 100 .0 yd × 100 .0 yd × ( 1 m) 3 ( 1.0936 ) 3 yd 3 × ( 10) 3 dm 3 1 m 3 = 1 .7840 × 10 7 dm 3 = 1 .78 × 10 7 L

    7500 gal MCHM × 3.785 L gal × 1 000 mL L × 0.9074 g MCHM mL = 2 .576 × 10 7 = 2 .6 × 10 7 g MCHM

    2 .576 × 10 7 g MCHM × 1 mol MCHM 128 .21 g MCHM × 1 1 .784 × 10 7 L = 0.01126 = 0.011 M MCHM

  2. A concentration of 1.0 × 10 –4 M is a decrease in concentration of approximately 100 times. It requires a total volume that is 100 times the volume of the first part of the river. If depth and width are constant, the length of the spill must increase by a factor of approximately 100. That is, the spill would need to cover 100(100 yd) = 10,000 yd of the river. For perspective, this is about 5.7 miles, a substantial distance.

  1. We can calculate a more precise distance. Use the dilution formula to calculate the volume of river required to produce a 1.0 × 10 –4 M solution, assuming uniform mixing and MCMH concentration throughout the length of the spill.

    0.01126 M MCHM × 1.784 × 10 7 L = 1.0 × 10 –4 M × ?L

    volume of river = 2.009 × 10 9 L = 2.0 × 10 9 L

    2.009 × 10 9 dm 3 × 1 m 3 ( 10 ) 3 dm 3 × ( 1.0936 ) 3 yd 3 1 m 3 × 1 100 yd × 1 7 ft × 3 ft 1 yd = 1.126 × 10 4 = 1.1 × 10 4 yd

    The total length of the spill would be 1.1 × 10 4 yd. The initial length of 100 yards (or 0.01 × 10 4 yd) is barely significant. The spill would need to spread 1.1 × 10 4 yards or 6.4 miles farther to achieve a “safe” concentration.

4.111

  1. By virtue of its –NH group, Ritalin is a weak base like NH 3 . Because it is a weak base, it is a weak electrolyte.
  2. The molecular formula of Ritalin is C 14 H 19 NO 2 . Molar mass = 233.31 g/mol

    10.0 mg ritalin × 1 × 10 3 g mg × 1 mol ritalin 233.31 g ritalin × 1 L 5.0 L = 8.572 × 10 6 = 8.6 × 10 6 M

  3. Six hours is two half-lives for Ritalin. One half-life reduces molarity by ½ and two half-lives reduce molarity by ½(½) or ¼ . The concentration of Ritalin after six hours will be (8.572 × 10 –6 M )/4 = 2.143 × 10 –6 M = 2.1 × 10 –6 M .

4.112  Ag + (aq) + Cl (aq) → AgCl(s)

0.2997 mol Ag + 1 L × 0.04258 L × 1 mol Cl 1 mol Ag + × 35.453 g Cl 1 mol Cl = 0.45242 = 0.4524 g Cl

25.00 mL seawater × 1 .025 g mL = 25.625 = 25.63 g seawater

mass % Cl = 0 .45242 g Cl 25.625 g seawater × 100 = 1.766 % Cl

4.113

  1. As O 4 3– ; +5
  2. Ag 3 P O 4 is silver phosphate; Ag 3 As O 4 is silver arsenate
  3. 0.0250 L soln × 0 .102 mol Ag + 1 L soln × 1 mol Ag 3 AsO 4 3 mol Ag + × 1 mol As 1 mol Ag 3 AsO 4 × 74.92 g As 1 mol As = 0.06368 = 0.0637 g As

    mass percent = 0 .06368 g As 1.22 g sample × 100 = 5.22 % As


4.114 Analyze .  Given 10 ppb AsO 4 3– , find mass Na 3 AsO 4 in 1.00 L of drinking water.

Plan .  Use the definition of ppb to calculate g AsO 4 3‒ in 1.0 L of water. Convert

g AsO 4 3‒ → g Na 3 AsO 4 using molar masses. Assume the density of H 2 O is 1.00 g/mL.

Solve . 1 billion = 1 × 10 9 ; 1 ppb = 1 g solute 1 × 10 9 g solution

1 g solute 1 × 10 9 g solution × 1 g solution 1 mL solution × 1 × 10 3 mL 1 L solution = g AsO 4 3 1 × 10 6 L H 2 O

10 ppb AsO 4 3 = 10 g AsO 4 1 × 10 6 L H 2 O × 1 L H 2 O = 1 .0 × 10 5 g As/L .

1.0 × 10 5 g AsO 4 3 × 1 mol AsO 4 3 138.92 g AsO 4 3 × 1 mol Na 3 AsO 4 1 mol As × 207.89 g Na 3 AsO 4 3 1 mol Na 3 AsO 4 = 1.5 × 10 5 g Na 3 AsO 4 in 1 .00 L H 2 O

4.115

  1. mol HCl initial NH 3 from air = mol HCl remaining = mol NaOH required for titration

    mol NaOH = 0 .0588 M × 0 .0131 L = 7 .703 × 10 4 = 7.70 × 10 4 mol NaOH = 7 .70 × 10 4 mol HCl remain

    mol HCl initial – mol HCl remaining = mol NH 3 from air

    (0.0105 M HCl × 0.100 L) – 7.703 × 10 –4 mol HCl = mol NH 3

    10.5 × 10 –4 mol HCl – 7.703 × 10 –4 mol HCl = 2.80 × 10 –4 = 2.8 × 10 –4 mol NH 3

    2.8 × 10 4 mol NH 3 × 17.03 g NH 3 1 mol NH 3 = 4.77 × 10 3 = 4.8 × 10 3 g NH 3

  2. ppm is defined as molecules of NH 3 /1 × 10 6 molecules in air.

    Calculate molecules NH 3 from mol NH 3 .

    2.80 × 10 4 mol NH 3 × 6.022 × 10 23 molecules 1 mol = 1.686 × 10 20 = 1.7 × 10 20 NH 3 molecules

    Calculate total volume of air processed, then g air using density, then molecules air using molar mass.

    10.0 L 1 min × 10.0 min × 1 .20 g air 1 L air × 1 mol air 29.0 g air × 6.022 × 10 23 molecules 1 mol = 2.492 × 10 24 = 2.5 × 10 24 air molecules

    ppm NH 3 = 1 .686 × 10 20 NH 3 molecules 2.492 × 10 24 air molecules × 1 × 10 6 = 68 ppm NH 3

  3. 68 ppm b 50 ppm. The manufacturer is not in compliance.

 

5 Thermochemistry

Visualizing Concepts

5.1

  1. Analyze/Plan . The exercise gives the charges and separation of two particles. Use Equation 5.2 to calculate electrostatic potential energy, E el . Solve .

    Ε el = κ Q 1 Q 2 d ; κ = 8.99 × 10 9 J-m/C 2 ; Q 1 = Q 2 = 2.0 × 10 –5 C; d = 1.0 cm

    Ε el = 8.99 × 10 9 J-m C 2 × 1 1.0 cm × 100 cm m × 2 .0 × 10 5 C × 2 .0 × 10 5 C = 359.6 = 3.6 × 10 2 J

  2. The spheres are both positively charged, so they will move away from each other. (Like charged particles repel, oppositely charged particles attract.)
  3. As the like charged spheres move apart, the electrostatic potential energy of the system is converted to kinetic energy. As the distance between them approaches infinity, potential energy approaches zero and the kinetic energy of each particle is 1.8 × 10 2 J, one half of the initial potential energy calculated in part (a).

    E k = 1/2 mv 2 ; v = (2 E k /m) 1/2 ; E k = ½(3.6 × 10 2 J) = 1.8 × 10 2 J

    v = ( 2 × 1.8 × 10 2 kg-m 2 s 2 × 1 1 .0 kg ) 1 / 2 = 18.97 = 19 m/s

5.2

  1. The caterpillar uses energy produced by its metabolism of food to climb the twig and increase its potential energy.
  2. Heat, q, is the energy transferred from a hotter to a cooler object. Without knowing the temperature of the caterpillar and its surroundings, we cannot predict the sign of q. It is likely that q is approximately zero, because a small creature like a caterpillar is unlikely to support a body temperature much different from its environmental temperature.
  3. Work, w, is the energy transferred when a force moves an object. When the caterpillar climbs the twig, it does work as its body moves against the force of gravity.
  4. No. The amount of work is independent of time and therefore independent of speed (assuming constant caterpillar speed).
  5. No. Potential energy depends only on the caterpillar’s position, so the change in potential energy depends only on the distance climbed, not on the speed of the climb.

5.3

  1. The internal energy, E, of the products is greater than that of the reactants, so the diagram represents an increase in the internal energy of the system.
  2. ΔE for this process is positive, (+).
  3. If no work is associated with the process, it is endothermic.

5.4

  1. For an endothermic process, the sign of q is positive; the system gains heat. This is true only for system (iii).
  2. In order for ΔE to be less than 0, there is a net transfer of heat or work from the system to the surroundings. The magnitude of the quantity leaving the system is greater than the magnitude of the quantity entering the system. In system (i), the magnitude of the heat leaving the system is less than the magnitude of the work done on the system. In system (iii), the magnitude of the work done by the system is less than the magnitude of the heat entering the system. None of the systems has ΔE < 0.
  3. In order for ΔE to be greater than 0, there is a net transfer of work or heat to the system from the surroundings. In system (i), the magnitude of the work done on the system is greater than the magnitude of the heat leaving the system. In system (ii), work is done on the system with no change in heat. In system (iii), the magnitude of the heat gained by the system is greater than the magnitude of the work done on the surroundings. ΔE > 0 for all three systems.

5.5

  1. No. This distance traveled to the top of a mountain depends on the path taken by the hiker. Distance is a path function, not a state function.
  2. Yes. Change in elevation depends only on the location of the base camp and the height of the mountain, not on the path to the top. Change in elevation is a state function, not a path function.

5.6

  1. State B
  2. ΔE AB = energy difference between State A and State B.

    ΔE AB = ΔE 1 + ΔE 2 or ΔE AB = ΔE 3 + ΔE 4

  3. ΔE CD = energy difference between State C and State D.

    ΔE CD = ΔE 2 – ΔE 4 or ΔE CD = ΔE 3 – ΔE 1

    (Note that the sign of ΔE depends on the definition of initial and final state, but the magnitude is the absolute value of the difference in energy.)

  4. The energy of State E is ΔE 1 + ΔE 4 , whereas the energy of State B is ΔE 1 + ΔE 2 . Because ΔE 4 > ΔE 2 , State E is above State B on the diagram; State E would be the highest energy on the diagram.

5.7

  1. You, part of the surroundings, do work on the air, part of the system. Energy is transferred to the system via work and the sign of w is (+).
  2. The body of the pump (the system) is warmer than the surroundings. Heat is transferred from the warmer system to the cooler surroundings, and the sign of q is (–).
  3. The sign of w is positive, and the sign of q is negative, so we cannot absolutely determine the sign of ΔE. It is likely that the heat lost is much smaller than the work done on the system, so the sign of ΔE is probably positive.

5.8

  1. The temperature of the system and surroundings will equalize, so the temperature of the hotter system will decrease, and the temperature of the colder surroundings will increase. The system loses heat by decreasing its temperature, so the sign of q sys is (–). The surrounding gains heat by increasing its temperature, so the sign of q surr is (+). From the system’s perspective, the process is exothermic because it loses heat.
  2. If neither volume nor pressure of the system changes, w = 0 and ΔE = q = ΔH. The change in internal energy is equal to the change in enthalpy.

5.9

  1. w = – PΔV. Because ΔV for the process is (–), the sign of w is (+).
  2. ΔE = q + w. At constant pressure, ΔH = q. If the reaction is endothermic, the signs of ΔH and q are (+). From (a), the sign of w is (+), so the sign of ΔE is (+). The internal energy of the system increases during the change. (This situation is described by the diagram (ii) in Exercise 5.4.)

5.10

  1. N 2 (g) + O 2 (g) → 2 NO(g). Because ΔV = 0, w = 0.
  2. The reaction of two elements to form one mole of a compound fits the definition of a formation reaction. Find the value for enthalpy of formation of NO(g) in Appendix C. ΔH = ΔH f = 90.37 kJ for production of 1 mol of NO(g).

5.11

  1. ΔH A = ΔH B + ΔH C . Diagram (i) indicates that reaction A can be written as the sum of reactions B and C.
  2. ΔH Z = ΔH X + ΔH Y . Diagram (ii) indicates that reaction Z can be written as the sum of reactions X and Y.
  3. Hess’s law states that the enthalpy change for a net reaction is the sum of the enthalpy changes of the component steps, regardless of whether the reaction actually occurs via this path.
  4. No. The enthalpy relationships are true because enthalpy is a state function, independent of path. Work is not a state function.

5.12  Because mass must be conserved in the reaction A → B, the component elements of A and B must be the same. Further, if Δ H f o > 0 for both A and B, the energies of both A and B are above the energies of their component elements on the energy diagram.

  1. The bold arrow shows the reaction as written; combination of the two thin arrows shows an alternate route from A to B.
    image
  2. Δ H rxn o = Δ H f o B Δ H f o A . If the overall reaction is exothermic, the sign of ΔH is (–) and Δ H f o A > Δ H f o B . This means that the enthalpy of A is the highest energy level on the diagram. This is the situation pictured in the diagram above, but nothing in the given information requires this arrangement. If the reaction is endothermic, Δ H f o B > Δ H f o A and the enthalpy of B would be the highest energy level on the diagram.

The Nature of Chemical Energy (Section 5.1)

5.13 Analyze/Plan . Use Equation 5.2 to calculate electrostatic potential energy, E el . The distances between the particles are given in the exercise. The charge of an electron and a proton are given in Section 5.1. Solve .

  1. Ε el = κ Q 1 Q 2 d d = 53 pm ; Q e = 1 . 60 × 10 –19 C ; Q p = 1 . 60 × 10 –19 C

    Ε el = 8.99 × 10 9 J-m C 2 × 53 pm × 1 pm 1 × 10 12 m × 1.60 × 10 19 C × 1.60 × 10 19 C = 4 .342 × 10 18 = 4 .3 × 10 18 J

  2. Ε el = 8.99 × 10 9 J-m C 2 × 1 .0 nm × 1 nm 1 × 10 9 m × 1.60 × 10 19 C × 1.60 × 10 19 C = 2 .301 × 10 19 = 2.3 × 10 19 J

    The change in potential energy is [–2.3 × 10 –19 J – (–4.3 × 10 –18 J )] = 4.1 × 10 –18 J

  3. The electrostatic potential energy of the system increases (becomes less negative) as the separation between the oppositely charged particles increases.

5.14

  1. Ε el = κ Q 1 Q 2 d ; d = 62 pm; Q p = 1 .60 × 10 –19 C

    Ε el = 8.99 × 10 9 J-m C 2 × 62 pm × 1 pm 1 × 10 12 m × ( 1.60 × 10 19 C) 2 = 3 .712 × 10 18 = 3.7 × 10 18 J

  2. Ε el = 8.99 × 10 9 J-m C 2 × 1 .0 nm × 1 nm 1 × 10 9 m × ( 1.60 × 10 19 C) 2 = 2 .301 × 10 19 = 2.3 × 10 19 J

    The change in potential energy is [2.3 × 10 –19 J – (3.7 × 10 –18 J )] = – 3.5 × 10 –18 J

  3. The electrostatic potential energy of the system decreases as the separation between the like charged particles increases.

5.15

  1. Analyze/Plan . Use the equation for electrostatic attractive force and the distance between particles given in the exercise. Find the charges of a proton and an electron in Section 5.1. Solve .

    F el = κ Q 1 Q 2 d 2 ; d = 1 .0 × 10 2 pm ;

    Q e = –1.60 × 10 –19 C; Q p = 1.60 × 10 –19 C ; 1 J = 1 N-m

    F el = 8 .99 × 10 9 J-m C 2 × ( 1 .0 × 10 2 pm) 2 × (1 pm) 2 (1 × 10 12 ) 2 m 2 × 1.60 × 10 19 C × 1.60 × 10 19 C = 2 .301 × 10 8 J/m = 2 .3 × 10 8 N
  2. Analyze/Plan . Use the formula for gravitational force between two particles and the distance given in the exercise. Find the masses of a proton and an electron on the inside cover of the text. Solve .

    F g = Gm 1 m 2 d 2 ; G = 6 .674 × 10 11 N-m 2 kg 2 ; d = 1 .0 × 10 2 pm;m e = 9 .109 × 10 –31 kg; m p = 1 .673×10 –27 kg


  1. F g = 6 .674 × 10 11 N-m 2 kg 2 × ( 1 .0 × 10 2 pm) 2 × (1 pm) 2 (1 × 10 12 ) 2 m 2 × 9.109 × 10 31 kg × 1.673 × 10 27 kg = 1 .0171 × 10 47 = 1 .0 × 10 47 N

  1. The magnitude of the electrostatic force of attraction is 2.3 × 10 –8 N. (The negative sign of the electrostatic force indicates attraction.) The gravitational force is 1.0 × 10 –47 N. The electrostatic attraction is 2.3 × 10 39 time larger. (This is almost 40 orders of magnitude.)

5.16

  1. F el = κ Q 1 Q 2 d 2 ; d = 75 pm;Q p = 1 .60×10 –19 C ; 1 J = 1 N-m

    F el = 8 .99 × 10 9 J-m C 2 × ( 75 pm) 2 × (1 pm) 2 (1 × 10 12 ) 2 m 2 × ( 1.60 × 10 19 C) 2 = 4 .091 × 10 8 J/m = 4 .1 × 10 8 N

  2. F g = Gm 1 m 2 d 2 ; G = 6 .674 × 10 11 N-m 2 kg 2 ; d = 75 pm;m p = 1 .673 × 10 –27 kg

    F g = 6 .674 × 10 11 N-m 2 kg 2 × ( 75 pm) 2 × (1 pm) 2 (1 × 10 12 ) 2 m 2 × ( 1.673 × 10 27 kg) 2 = 3 .321 × 10 44 = 3 .3 × 10 44 N

  3. The magnitude of the electrostatic force of repulsion is 4.1 × 10 –8 N. (The positive sign of the electrostatic force indicates repulsion.) The gravitational force is 3.3 × 10 –44 N. If the two protons are allowed to move, the electrostatic repulsion is much greater than gravitational attraction and the protons move apart.

5.17 Analyze/Plan . We must find the work required to completely separate two oppositely charged particles. Work is the energy required to move an object against a force (Section 1.4). At infinite separation, the electrostatic potential energy of the pair of ions is zero. The magnitude of the work required is equal to the electrostatic potential energy, E el , of the pair of ions. Use Equation 5.2 to calculate E el and work. The charges of the ions and the distance between them are given in the exercise. Solve .

Ε el = κ Q 1 Q 2 d ; d = 0 .50 nm;Q Cl = –1 .6 × 10 –19 C; Q Na = 1 .6 × 10 –19 C

Ε el = 8.99 × 10 9 J-m C 2 × 0 .50 nm × 1 nm 1 × 10 9 m × 1.6 × 10 19 C × 1.6 × 10 19 C = 4 .603 × 10 19 = 4 .6 × 10 19 J

The sign of E el is negative, so the work required to separate the ions is 4.6 × 10 –19 J.

5.18 Ε el = κ Q 1 Q 2 d ; d = 0 .35 nm;Q O = –3 .2 × 10 –19 C; Q Mg = 3 .2 × 10 –19 C

Ε el = 8.99 × 10 9 J-m C 2 × 0 .35 nm × 1 nm 1 × 10 9 m × 3.2 × 10 19 C × 3.2 × 10 19 C = 2 .630 × 10 18 = 2 .6 × 10 18 J

The sign of E el is negative, so the work required to separate the ions is 2.6 × 10 –18 J.


5.19

  1. Gravity; work is done because the force of gravity is opposed and the pencil is lifted a distance above the desk.
  2. Spring force; work is done because the force of the coiled spring is opposed as the spring is compressed over a distance.

5.20

  1. Electrostatic attraction; no work is done because the particles are held apart at a constant distance.
  2. Magnetic attraction; work is done because the nail is moved a distance in opposition to the force of magnetic attraction.

The First Law of Thermodynamics (Section 5.2)

5.21

  1. Matter cannot leave a closed system. Energy in the form of heat or work can be transferred between a closed system and the surroundings.
  2. Neither matter nor energy can leave or enter an isolated system.
  3. Any part of the universe not part of the system is called the surroundings.

5.22

  1. The liquid is an open system because it exchanges both matter and energy with the surroundings. Matter exchange occurs when solution flows into and out of the apparatus. The apparatus is not insulated, so energy exchange also occurs. Closed systems exchange energy but not matter, whereas isolated systems exchange neither.
  2. If the inlet and outlet are closed, the system can exchange energy but not matter with the surroundings; it becomes a closed system.

5.23

  1. According to the first law of thermodynamics, energy is conserved.
  2. The total internal energy (E) of a system is the sum of all the kinetic and potential energies of the system components.
  3. The internal energy of a closed system (where no matter exchange with surroundings occurs) increases when work is done on the system by the surroundings and/or when heat is transferred to the system from the surroundings (the system is heated).

5.24

  1. ΔE = q + w
  2. The quantities q and w are negative when the system loses heat to the surroundings (it cools) or does work on the surroundings.

5.25 Analyze .  Given: heat and work. Find: magnitude and sign of ΔE.

Plan .  In each case, evaluate q and w in the expression ΔE = q + w. For an exothermic process, q is negative; for an endothermic process, q is positive. Solve .

  1. q = 0.763 kJ, w = –840 J = –0.840 kJ. ΔE = 0.763 kJ – 0.840 kJ = – 0.077 kJ. The process is endothermic.
  2. q is negative because the system releases heat, and w is positive because work is done on the system. ΔE = –66.1 kJ + 44.0 kJ = –22.1 kJ. The process is exothermic.

5.26  In each case, evaluate q and w in the expression ΔE = q + w. For an exothermic process, q is negative; for an endothermic process, q is positive.

  1. q is negative and w is positive. ΔE = –0.655 kJ + 0.382 kJ = –0.273 kJ. The process is exothermic.
  2. q is positive and w is essentially zero. ΔE = 322 J. The process is endothermic.

5.27 Analyze .  How do the different physical situations (cases) affect the changes to heat and work of the system upon addition of 100 J of energy?

Plan . Use the definitions of heat and work and the First Law to answer the questions.

Solve . If the piston is allowed to move, case (1), the heated gas will expand and push the piston up, doing work on the surroundings. If the piston is fixed, case (2), most of the electrical energy will be manifested as an increase in heat of the system.

  1. Because little or no work is done by the system in case (2), the gas will absorb most of the energy as heat; the case (2) gas will have the higher temperature.
  2. In Case 1, w is negative because work is done on the surroundings by expansion. Because the transfer of electrical energy is never completely efficient and some energy will be transferred as heat, q is positive. In Case 2, w is zero because no work (expansion) is done. The value of q is positive because all energy is transferred as heat.
  3. ΔE is greater for case (2) because the entire 100 J increases the internal energy of the system, rather than a part of the energy doing work on the surroundings.

5.28 Ε el = κ Q 1 Q 2 r For two oppositely charged particles, the sign of E el is negative; the closer the particles, the greater the magnitude of E el .

  1. The potential energy of oppositely charged spheres increases (becomes less negative) as the particles are separated (r increases).
  2. ΔE for the process is positive; the internal energy of the system increases as the oppositely charged particles are separated.
  3. Work is done on the system to separate the particles so w is positive. Mechanical separation of macroscopic charged spheres involves no heat transfer. Work alone accounts for the change in energy of the system.

5.29

  1. A state function is a property of a system that depends only on the physical state (pressure, temperature, etc.) of the system, not on the route used by the system to get to the current state.
  2. Internal energy and enthalpy are state functions; heat is not a state function.
  3. Volume is a state function. The volume of a system depends only on conditions (pressure, temperature, amount of substance), not the route or method used to establish that volume.

5.30

  1. Independent. Potential energy is a state function.
  2. Dependent. Some of the energy released could be employed in performing work, as is done in the body when sugar is metabolized; heat is not a state function.

  1. Dependent. The work accomplished depends on whether the gasoline is used in an engine, burned in an open flame, or in some other manner. Work is not a state function.

Enthalpy (Sections 5.3 and 5.4)

5.31 Analyze . Given, P = 1.0 atm, ΔV = +0.50 L. Find work involved, in J.

Plan . This change is P – V work done at constant P. w = –PΔV. 1 L-atm = 101.3 J

Solve . w = –1.0 atm(0.50 L) = –0.50 L-atm; 0.50 L-atm × 101.3 J/L-atm = –50.65 = –51 J

The negative sign indicates that work is done by the system on the surroundings.

5.32  P = 0.857 atm. ΔV = 1.26 L – 5.00 L = –3.74 L

w = –0.857 atm(– 3.74 L) = 3.2052 = 3.21 L-atm;

3.2052 L-atm × 101.3 J/L-atm = 324.69 = 325 J

5.33

  1. Change in enthalpy (ΔH) is usually easier to measure than change in internal energy (ΔE) because, at constant pressure, ΔH = q p . The heat flow associated with a process at constant pressure can easily be measured as a change in temperature. Measuring ΔE requires a means to measure both q and w.
  2. H describes the enthalpy of a system at a certain set of conditions; the value of H depends only on these conditions. q describes energy transferred as heat, an energy change, which, in the general case, does depend on how the change occurs. We can equate change in enthalpy, ΔH, with heat, q p , only for the specific conditions of constant pressure and exclusively P-V work.
  3. If ΔH is positive, the enthalpy of the system increases, and the process is endothermic.

5.34

  1. When a process occurs under constant external pressure and only P–V work occurs, the enthalpy change (ΔH) equals the amount of heat transferred. ΔH = q p .
  2. ΔH = q p . If the system releases heat, q and ΔH are negative, and the enthalpy of the system decreases.
  3. If ΔH = 0, q p = 0 and ΔE = w.

5.35

  1. At constant pressure, ΔE = ΔH – PΔV. To calculate ΔE, more information about the conditions of the reaction must be known. For an ideal gas at constant pressure and temperature, PΔV = RTΔn. We know the value of Δn = –3 from the chemical reaction. We must know either the temperature, T, or the values of P and ΔV to calculate ΔE from ΔH.
  2. ΔE is larger than ΔH.
  3. Because the value of Δn is negative, the quantity (–PΔV) is positive. We add a positive quantity to ΔH to calculate ΔE, so ΔE must be larger.

5.36

  1. At constant volume (ΔV = 0), ΔE = q v .
  2. ΔE will be larger than ΔH.

  1. According to the definition of enthalpy, H = E + PV, so ΔH = ΔE + Δ(PV). For an ideal gas at constant temperature and volume, ΔPV = VΔP = RTΔn. For this reaction, there are 2 mol of gaseous product and 3 mol of gaseous reactants, so Δn = –1. Thus VΔP or Δ(PV) is negative. Because ΔH = ΔE + Δ(PV), the negative Δ(PV) term means that ΔE is larger or less negative than ΔH.

5.37 Analyze/Plan .  q = 824 J = 0.824 kJ (heat is absorbed by the system), w = 0.65 kJ (work is done on the system). Solve .

ΔE = q + w = 0.824 kJ + 0.65 kJ = 1.47 kJ. ΔH = q = 0.824 kJ (at constant pressure).

Check .  The reaction is endothermic.

5.38  The gas is the system. If 0.49 kJ of heat is added, q = +0.49 kJ. Work done by the system decreases the overall energy of the system, so w = –214 J = –0.214 kJ .

ΔE = q + w = 0.49 kJ – 0.214 kJ = 0.276 kJ. ΔH = q = 0.49 kJ (at constant pressure).

5.39

  1. C 2 H 5 OH(l) + 3 O 2 (g) → 3 H 2 O(g) + 2 CO 2 (g)  ΔH = –1235 kJ
  2. Analyze .  How are reactants and products arranged on an enthalpy diagram?

    Plan .  The substances (reactants or products, collectively) with higher enthalpy are shown on the upper level, and those with lower enthalpy are shown on the lower level.

    image

    Solve .  For this reaction, ΔH is negative, so the products have lower enthalpy and are shown on the lower level; reactants are on the upper level. The arrow points in the direction of reactants to products and is labeled with the value of ΔH.

5.40

  1. Ca(OH) 2 (s) → CaO(s) + H 2 O(g)
  2. ΔH = 109 kJ
    image

5.41 Analyze/Plan .  Consider ΔH for the exothermic reaction as written. Solve.

  1. ΔH = – 284.6 kJ/2 mol O 3 (g) = – 142.3 kJ/mol O 3 (g)
  2. Because ΔH is negative, the reactants, 2 O 3 (g) has the higher enthalpy.

5.42 Plan .  Consider the sign of an enthalpy change that would convert one of the substances into the other. Solve .

  1. CO 2 (s) → CO 2 (g). This change is sublimation, which is endothermic, +ΔH. CO 2 (g) has the higher enthalpy.
  2. H 2 → 2 H. Breaking the H–H bond requires energy, so the process is endothermic, +ΔH. Two moles of H atoms have higher enthalpy.

  1. H 2 O(g) → H 2 (g) + 1/2 O 2 (g). Decomposing H 2 O into its elements requires energy and is endothermic, +ΔH. One mole of H 2 (g) and 0.5 mol O 2 (g) at 25 °C have the higher enthalpy.
  1. N 2 (g) at 100 °C → N 2 (g) at 300 °C. An increase in the temperature of the sample requires that heat is added to the system, +q and +ΔH. N 2 (g) at 300 °C has the higher enthalpy.

5.43 Analyze/Plan .  Follow the strategy in Sample Exercise 5.4. Solve .

  1. Exothermic (ΔH is negative)
  2. 3. 55 g Mg × 1 mol Mg 24.305 g Mg × 1204 kJ 2 mol Mg = 87.9 kJ heat transferred

    Check .  The units of kJ are correct for heat. The negative sign indicates heat is evolved.

  3. 234 kJ × 2 mol MgO 1204 kJ × 40.30 g MgO 1 mol Mg = 15.7 g MgO produced

    Check .  Units are correct for mass. (200 × 2 × 40/1200) ≈ (16,000/1200) > 10 g

  4. 2 MgO(s) → 2 Mg(s) + O 2 (g) ΔH = +1204 kJ

    This is the reverse of the reaction given above, so the sign of ΔH is reversed.

    4 0 .3 g MgO × 1 mol MgO 40.30 g MgO × 1204 kJ 2 mol MgO = + 602 kJ heat absorbed

    Check .  40.3 g MgO is just 1 mol MgO, so the calculated value is the heat absorbed per mol of MgO, 1204 kJ/2 mol MgO = 602 kJ.

5.44

  1. The sign of ΔH is positive, so the reaction is endothermic.
  2. 24.0 g CH 3 OH × 1 mol CH 3 OH 32.04 g CH 3 OH × 2 52 .8 kJ 2 mol CH 3 OH = 9 4.7 kJ heat absorbed
  3. 8 2 .1 kJ × 2 mol CH 4 2 52 .8 kJ × 1 6 .04 g CH 4 1 mol CH 4 = 10.4 g CH 4 produced
  4. The sign of ΔH is reversed for the reverse reaction: ΔH = –252.8 kJ

    3 8 .5 g CH 4 × 1 mol CH 4 16 .04 g CH 4 × 252.8 kJ 2 mol CH 4 = 303 kJ heat released

5.45 Analyze .  Given: balanced thermochemical equation, various quantities of substances and/or enthalpy. Plan . Enthalpy is an extensive property; it is “stoichiometric.” Use the mole ratios implicit in the balanced thermochemical equation to solve for the desired quantity. Use molar masses to change mass to moles and vice versa where appropriate. Solve .

  1. 0.450 mol AgCl × 65.5 kJ 1 mol AgCl = 29.5 kJ

    Check .  Units are correct; sign indicates heat evolved.

  2. 9.00 g AgCl × 1 mol AgCl 143.3 g AgCl × 65.5 kJ 1 mol AgCl = 4.11 kJ

    Check .  Units correct; sign indicates heat evolved.


  1. 9. 25 × 10 4 mol AgCl × + 65.5 kJ 1 mol AgCl = 0.0606 kJ = 6 0.6 J

    Check .  Units correct; sign of ΔH reversed; sign indicates heat is absorbed during the reverse reaction.

5.46

  1. 1. 36 mol O 2 × 89.4 kJ 3 mol O 2 = 40.53 = 40.5 kJ
  2. 1 0 .4 g KCl × 1 mol KCl 74.55 g KCl × 89.4 kJ 2 mol KCl = 6.2358 = 6.24 kJ
  3. Because the sign of ΔH is reversed for the reverse reaction, it seems reasonable that other characteristics would be reversed, as well. If the forward reaction proceeds spontaneously, the reverse reaction is probably not spontaneous. Also, we know from experience that KCl(s) does not spontaneously react with atmospheric O 2 (g), even at elevated temperature.

5.47 Analyze .  Given: balanced thermochemical equation. Plan . Follow the guidelines given in Section 5.4 for evaluating thermochemical equations. Solve .

  1. When a chemical equation is reversed, the sign of ΔH is reversed.

    CO 2 (g) + 2 H 2 O(l) → CH 3 OH(l) + 3/2 O 2 (g)    ΔH = +726.5 kJ

  2. Enthalpy is extensive. If the coefficients in the chemical equation are multiplied by 2 to obtain all integer coefficients, the enthalpy change is also multiplied by 2.

    2 CH 3 OH(l) + 3 O 2 (g) → 2 CO 2 (g) + 4 H 2 O(l) ΔH = 2(–726.5) kJ = –1453 kJ

  3. The exothermic forward reaction is more likely to be thermodynamically favored.
  4. Decrease. Vaporization (liquid → gas) is endothermic. If the product were H 2 O(g), the reaction would be more endothermic and would have a smaller negative ΔH. (Depending on temperature, the enthalpy of vaporization for 2 mol H 2 O is about +88 kJ, not large enough to cause the overall reaction to be endothermic.)

5.48

  1. 3 C 2 H 2 (g) → C 6 H 6 (l)   ΔH = –630 kJ
  2. C 6 H 6 (l) → 3 C 2 H 2 (g)   ΔH = +630 kJ

    ΔH for the formation of 3 mol of acetylene is 630 kJ. ΔH for the formation of 1 mol of C 2 H 2 is then 630 kJ/3 = 210 kJ.

  3. The exothermic reverse reaction is more likely to be thermodynamically favored.
  4. image

    If the reactant is in the higher enthalpy gas phase, the overall ΔH for the reaction decreases.


Calorimetry (Section 5.5)

The specific heat of water to four significant figures, 4.184 J/g-K, will be used in many of the following exercises; temperature units of K and °C will be used interchangeably.

5.49

  1. J/mol-K or J/mol-°C. Heat capacity is the amount of heat in J required to raise the temperature of an object or a certain amount of substance 1 °C or 1 K. Molar heat capacity is the heat capacity of one mole of substance.
  2. J g- o C or J g-K Specific heat is a particular kind of heat capacity where the amount of substance is 1 g.
  3. To calculate heat capacity from specific heat, the mass of the particular piece of copper pipe must be known.

5.50 Analyze .  Both objects are heated to 100 °C. The two hot objects are placed in the same amount of cold water at the same temperature. Object A raises the water temperature more than object B. Plan . Apply the definition of heat capacity to heating the water and heating the objects to determine which object has the greater heat capacity. Solve .

  1. Both beakers of water contain the same mass of water, so they both have the same heat capacity. Object A raises the temperature of its water more than object B, so more heat was transferred from object A than from object B. Because both objects were heated to the same temperature initially, object A must have absorbed more heat to reach the 100 °C temperature. The greater the heat capacity of an object, the greater the heat required to produce a given rise in temperature. Thus, object A has the greater heat capacity.
  2. Because no information about the masses of the objects is given, we cannot compare or determine the specific heats of the objects.

5.51 Plan .  Manipulate the definition of specific heat to solve for the desired quantity, paying close attention to units. C s = q/(m × ΔT). Solve .

  1. 4 .184 J 1 g-K or 4 .184 J 1 g- o C
  2. 4.184 J 1 g-°C × 18.02 g H 2 O 1 mol H 2 O = 75.40 J mol-°C
  3. 185 g H 2 O × 4 .184 J 1 g-°C = 774 J/°C
  4. 10.00 kg H 2 O × 1000 g 1 kg × 4.184 J 1 g-°C × 1 kJ 1000 J × ( 46.2 ° C 24.6 o C) = 904 kJ

    Check .  (10 × 4 × 20) ≈ 800 kJ; the units are correct. Note that the conversion factors for kg → g and J → kJ cancel. An equally correct form of specific heat would be kJ/kg - ° C

5.52

  1. In Table 5.2, Hg(l) has the smallest specific heat, so it will require the smallest amount of energy to heat 50.0 g of the substance 10 K.
  2. 50.0 g Hg(l) × 10 K × 0 .14 J g-K = 70 J

5.53 Analyze/Plan .  Follow the logic in Sample Exercise 5.5. Solve .

  1. 8 0 .0 g C 8 H 18 × 2. 22 J g-K × ( 25.0 o C 10.0 o C) = 2 .66 × 10 3 J (or 2 .66 kJ)
  2. Plan. Calculate the molar heat capacity of octane and compare it with the molar heat capacity of water, 75.40 J/mol- o C, as calculated in Exercise 5.51(b). Solve.

    2 .22 J g-K × 114.2 g C 8 H 18 1 mol C 8 H 18 = 253.58 J mol-K = 254 J mol-K

    The molar heat capacity of C 8 H 18 (l), 254 J/mol-K, is greater than that of H 2 O(l), so it will require more heat to increase the temperature of octane than to increase the temperature of water.

5.54

  1. specific heat = J 1 g- o C = 322 J 100 .0 g × (50 o C 25 o C) = 0.1288 = 0 .13 J 1 g- o C
  2. In general, the greater the heat capacity, the more heat is required to raise the temperature of 1 gram of substance 1 o C. The specific heat of gold is 0.13 J/g- o C, whereas that of iron is 0.45 J/g- o C (Table 5.2). For gold and iron blocks with equal mass, same initial temperature and same amount of heat added, the one with the lower specific heat, gold, will require less heat per o C and have the higher final temperature.
  3. 0 .1288 J 1 g- o C × 196 .97 g Au 1 mol Au = 25.37 = 25 J mol- o C

5.55 Analyze .  Because the temperature of the water increases, the dissolving process is exothermic and the sign of ΔH is negative. The heat lost by the NaOH(s) dissolving equals the heat gained by the solution.

Plan/Solve .  Calculate the heat gained by the solution. The temperature change is 37.8 – 21.6 = 16.2 °C. The total mass of solution is (100.0 g H 2 O + 6.50 g NaOH) = 106.5 g.

106.5 g solution × 4 .184 J 1 g- o C × 16.2 o C × 1 kJ 1000 J = 7.2187 = 7.22 kJ

This is the amount of heat lost when 6.50 g of NaOH dissolves.

The heat loss per mole NaOH is

7.2187 kJ 6. 50 g NaOH × 40.00 g NaOH 1 mol NaOH = 44.4 kJ/mol Δ H = q p = 44.4 kJ/mol NaOH

Check .  (–7/7 × 40) ≈ –40 kJ; the units and sign are correct.

5.56

  1. Follow the logic in Solution 5.55. The total mass of the solution is (60.0 g H 2 O + 4.25 g NH 4 NO 3 ) = 64.25 = 64.3 g. The temperature change of the solution is 22.0 – 16.9 = –5.1 °C. The heat lost by the surroundings is

    64.25 g solution × 4 .184 J 1 g-°C × 5.1 o C × 1 kJ 1000 J = 1.371 = 1.4 kJ

    That is, 1.4 kJ is absorbed when 4.25 g NH 4 NO 3 (s) dissolves.

    + 1.371 kJ 4 .25 NH 4 NO 3 × 80.04 g NH 4 NO 3 1 mol NH 4 NO 3 = + 25.82 = + 26 kJ/mol NH 4 NO 3

  2. This process is endothermic because the temperature of the surroundings decreases, indicating that heat is absorbed by the system.

5.57 Analyze/Plan .  Follow the logic in Sample Exercise 5.7. Solve .

q bomb = – q rxn ; ΔT = 30.57 °C – 23.44 °C = 7.13 °C

q bomb = 7.854 kJ 1 o C × 7 .13 o C = 56 .00 = 56 .0 kJ

At constant volume, q v = ΔE. ΔE and ΔH are very similar.

Δ H r x n Δ E rxn = q rxn = q bomb = 56.0 kJ 2 .200 g C 6 H 4 O 2 = 25.454 = 25.5 kJ/g  C 6 H 4 O 2

Δ H rxn = 25.454 kJ 1 g C 6 H 4 O 2 × 108.1 g C 6 H 4 O 2 1 mol C 6 H 4 O 2 = 2.75 × 10 3 kJ/mol C 6 H 4 O 2

5.58

  1. C 6 H 5 OH(s) + 7 O 2 (g) → 6 CO 2 (g) + 3 H 2 O(l)
  2. q bomb = –q rxn ; ΔT = 26.37 °C – 21.36 °C = 5.01 °C

    q bomb = 11.66 kJ 1 o C × 5. 01 o C = 58 .417 = 58 .4 kJ

    At constant volume, q v = ΔE. ΔE and ΔH are very similar.

    Δ H rxn Δ E rxn = q rxn = q bomb = 58.417 kJ 1 .800 g C 6 H 5 OH = 32.454 = 32.5 kJ/g C 6 H 5 OH

    Δ H rxn = 32.454 kJ 1 g C 6 H 5 OH × 94.11 g C 6 H 5 OH 1 mol C 6 H 5 OH = 3.054 × 10 3 kJ mol C 6 H 5 OH = 3 . 05 × 10 3 kJ/mol C 6 H 5 OH

5.59 Analyze .  Given: specific heat and mass of glucose, ΔT for calorimeter. Find: heat capacity, C, of calorimeter. Plan . All heat from the combustion raises the temperature of the calorimeter. Calculate heat from combustion of glucose, divide by ΔT for calorimeter to get kJ/°C. ΔT = 24.72 °C – 20.94 °C = 3.78 °C Solve .

  1. C total = 3.500 g glucose × 15 .57 kJ 1 g glucose × 1 3.78 C = 14.42 = 14.4 kJ/C
  2. Qualitatively, assuming the same exact initial conditions in the calorimeter, twice as much glucose produces twice as much heat, which raises the calorimeter temperature by twice as many °C. Quantitatively,

    7.000 g glucose × 15 .57 kJ 1 g glucose × 1 o C 14 .42 kJ = 7.56 o C

Check .  Units are correct. ΔT is twice as large as in part (a). The result has 3 sig figs, because the heat capacity of the calorimeter is known to 3 sig figs.

5.60

  1. C = 2 .760 g C 6 H 5 COOH × 26 .38 kJ 1 g C 6 H 5 COOH × 1 8.33 o C = 8.74055 = 8.74 kJ/ o C
  2. 8.74055 kJ o C × 4.95 o C × 1 1 .440 g sample = 30.046 = 30.0 kJ/g sample
  3. If water is lost from the calorimeter, the heat capacity of the calorimeter decreases.

Hess’s Law (Section 5.6)

5.61  Yes, because internal energy is a state function. Hess’s Law works for any state function.

5.62

  1. Analyze/Plan .  Arrange the reactions so that in the overall sum, B appears in both reactants and products and can be canceled. This is a general technique for using Hess’s Law. Solve .

    A B Δ H = + 30 kJ B C _ Δ H = + 6 0 kJ _ A C Δ H = + 90 kJ

    Check .  The process of A forming C can be described as A forming B and B forming C.

  2. image

5.63 Analyze/Plan .  Follow the logic in Sample Exercise 5.8. Manipulate the equations so that “unwanted” substances can be canceled from reactants and products. Adjust the corresponding sign and magnitude of ΔH. Solve .

P 4 O 6 (s) P 4 (s)+3 O 2 (g) P 4 (s)+5 O 2 (g) P 4 O 10 (s) ΔH=  1640 .1 kJ ΔH=-2940 .1 kJ _ P 4 O 6 (s)+2 O 2 (g) P 4 O 10 (s) ΔH=- 1300 .0 kJ

Check .  We have obtained the desired reaction.

5.64 2 C(s) + O 2 (g) + 4 H 2 ( g ) 2 CH 3 OH ( g ) 2 CO(g) O 2 ( g ) + 2 C ( s ) Δ H = 402.4 kJ Δ H = 221.0 kJ _ 2 CO(g) + 4 H 2 (g) 2 CH 3 OH ( g ) Δ H = 181.4 kJ CO ( g ) + 2 H 2 ( g ) CH 3 OH ( g ) Δ H = ( 181.4 ) / 2 = 90.7 kJ

5.65 Analyze/Plan .  Follow the logic in Sample Exercise 5.9. Manipulate the equations so that “unwanted” substances can be canceled from reactants and products. Adjust the corresponding sign and magnitude of ΔH. Solve .

C 2 H 4 ( g ) 2 H 2 ( g ) + 2 C(s) 2 C(s) + 4 F 2 ( g ) 2 CF 4 ( g ) 2 H 2 ( g ) + 2 F 2 ( g ) 4 HF(g) Δ H = 52.3 kJ Δ H = 2 ( 680 kJ) Δ H = 2 ( 537 kJ) _ C 2 H 4 ( g ) + 6 F 2 ( g ) 2 CF 4 ( g ) + 4 HF(g) Δ H = 2 .49 × 10 3 kJ

Check .  We have obtained the desired reaction.


5.66 N 2 O ( g ) N 2 ( g ) + 1/2 O 2 (g) NO 2 (g) NO(g) + 1/2 O 2 (g) N 2 ( g ) + O 2 ( g ) 2 NO(g) Δ H = 1 / 2 ( 163.2 kJ) Δ H = 1 / 2 ( 113.1 kJ) Δ H = 180.7 kJ _ N 2 O ( g ) + NO 2 ( g ) 3 NO ( g ) Δ H = 155.7 kJ

Enthalpies of Formation (Section 5.7)

5.67

  1. Standard conditions for enthalpy changes are usually P = 1 atm and T = 298 K. For the purpose of comparison, standard enthalpy changes, ΔHº, are tabulated for reactions at these conditions.
  2. Enthalpy of formation , ΔH f , is the enthalpy change that occurs when a compound is formed from its component elements.
  3. Standard enthalpy of formation , Δ H f o , is the enthalpy change that accompanies formation of 1 mole of a substance from elements in their standard states.

5.68

  1. The standard enthalpy of formation for any element in its standard state is zero. Elements in their standard states are the reference point for the enthalpy of formation scale.
  2. 12 C(s) + 11 H 2 (g) + 11/2 O 2 (g) → C 12 H 22 O 11 (s)

5.69

  1. 1/2 N 2 (g) + O 2 (g) → NO 2 (g) Δ H f o = 33.84 kJ
  2. S(s) + 3/2 O 2 (g) → SO 3 (g) Δ H f o = 395.2 kJ
  3. Na(s) + 1/2 Br 2 (l) → NaBr(s) Δ H f o = 361.4 kJ
  4. Pb(s) + N 2 ( g ) + 3 O 2 ( g ) Pb ( NO 3 ) 2 ( s ) Δ H f o = 451.9 kJ

5.70

  1. H 2 (g) + O 2 (g) → H 2 O 2 (g) Δ H f o = 136.10 kJ
  2. Ca(s) + C(s) + 3/2 O 2 (g) → CaCO 3 (s) Δ H f o = 1207.1 kJ
  3. 1/4 P 4 (s) + 1/2 O 2 (g) + 3/2 Cl 2 (g) → POCl 3 (l) Δ H f o = 597.0 kJ
  4. 2 C(s) + 3 H 2 (g) + 1/2 O 2 (g) → C 2 H 5 OH(l) Δ H f o = 277.7 kJ

5.71 Plan . Δ H rxn o = Σ n Δ H f o (products) Σ n Δ H f o (reactants) .

Be careful with coefficients, states, and signs. Solve .

Δ H rxn o = Δ H f o Al 2 O 3 ( s ) + 2 Δ H f o Fe(s) Δ H f o Fe 2 O 3 ( s ) 2 Δ H f o Al(s) Δ H rxn o = ( 1669.8 kJ) + 2 ( 0 ) ( 822.16 kJ) 2 ( 0 ) = 847.6 kJ

5.72  Use heats of formation to calculate ΔHº for the combustion of butane.

C 3 H 8 ( g ) + 5 O 2 ( g ) 3 CO 2 ( g ) + 4 H 2 O(l) Δ H rxn o = 3 Δ H f o CO 2 ( g ) + 4 Δ H f o H 2 O(l) Δ H f o C 3 H 8 ( g ) 5 Δ H f o O 2 ( g ) Δ H rxn o = 3 ( 393.5 kJ) + 4( 285.83 kJ) ( 103.85 kJ) 5 ( 0 ) = 2219.97 = 2220.0 kJ/mol C 3 H 8 10.00 g C 3 H 8 × 1 mol C 3 H 8 4 4 .096 g C 3 H 8 × 2219.97 kJ 1 mol C 3 H 8 = 503.4 kJ


5.73 Plan . Δ H rxn o = Σ n Δ H f o (products) Σ n Δ H f o (reactants) . Be careful with coefficients, states, and signs. Solve .

  1. ΔH rxn o = 2 ΔH f o SO 3 (g) 2 ΔH f o SO 2 (g) ΔH f o O 2 (g) = 2 ( –395 .2 kJ ) – 2 ( –296 .9 kJ ) – 0 = –196 .6 kJ
  2. Δ H rxn o = Δ H f o MgO(s) + Δ H f o H 2 O(l) Δ H f o Mg(OH) 2 ( s ) = –601 .8 kJ + ( –285 .83 kJ ) ( –924 .7 kJ ) = 37 .1 kJ
  3. Δ H rxn o = 4 Δ H f o H 2 O ( g ) + Δ H f o N 2 (g) Δ H f o N 2 O 4 ( g ) 4 Δ H f o H 2 ( g ) = 4 ( –241 .82 kJ ) + 0 – ( 9 .66 kJ ) – 4 ( 0 ) = –976 .94 kJ
  4. Δ H rxn = Δ H f SiO 2 ( s ) + 4 Δ H f HCl(g) Δ H f SiCl 4 ( l ) 2 Δ H f H 2 O ( l ) = –910 .9 kJ + 4 ( –92 .30 kJ ) ( –640 .1 kJ ) – 2 ( –285 .83 kJ ) = –68 .3 kJ

5.74

  1. Δ H rxn o = Δ H f o CaCl 2 ( s ) + Δ H f o H 2 O(g) Δ H f o CaO ( s ) 2 Δ H f o HCl ( g ) = 795 .8 kJ + (-241 .82 kJ) ( 635 .5 kJ)-2(-92 .30 kJ) = –217 .5 kJ
  2. Δ H rxn o = 2 Δ H f o Fe 2 O 3 ( s ) 4 Δ H f o FeO ( s ) Δ H f o O 2 ( g ) = 2 ( 822. 16 kJ) 4( 271 .9 kJ) ( 0 ) = 556 .7 kJ
  3. Δ H rxn = Δ H f Cu 2 O ( s ) + Δ H f NO 2 (g) 2 Δ H f CuO ( s ) Δ H f NO ( g ) = –170 .7 kJ + ( 33 .84 kJ ) 2( 156 .1 kJ) ( 90 .37 kJ ) = 85 .0 kJ
  4. Δ H rxn o = 2 Δ H f o N 2 H 4 ( g ) + 2 Δ H f o H 2 O(l) 4 Δ H f o NH 3 ( g ) Δ H f o O 2 ( g ) = 2 ( 95 .40 kJ ) + 2 ( –285 .83 kJ ) – 4 ( –46 .19 kJ ) ( 0 ) = 196 .10 kJ

5.75 Analyze .  Given: combustion reaction, enthalpy of combustion, enthalpies of formation for most reactants and products. Find: enthalpy of formation for acetone.

Plan . Rearrange the expression for enthalpy of reaction to calculate the desired enthalpy of formation. Solve .

Δ H rxn = 3 Δ H f CO 2 ( g ) + 3 Δ H f H 2 O(l) Δ H f C 3 H 6 O(l) 4 Δ H f O 2 ( g ) 1790 kJ = 3 ( 393.5 kJ ) + 3 ( 285.83 kJ ) Δ H f C 3 H 6 O(l) 4 ( 0 ) Δ H f C 3 H 6 O(l) = 3 ( 393.5 kJ ) + 3 ( 285.83 kJ) + 1790 kJ = 248 kJ

5.76 Δ H rxn = Δ H f Ca(OH) 2 ( s ) + Δ H f C 2 H 2 ( g ) 2 Δ H f H 2 O(l) Δ H f CaC 2 ( s )

127. 2 kJ = 986.2 kJ + 226 .77 kJ 2 ( 285.83 kJ) Δ H f CaC 2 ( s ) Δ H f for CaC 2 ( s ) = 60.6 kJ

5.77

  1. C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g) ΔH° = –5064.9 kJ
  2. Plan .  Follow the logic in Solution 5.75 and 5.76. Solve .

    Δ H rxn = 8 Δ H f CO 2 ( g ) + 9 Δ H f H 2 O(g) Δ H f C 8 H 18 ( l ) 25 / 2 Δ H f O 2 ( g ) 5064.9 kJ = 8 ( 393.5 kJ ) + 9 ( 241.82 kJ ) Δ H f C 8 H 18 ( l ) 25 / 2 ( 0 ) Δ H f C 8 H 18 ( l ) = 8 ( 393.5 kJ) + 9 ( 241.82 kJ) + 5064 .9 kJ = 259.5 kJ


5.78

  1. C 4 H 10 O(l) + 6 O 2 (g) → 4 CO 2 (g) + 5 H 2 O(l) ΔH° = –2723.7 kJ
  2. Δ H rxn = 4 Δ H f CO 2 ( g ) + 5 Δ H f H 2 O(l) Δ H f C 4 H 10 O ( l ) 6 Δ H f O 2 ( g )

    2 723.7 = 4 ( 393.5 kJ ) + 5 ( 285.83 kJ ) Δ H f C 4 H 10 O ( l ) 6 ( 0 ) Δ H f C 4 H 10 O ( l ) = 4 ( 393.5 kJ ) + 5 ( 285.83 kJ ) + 2 723.7 kJ = 2 79.45 = 279.5 kJ

5.79

  1. C 2 H 5 OH(l) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O(g)
  2. Δ H rxn = 2 Δ H f CO 2 ( g ) + 3 Δ H f H 2 O(g) Δ H f C 2 H 5 OH(l) 3 Δ H f O 2 ( g )

    = 2(–393.5 kJ) + 3(–241.82 kJ) – (–277.7 kJ) – 3(0) = –1234.76 = –1234.8 kJ

  3. Plan . The enthalpy of combustion of ethanol [from part (b)] is –1234.8 kJ/mol. Change mol to mass using molar mass, then mass to volume using density. Solve .

    1234 .76 kJ mol C 2 H 5 OH × 1 mol C 2 H 5 OH 46 .06844 g × 0 .789 g mL × 1000 mL L = 21,147 = 2 .11 × 10 4 kJ/L

    Check . (1200/50) ≈ 25; 25 × 800 ≈ 20,000

  4. Plan . The enthalpy of combustion corresponds to any of the molar amounts in the equation as written. Production of –1234.76 kJ also produces 2 mol CO 2 . Use this relationship to calculate mass CO 2 /kJ.

    2 mol CO 2 1234 .76 kJ × 44 .0095 g CO 2 mol = 0 .071284 g CO 2 /kJ emitted

    Check . The negative sign associated with enthalpy indicates that energy is emitted.

5.80

  1. CH 3 OH(l) + 3/2 O 2 (g) → CO 2 (g) + 2 H 2 O(g)
  2. Δ H rxn = Δ H f CO 2 ( g ) + 2 Δ H f H 2 O(g) Δ H f CH 3 OH(l) 3 / 2 Δ H f O 2 ( g )

    = –393.5 kJ + 2(–241.82 kJ) –(–238.6 kJ) – 3/2(0) = –638.54 = –638.5 kJ

  3. 638 .54 kJ mol CH 3 OH × 1 mol CH 3 OH 32 .04 g × 0 .791 g mL × 1000 mL L = 1 .58 × 10 4 kJ/L produced
  4. 1 mol CO 2 638 .54 kJ × 44 .0095 g CO 2 mol = 0 .06892 g CO 2 /kJ emitted

Bond Enthalpies (Section 5.8)

5.81

  1. + ΔH; energy must be supplied to separate oppositely charged ions.
  2. – ΔH; energy is released when a chemical bond is formed.
  3. + ΔH; energy must be supplied to separate a negatively charged electron from a neutral atom.
  4. + ΔH; energy must be supplied to melt a solid.

5.82

  1. – ΔH; the reactants have four N–O bonds (some of them multiple bonds) while the products have these same N–O bonds plus an N–N bond. Overall the reaction involves formation of a new chemical bond and enthalpy decreases.
  2. – ΔH; energy is released when a chemical bond is formed.
  3. – ΔH; energy is released when oppositely charged ions form ionic bonds.
  4. + ΔH; energy must be supplied to break a chemical bond.

5.83 Analyze . Given: structural formulas. Find: enthalpy of reaction.

Plan. Count the number and kinds of bonds that are broken and formed by the reaction. Use bond enthalpies from Table 5.4 and Equation 5.32 to calculate the overall enthalpy of reaction, ΔH. Solve .

  1. ΔH = D(H–H) + D(Br–Br) – 2 D(H–Br)

    = 436 kJ + 193 kJ – 2(366 kJ) = –103 kJ

  2. ΔH = 6 D(C–H) + 2 D(C–O) + 2 D(O–H) + 3 D(O=O)

    – 4 D(C=O) – 8 D(O–H)

    = 6 D(C–H) + 2 D(C–O) + 3 D(O=O) – 4 D(C=O) – 6 D(O–H)

    ΔH = 6(413) + 2(358) + 3(495) – 4(799) – 6(463) = –1295 kJ

5.84

  1. ΔH = 3 D(C–Br) + D(C–H) + D(Cl–Cl) – 3 D(C–Br) – D(C–Cl) – D(H–Cl)

    = D(C–H) + D(Cl–Cl) – D(C–Cl) – D(H–Cl)

    ΔH = 413 + 242 – 328 – 431 = –104 kJ

  2. ΔH = 4 D(C–H) + 2 D(O=O) – 2 D(C=O) – 4 D(O–H)

    ΔH = 4(413) + 2(495) – 2(799) – 4(463) = – 808 kJ

5.85

  1. Plan . Δ H rxn o = Σ n Δ H f o (products) Σ n Δ H f o (reactants). Be careful with coefficients, states, and signs. Solve .

    Δ H rxn o = 2 Δ H f o Br ( g ) Δ H f o Br 2 ( g )

    = 2(111.8) – 30.71 = 192.9 kJ

    This reaction is just the breaking of a Br–Br single bond to form Br atoms; reactants and products are all in the gas phase. The enthalpy of reaction represents the bond enthalpy D(Br–Br), 193 kJ.

  2. The value of D(Br–Br) in Table 5.4 is 193 kJ, the same as the enthalpy calculated in part (a). The difference between the two values is zero (to three significant figures).

5.86

  1. The relevant reaction is N 2 (g) → 2 N(g).

    Δ H rxn o = 2 Δ H f o N ( g ) Δ H f o N 2 ( g ) = 2 ( 472 .7 ) – 0 = 945 .4 kJ

    Our estimate for D(N≡N) is 945.4 kJ = 945 kJ.


  1. Calculate the overall enthalpy change for the reaction using standard enthalpies of formation. Use this value for ΔH and bond enthalpies from Table 5.4 to estimate the enthalpy of the nitrogen-nitrogen bond in N 2 H 4 .

    Δ H rxn o = 2 Δ H f o NH 3 ( g ) Δ H f o N 2 H 4 ( g ) Δ H f o H 2 ( g ) = 2(–46 .19) – 95 .40 – 0 = –187 .78 = –188 kJ

    ΔH = 4 D(N–H) + D(N–N) + D(H–H) – 6 D(N–H)

    ΔH = D(N–N) + D(H–H) – 2 D(N–H)

    D(N–N) = ΔH – D(H–H) + 2 D(N–H)

    D(N–N) = (–188) – (436) + 2(391) = 158 kJ

  1. The nitrogen-nitrogen bond in N 2 H 4 has an enthalpy of 158 kJ; in N 2 the enthalpy is 945 kJ. We are comparing the same pair of bonded atoms, so it is safe to say that the bond in N 2 H 4 is weaker than the bond in N 2 .

5.87

  1. ΔH = 2 D(H–H) + D(O=O) – 4 D(O–H) = 2(436) + 495 – 4(463) = –485 kJ
  2. The estimate from part (a) is less negative or larger than the true reaction enthalpy. When we use bond enthalpies to estimate reaction enthalpies, we assume all reactants and products are gases. We have estimated the enthalpy change for production of H 2 O(g). Because condensation, [(g) → (l)], is exothermic, we expect ΔH for production of liquid water to be more negative or smaller than the value we estimated in part (a).
  3. Δ H rxn o = 2 Δ H f o H 2 O ( l ) 2 Δ H f o H 2 ( g ) Δ H f o O 2 ( g ) = 2 ( – 285 .83 ) – 2 ( 0 ) – 0 = –571 .66 = – 572 kJ

    As predicted in part (b), the true enthalpy of reaction is more negative than the result calculated using bond enthalpies.

5.88

  1. ΔH = D(H–H) + D(I–I) – 2 D(H–I) = 436 + 151 – 2(299) = – 11 kJ
  2. When we use bond enthalpies to estimate reaction enthalpies, we assume all reactants and products are gases. In this case, one of the reactants, I 2 , is a solid. Because sublimation, [(s) → (g)], is endothermic, the actual starting reactants have a lower enthalpy than gas phase reactants. Energy must be supplied to get to the estimated starting point. The actual enthalpy change for the reactants as written will be more endothermic (more positive, larger) than the value we estimated in part (a).
  3. Δ H rxn o = 2 Δ H f o HI ( g ) Δ H f o H 2 ( g ) Δ H f o I 2 ( s ) = 2 ( 25 .94 ) – 0 – 0 = 51 .88 = 52 kJ

Foods and Fuels (Section 5.9)

5.89

  1. Fuel value is the amount of energy produced when 1 gram of a substance (fuel) is combusted.
  2. The fuel value of fats is 9 kcal/g and of carbohydrates is 4 kcal/g. Therefore, 5 g of fat produce 45 kcal, whereas 9 g of carbohydrates produce 36 kcal; 5 g of fat are a greater energy source.
  3. These products of metabolism are expelled as waste, H 2 O(l) primarily in urine and feces, and CO 2 (g) as gas when breathing.

5.90

  1. One gram of fat produces more energy than one gram of carbohydrates when metabolized.
  2. For convenience, assume 100 g of chips.

    12 g protein × 17 kJ 1 g protein × 1 Cal 4 .184 kJ = 48.76 = 49 Cal

    14 g fat × 38 kJ 1 g fat × 1 Cal 4 .184 kJ = 127.15 = 130 Cal

    74 g carbohydrates × 17 kJ 1 g carbohydrates × 1 Cal 4 .184 kJ = 300.67 = 301 Cal

    total Cal = (48.76 + 127.15 + 300.67) = 476.58 = 480 Cal

    % Cal from fat = 127 .15 Cal fat 476 .58 total Cal × 100 = 26.68 = 27 %

    (Because the conversion from kJ to Cal was common to all three components, we would have determined the same percentage by using kJ.)

  3. 25 g fat × 38 kJ g fat = x g protein × 17 kJ g protein ; x = 56 g protein

5.91

  1. Plan .  Calculate the Cal (kcal) from each nutritional component of the soup, then sum. Solve .

    2. 5 g fat × 38 kJ 1 g fat = 95.0 or 0 .95 × 10 2 kJ

    1 4 g carbohydrates × 17 kJ 1 g carbohydrate = 238 or 2 .4 × 10 2 kJ

    7 g protein × 17 kJ 1 g protein = 119 or 1 × 10 2 kJ

    total energy = 95.0 kJ + 238 kJ + 119 kJ = 452 or 5 × 10 2 kJ

    4 52 kJ × 1 kcal 4 .184 kJ × 1 Cal 1kcal = 108.03 or 1 × 10 2 Cal/serving

    Check .  100 Cal/serving is a reasonable result; units are correct. The data and the result have 1 sig fig.

  2. Sodium does not contribute to the calorie content of the food, because it is not metabolized by the body; it enters and leaves as Na + .

5.92  Calculate the fuel value in a pound of M&M® candies.

96 g fat × 38 kJ 1 g fat = 3648 kJ = 3 .6 × 10 3 kJ

320 g carbohydrate × 17 kJ 1 g carbohydrate = 5440 kJ = 5 .4 × 10 3 kJ

21 g protein × 17 kJ 1 g protein = 357 kJ = 3 .6 × 10 2 kJ


total fuel value = 3648 kJ + 5440 kJ + 357 kJ = 9445 kJ = 9.4 × 10 3 kJ/lb

9445 kJ lb × 1 lb 453 .6 g × 42 g serving = 874.5 kJ = 8 .7 × 10 2 kJ/serving

874.5 kJ serving × 1 kcal 4 .184 kJ × 1 Cal 1 kcal = 209.0 Cal = 2 .1 × 10 2 Cal/serving

Check .  210 Cal is the approximate food value of a candy bar, so the result is reasonable.

5.93 Plan .  g → mol → kJ → Cal Solve .

16.0 g C 6 H 12 O 6 × 1 mol C 6 H 12 O 6 180.2 g C 6 H 12 O 6 × 2812 kJ mol C 6 H 12 O 6 × 1 Cal 4 .184 kJ = 59.7 Cal

Check .  60 Cal is a reasonable result for most of the food value in an apple.

5.94 177 mL × 1 .0 g wine 1 mL × 0.106 g ethanol 1 g wine × 1 mol ethanol 46 .1 g ethanol × 1367 kJ 1 mol ethanol × 1 Cal 4 .184 kJ = 133 = 1.3 × 10 2 Cal

Check .  A “typical” 6 oz. glass of wine has 150–250 Cal, so this is a reasonable result. Note that alcohol is responsible for most of the food value of wine.

5.95 Plan .  Use enthalpies of formation to calculate molar heat (enthalpy) of combustion using Hess’s Law. Use molar mass to calculate heat of combustion per kg of hydrocarbon. Solve .

Propyne: C 3 H 4 (g) + 4 O 2 (g) → 3 CO 2 (g) + 2 H 2 O(g)

  1. ΔH rxn o = 3 ( –393 .5 kJ ) + 2 ( –241 .82 kJ ) ( 185 .4 kJ ) – 4 ( 0 ) = –1849 .5 = –1850 kJ/molC 3 H 4
  2. 1849.5 kJ 1 mol C 3 H 4 × 1 mol C 3 H 4 40.065 g C 3 H 4 × 1000 g C 3 H 4 1 kg C 3 H 4 = 4.616 × 10 4 kJ/kg C 3 H 4

Propylene: C 3 H 6 (g) + 9/2 O 2 (g) → 3 CO 2 (g) + 3 H 2 O(g)

  1. ΔH rxn o =3 ( –393 .5 kJ ) + 3 ( –241 .82 kJ ) ( 20 .4 kJ ) –9/2 ( 0 ) = –1926 .4 = –1926 kJ/molC 3 H 6
  2. 1926.4 kJ 1 mol C 3 H 6 × 1 mol C 3 H 6 42.080 g C 3 H 6 × 1000 g C 3 H 6 1 kg C 3 H 6 = 4.578 × 10 4 kJ/kg C 3 H 6

Propane: C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(g)

  1. Δ H rxn = 3 ( –393 .5 kJ ) + 4 ( –241 .82 kJ ) ( –103 .8 kJ ) –5 ( 0 ) = –2044 .0 = –2044 kJ/molC 3 H 8
  2. 2044.0 kJ 1 mol C 3 H 8 × 1 mol C 3 H 8 44.096 g C 3 H 8 × 1000 g C 3 H 8 1 kg C 3 H 8 = 4.635 × 10 4 kJ/kg C 3 H 8
  3. These three substances yield nearly identical quantities of heat per unit mass, but propane is marginally higher than the other two.

5.96 Δ H rxn = Δ H f CO 2 ( g ) + 2 Δ H f H 2 O(g) Δ H f CH 4 ( g ) 2 Δ H f O 2 ( g ) = –393 .5 kJ + 2 ( –241 .82 kJ ) ( –74 .8 kJ ) – 2 ( 0 ) kJ = –802 .3 kJ

Δ H rxn = Δ H f CF 4 ( g ) + 4 Δ H f HF(g) Δ H f CH 4 ( g ) 4 Δ H f F 2 ( g ) = –679 .9 kJ + 4 ( –268 .61 kJ ) ( –74 .8 kJ ) – 4 ( 0 ) kJ = –1679 .5 kJ

The second reaction is twice as exothermic as the first. The “fuel values” of hydrocarbons in a fluorine atmosphere are approximately twice those in an oxygen atmosphere. Note that the difference in ΔH° values for the two reactions is in the Δ H f º for the products, because the Δ H f º for the reactants is identical.

5.97 Analyze/Plan. Given population, Cal/person/day and kJ/mol glucose, calculate kg glucose/yr. Calculate kJ/yr, then kg/yr. 1 billion = 1 × 10 9 . 365 day = 1 yr. 1 Cal = 1 kcal, 4.184 kJ = 1 kcal = 1 Cal. Solve.

7.0 × 10 9 persons × 1500 Cal person-day × 3 65 day 1 yr × 4 .184 kJ 1 Cal = 1.6035 × 10 16 = 1.6 × 10 16 kJ/yr

1. 6035 × 10 16 kJ yr × 1 mol C 6 H 12 O 6 2803 kJ × 180.2 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 × 1 kg 1000 g = 1. 0 × 10 12 kg C 6 H 12 O 6 /yr

Check. 1 × 10 12 kg is 1 trillion kg of glucose.

5.98

  1. Use density to change L to g, molar mass to change g to mol, heat of combustion to change mol to kJ. Ethanol is C 2 H 5 OH, gasoline is C 8 H 18 . From Exercise 5.79 (c), heat of combustion of ethanol is –1234.8 kJ/mol.

    1. 0 L C 2 H 5 OH × 1000 mL 1 L × 0. 79 g 1 mL × 1 mol C 2 H 5 OH 4 6 .07 g × 1 234 .8 kJ 1 mol C 2 H 5 OH = 21,174 = 2 .1 × 10 4 kJ/LC 2 H 5 OH

    1. 0 L C 8 H 18 × 1000 mL 1 L × 0. 70 g 1 mL × 1 mol C 8 H 18 1 14 .23 g C 8 H 18 × 5 400 kJ 1 mol C 8 H 18 = 33 , 091 = 3.3 × 10 4 kJ/L C 8 H 18

  2. If density and heat of combustion of E85 are weighted averages of the values for the pure substances, than energy per liter E85 is also a weighted average of energy per liter for the two substances.

    kJ/L E85 = 0.15(kJ/L C 8 H 18 ) + 0.85(kJ/L C 2 H 5 OH)

    kJ/L E85 = 0.15(33,091 kJ) + 0.85(21,174 kJ) = 22,962 = 2.3 × 10 4 kJ/L E85

  3. Whether comparing gal or L, all conversion factors for the two fuels cancel, so we can apply the energy ratio directly to the volume under consideration.

    The energy ratio for E85 to gasoline is (22,962/33,091) = 0.6939 = 0.69

    10 gal gas × kJ from E85 0 .6939 kJ from gas = 14.41 = 14 gal E85
  4. If the E85/gasoline energy ratio is 0.69, the cost ratio must be 0.69 or less to “break-even” on price. 0.69($3.88) = $2.68/gal E85

    Check. 10 gal gas($3.88/gal) = $39; 14.4 gal E85($2.68/gal) = $39.

Additional Exercises

5.99  Like the combustion of H 2 (g) and O 2 (g) described in Section 5.4, the reaction that inflates airbags is spontaneous after initiation. Spontaneous reactions are usually exothermic, –ΔH. The airbag reaction occurs at constant atmospheric pressure, ΔH = q p ; both are likely to be large and negative. When the bag inflates, work is done by the system on the surroundings, so the sign of w is negative.

5.100  Freezing is an exothermic process (the opposite of melting, which is clearly endothermic). When the system, the soft drink, freezes, it releases energy to the surroundings, the can. Some of this energy does the work of splitting the can. (When water freezes, it expands. It is specifically this expansion that does the work of splitting the can.)

5.101

  1. No work is done when the gas expands.
  2. No work is done because the evacuated flask is truly empty. There is no surrounding substance to be “pushed back.”
  3. ΔE = q + w. From part (b), no work is done when the gas expands. The flasks are perfectly insulated, so no heat flows. ΔE = 0 + 0 = 0. The answer is a bit surprising, because a definite change occurred that required no work or heat transfer and consequently involved no energy change.

5.102

  1. q = 0, w > 0 (work done to system), ΔE > 0
  2. Because the system (the gas) is losing heat, the sign of q is negative.

    Two interpretations of the final state in (b) are possible. If the final state in (b) is identical to the final state in (a), ΔE(a) = ΔE(b). If the final volumes are identical, case (b) requires either more (non-PV) work or heat input to compress the gas because some heat is lost to the surroundings. (The moral of this story is that the more energy lost by the system as heat, the greater the work on the system required to accomplish the desired change.)

    Alternatively, if w is identical in the two cases and q is negative for case (b), then ΔE(b) < ΔE(a). Assuming identical final volumes, the final temperature and pressure in (b) are slightly lower than those values in (a).

5.103  ΔE = q + w = +38.95 kJ – 2.47 kJ = +36.48 kJ

ΔH = q p = +38.95 kJ

5.104  If a function sometimes depends on path, then it is simply not a state function. Enthalpy is a state function, so ΔH for the two pathways leading to the same change of state pictured in Figure 5.10 must be the same. However, q is not the same for the both. Our conclusion must be that ΔH ≠ q for these pathways. The condition for ΔH = q p (other than constant pressure) is that the only possible work on or by the system is pressure-volume work. Clearly, the work being done in this scenario is not pressure-volume work, so ΔH ≠ q, even though the two changes occur at constant pressure.

5.105  Find the heat capacity of 1.7 × 10 3 gal H 2 O.

C H 2 O = 1.7 × 10 3 gal H 2 O × 4 qt 1 gal × 1 L 1 .057 qt × 1 × 10 3 cm 3 1 L × 1 g 1 cm 3 × 4.184 J 1 g-°C = 2.692 × 10 7 J/ C = 2.7 × 10 4 kJ / C; than,


Additional Exercises

5.99  Like the combustion of H 2 (g) and O 2 (g) described in Section 5.4, the reaction that inflates airbags is spontaneous after initiation. Spontaneous reactions are usually exothermic, –ΔH. The airbag reaction occurs at constant atmospheric pressure, ΔH = q p ; both are likely to be large and negative. When the bag inflates, work is done by the system on the surroundings, so the sign of w is negative.

5.100  Freezing is an exothermic process (the opposite of melting, which is clearly endothermic). When the system, the soft drink, freezes, it releases energy to the surroundings, the can. Some of this energy does the work of splitting the can. (When water freezes, it expands. It is specifically this expansion that does the work of splitting the can.)

5.101

  1. No work is done when the gas expands.
  2. No work is done because the evacuated flask is truly empty. There is no surrounding substance to be “pushed back.”
  3. ΔE = q + w. From part (b), no work is done when the gas expands. The flasks are perfectly insulated, so no heat flows. ΔE = 0 + 0 = 0. The answer is a bit surprising, because a definite change occurred that required no work or heat transfer and consequently involved no energy change.

5.102

  1. q = 0, w > 0 (work done to system), ΔE > 0
  2. Because the system (the gas) is losing heat, the sign of q is negative.

    Two interpretations of the final state in (b) are possible. If the final state in (b) is identical to the final state in (a), ΔE(a) = ΔE(b). If the final volumes are identical, case (b) requires either more (non-PV) work or heat input to compress the gas because some heat is lost to the surroundings. (The moral of this story is that the more energy lost by the system as heat, the greater the work on the system required to accomplish the desired change.)

    Alternatively, if w is identical in the two cases and q is negative for case (b), then ΔE(b) < ΔE(a). Assuming identical final volumes, the final temperature and pressure in (b) are slightly lower than those values in (a).

5.103  ΔE = q + w = +38.95 kJ – 2.47 kJ = +36.48 kJ

ΔH = q p = +38.95 kJ

5.104  If a function sometimes depends on path, then it is simply not a state function. Enthalpy is a state function, so ΔH for the two pathways leading to the same change of state pictured in Figure 5.10 must be the same. However, q is not the same for the both. Our conclusion must be that ΔH ≠ q for these pathways. The condition for ΔH = q p (other than constant pressure) is that the only possible work on or by the system is pressure-volume work. Clearly, the work being done in this scenario is not pressure-volume work, so ΔH ≠ q, even though the two changes occur at constant pressure.

5.105  Find the heat capacity of 1.7 × 10 3 gal H 2 O.

C H 2 O = 1.7 × 10 3 gal H 2 O × 4 qt 1 gal × 1 L 1 .057 qt × 1 × 10 3 cm 3 1 L × 1 g 1 cm 3 × 4.184 J 1 g-°C = 2.692 × 10 7 J/ C = 2.7 × 10 4 kJ / C; than,


  1. 2.692 × 10 7 J 1 C o × 1 g- C o 0 .85 J × 1 kg 1 × 10 3 g × 1 brick 1 .8 kg = 1.8 × 10 4 or 18,000 bricks

    Check .  (1.7 × ~16 × 10 6 )/(~1.6 × 10 3 ) ≈ 17 × 10 3 bricks; the units are correct.

5.106

  1. q Cu = 0.385 J g-K × 121.0 g Cu × ( 30.1 o C 100.4 o C) = 3274.9 = 3.27 × 10 3 J

    The negative sign indicates the 3.27 × 10 3 J are lost by the Cu block.

  2. q H 2 O = 4.184 J g-K × 150.0 g H 2 O × ( 30.1 o C 25.1 o C) = 3138 = 3.1 × 10 3 J

    The positive sign indicates that 3.14 × 10 3 J are gained by the H 2 O.

  3. The difference in the heat lost by the Cu and the heat gained by the water is 3.275 × 10 3 J – 3.138 × 10 3 J = 0.137 × 10 3 J = 1 × 10 2 J. The temperature change of the calorimeter is 5.0 o C. The heat capacity of the calorimeter in J/K is

    0.137 × 10 3 J × 1 5.0 o C = 27.4 = 3 × 10 1 J/K .

    Because q H 2 O is known to 1 decimal place, the difference has 1 decimal place and the result has 1 sig fig.

    If the rounded results from (a) and (b) are used,

    C calorimeter = 0.2 × 10 3 J 5.0 o C = 4 × 10 1 J/K .
  4. q H 2 O = 3.275 × 10 3 J = 4.184 J g-K × 150.0 g × ( Δ T)

    ΔT = 5.22 °C; T f = 25.1 °C + 5.22 °C = 30.3 °C

5.107

  1. From the mass of benzoic acid that produces a certain temperature change, we can calculate the heat capacity of the calorimeter.

    0.235 g ben z oic acid 1.642 C o change observed × 26.38 kJ 1 g benzoic acid = 3.7755 = 3.78 kJ/ C o

    Now we can use this experimentally determined heat capacity with the data for caffeine.

    1.525 C o rise 0.265 g caffeine × 3.7755 kJ 1 C o × 194.2 g caffeine 1 mol caffeine = 4.22 × 10 3 kJ/mol caffeine
  2. The overall uncertainty is approximately equal to the sum of the uncertainties due to each effect. The uncertainty in the mass measurement is 0.001/0.235 or 0.001/0.265, about 1 part in 235 or 1 part in 265. The uncertainty in the temperature measurements is 0.002/1.642 or 0.002/1.525, about 1 part in 820 or 1 part in 760. Thus the uncertainty in heat of combustion from each measurement is

    4220 235 = 18 kJ; 4220 265 = 16 kJ; 4220 820 = 5 kJ; 4220 760 = 6 kJ

    The sum of these uncertainties is 45 kJ. In fact, the overall uncertainty is less than this because independent errors in measurement do tend to partially cancel.


5.108

  1. Mg(s) + 2 H 2 O(l) → Mg(OH) 2 (s) + H 2 (g) ΔH rxn = ΔH f Mg(OH) 2 (s) + ΔH f o H 2 (g) 2 ΔH f H 2 O(l) ΔH f Mg(s)

    = –924.7 kJ + 0 – 2(°285.83 kJ) – 0 = –353.04 = –353.0 kJ

  2. Use the specific heat of water, 4.184 J/g-°C, to calculate the energy required to heat the water. Use the density of water at 25 °C to calculate the mass of H 2 O to be heated. (The change in density of H 2 O going from 21 °C to 79 °C does not substantially affect the strategy of the exercise.) Then use the ‘heat stoichiometry’ in (a) to calculate mass of Mg(s) needed.

    75 mL × 0.997 g H 2 O mL × 4.184 J g- o C × 58 o C × 1 kJ 1000 J = 18.146 kJ = 18 kJ required

    1 8.146 kJ × 1 mol Mg 353 .04 kJ × 24 .305 g Mg 1 mol Mg = 1 .249 g = 1 .2 g Mg needed

5.109

  1. For comparison, balance the equations so that 1 mole of CH 4 is burned in each.

    CH 4 (g) + O 2 (g) → C(s) + 2 H 2 O(l)

    CH 4 (g) + 3/2 O 2 (g) → CO(g) + 2 H 2 O(l)

    CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)

  2. ΔH rxn = ΔH f C (s) + 2 ΔH f H 2 O(l) ΔH f CH 4 ( g ) ΔH f O 2 ( g ) = 0 + 2 ( 285.83 kJ ) ( 74.8 ) 0 = 496.9 KJ ΔH rxn = ΔH f CO(g) + 2 ΔH f H 2 O(l) ΔH f CH 4 ( g ) 3 / 2 ΔH f O 2 ( g ) = ( 110.5 kJ ) + 2 ( 285.83 KJ ) ( 74.8 KJ ) 3 / 2 ( 0 ) = 607.4 kJ ΔH rxn = ΔH f CO 2 (g) + 2 ΔH f H 2 O(l) ΔH f CH 4 ( g ) 2 ΔH f O 2 ( g ) = = 393.5 kJ + 2( 285.83 kJ) ( 74 .8 kJ) 2 ( 0 ) = 890.4 kJ
  3. Assuming that O 2 (g) is present in excess, the reaction that produces CO 2 (g) represents the most negative ΔH per mole of CH 4 burned. More of the potential energy of the reactants is released as heat during the reaction to give products of lower potential energy. The reaction that produces CO 2 (g) is the most “downhill” in enthalpy.

5.110

2 [ CH 4 ( g ) + 2 O 2 ( g ) CO 2 ( g ) + 2 H 2 O ( l ) ] C 2 H 6 ( g ) C 2 H 4 ( g ) + H 2 (g) 1 / 2 [ 4 CO 2 ( g ) + 6 H 2 O(l) 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ] 1 / 2 [2 H 2 O(l) 2 H 2 ( g ) + O 2 ( g ) ] Δ H ° = 2 ( 890.3 kJ) Δ H ° = 136.3 kJ Δ H ° = 1 / 2 ( 3120.8 kJ) Δ H ° = 1 / 2 ( 571.6 kJ) ¯ 2 CH 4 ( g ) C 2 H 4 ( g ) + 2 H 2 (g) Δ H ° = 201.9 kJ


5.111 For nitroethane:

1368 kJ 1 mol C 2 H 5 NO 2 × 1 mol C 2 H 5 NO 2 75.072 g C 2 H 5 NO 2 × 1.052 g C 2 H 5 NO 2 1 cm 3 = 19.17 kJ/cm 3

For ethanol:

1367 kJ 1 mol C 2 H 5 OH × 1 mol C 2 H 5 OH 46.069 g C 2 H 5 OH × 0.789 g C 2 H 5 OH 1 cm 3 = 23.4 kJ/cm 3

For methylhydrazine:

1307 kJ 1 mol CH 6 N 2 × 1 mol CH 6 N 2 46.072 g CH 6 N 2 × 0.874 g CH 6 N 2 1 cm 3 = 24.8 kJ/cm 3

Thus, methylhydrazine would provide the most energy per unit volume, with ethanol a close second.

5.112

  1. 3 C 2 H 2 (g) → C 6 H 6 (l)

    Δ H rxn = Δ H f C 6 H 6 ( l ) 3 Δ H f C 2 H 2 ( g ) = 49.0 kJ 3 ( 226.77 kJ) = 631.31 = 631.3 kJ

  2. Because the reaction is exothermic (ΔH is negative), the reactant, 3 moles of C 2 H 2 (g), has more enthalpy than the product, 1 mole of C 6 H 6 (l).
  3. The fuel value of a substance is the amount of heat (kJ) produced when 1 gram of the substance is burned. Calculate the molar heat of combustion (kJ/mol) and use this to find kJ/g of fuel.

    C 2 H 2 (g) + 5/2 O 2 (g) → 2 CO 2 (g) + H 2 O(l)

    ΔH rxn = 2 Δ H f CO 2 ( g ) + Δ H f H 2 O(l) Δ H f C 2 H 2 ( g ) 5 / 2 Δ H f O 2 ( g ) = 2 ( 393.5 kJ ) + ( 285.83 kJ ) 226.77 kJ 5 / 2 ( 0 ) = 1299.6 kJ/mol C 2 H 2

    3267.5 kJ 1 mol C 6 H 6 × 1 mol C 6 H 6 78.114 g C 6 H 6 = 41.830 = 42 kJ/g C 6 H 6

    C 6 H 6 ( 1 ) + 15 / 2 O 2 ( g ) 6 CO 2 ( g ) + 3 H 2 O(1)

    Δ H rxn = 6 Δ H f CO 2 ( g ) + 3 Δ H f H 2 O(1) C 6 H 6 ( 1 ) 15 / 2 Δ H f O 2 ( g ) = 6 ( 36.5 kJ ) + 3 ( 285.83 kJ ) 49.0 kJ 15 / 2 ( 0 ) = 3267.5 kJ/mol C 6 H 6

    3267.5 kJ 1 mol C 6 H 6 × 1 mol C 6 H 6 78.114 g C 6 H 6 = 41.830 = 42 kJ/g C 6 H 6

5.113  The reaction for which we want ΔH is:

4 NH 3 (l) + 3 O 2 (g) → 2 N 2 (g) + 6 H 2 O(g)

Before we can calculate ΔH for this reaction, we must calculate ΔH f for NH 3 (l).

We know that ΔH f for NH 3 (g) is –46.2 kJ/mol, and that for NH 3 (l) → NH 3 (g), ΔH = 23.2 kJ/mol

Thus, ΔH vap = ΔH f NH 3 (g) – ΔH f NH 3 (l).

23.2 kJ = –46.2 kJ –ΔH f NH 3 (l); ΔH f NH 3 (l) = –69.4 kJ/mol


Then for the overall reaction, the enthalpy change is:

ΔH rxn = 6 ΔH f H 2 O(g) + 2 ΔH f N 2 (g) –4 ΔH f NH 3 (l) –3 ΔH f O 2

= 6(–241.82 kJ) + 2(0) –4(–69.4 kJ) –3(0) = –1173.3 kJ

1173.3 kJ 4 mol NH 3 × 1 mol NH 3 17.0 g NH 3 = 0.81 g NH 3 1 cm 3 × 1000 cm 3 1 L = 1.4 × 10 4 kJ L NH 3

(This result has 2 significant figures because the density is expressed to 2 figures.)

2 CH 3 OH(l) + 3 O 2 (g) → 2 CO 2 (g) + 4 H 2 O(g)

ΔH = 2(–393.5 kJ) + 4(–241.82 kJ) – 2(–239 kJ) – 3(0) = –1276 kJ

1276 kJ 2 mol CH 3 OH × 1 mol CH 3 OH 32.04 g CH 3 OH × 0.792 g CH 3 OH 1 cm 3 × 1000 cm 3 1 L = 1.58 × 10 4 kJ L CH 3 OH

In terms of heat obtained per unit volume of fuel, methanol is a slightly better fuel than liquid ammonia.

5.114 1,3-butadiene , C 4 H 6 , MM = 54.092 g/mol

  1. C 4 H 6 (g) + 11/2 O 2 (g) → 4 CO 2 (g) + 3 H 2 O(l)

    Δ H rxn = 4 Δ H f CO 2 ( g ) + 3 Δ H f H 2 O(l) Δ H f C 4 H 6 ( g ) 11 / 2 Δ H f O 2 ( g ) = 4 ( 393.5 kJ ) + 3 ( 285.83 kJ ) 111.9 kJ + 11 / 2 ( 0 ) = 2543. kJ/mol C 4 H 6

  2. 2543.4 kJ 1 mol C 4 H 6 × 1 mol C 4 H 6 54 .092 g = 47.020 47 kJ/g
  3. % H = 6 ( 1.008 ) 54.092 × 100 = 11.18 % H

1-butene , C 4 H 8 , MM = 56.108 g/mol

  1. C 4 H 8 (g) + 6 O 2 (g) → 4 CO 2 (g) + 4 H 2 O(l)

    Δ H rxn = 4 Δ H f CO 2 ( g ) + 4 Δ H f H 2 O(l) Δ H f C 4 H 8 ( g ) 6 Δ H f O 2 ( g )

    = 4(–393.5 kJ) + 4(–285.83 kJ) – 1.2 kJ – 6(0) = –2718.5 kJ/mol C 4 H 8

  2. 2718.5 kJ 1 mol C 4 H 8 × 1 mol C 4 H 8 56 .108 g C 4 H 8 = 48.451 48 kJ/g
  3. % H = 8 ( 1.008 ) 56.108 × 100 = 14.37 % H

n-butane , C 4 H 10 (g), MM = 58.124 g/mol

  1. C 4 H 10 (g) + 13/2 O 2 (g) → 4 CO 2 (g) + 5 H 2 O(l)

    Δ H rxn = 4 Δ H f CO 2 ( g) + 5 Δ H f H 2 O(l) Δ H f C 4 H 10 ( g ) 13 / 2 Δ H f O 2 ( g ) = 4 ( 393.5 kJ ) + 5 ( 285.83 kJ ) ( 124. kJ ) 13 / 2 ( 0 ) = 2718.5 kJ/mol C 4 H 10


  1. 2878.5 kJ 1 mol C 4 H 10 × 1 mol C 4 H 10 58 .124 g C 4 H 10 = 49.523 50 kJ/g
  1. % H = 10 ( 1.008 ) 58.124 × 100 = 17.34 % H
  1. It is certainly true that as the mass % H increases, the fuel value (kJ/g) of the hydrocarbon increases, given the same number of C atoms. A graph of the data in parts (b) and (c) (see below) suggests that mass % H and fuel value are directly proportional when the number of C atoms is constant.
    image

5.115  ΔE p = m×g×d. Be careful with units. 1 J = 1 kg-m 2 / s 2

201 lb × 1 kg 2 .205 lb × 9.81 m s 2 × 45 ft time × 1 yd 3 ft × 1 m 1 .0936 yd × 20 times = 2 .453 × 10 5 kg-m 2 /s 2 = 2 .453 × 10 5 J = 2 .5 × 10 2 kJ

1 Cal = 1 kcal = 4.184 kJ

2 .453 × 10 2 kJ × 1 Cal 4 .184 kJ = 58 .63 = 59 Cal

No, if all work is used to increase the man’s potential energy, 20 rounds of stair-climbing will not compensate for one extra order of 245 Cal fries. In fact, more than 59 Cal of work will be required to climb the stairs, because some energy is required to move limbs and some energy will be lost as heat.

5.116 Plan .  Use dimensional analysis to calculate the amount of solar energy supplied per m 2 in 1 h. Use stoichiometry to calculate the amount of plant energy used to produce sucrose per m 2 in 1 h. Calculate the ratio of energy for sucrose to total solar energy, per m 2 per h.

Solve .  1 W = 1 J/s, 1 kW = 1 kJ/s

1 . 0 kW m 2 = 1 . 0 kJ / s m 2 = 1 . 0 kJ m 2 -s × 6 0 s 1 min × 6 0 min 1 h = 3 . 6 × 1 0 3 kJ m 2 -h

5 6 4 5 kJ mol sucrose × 1 mol sucrose 3 4 2 . 3 gsucrose × 0 . 2 0 g sucrose m 2 -h = 3 . 2 9 8 = 3 . 3 kJ / m 2 -h for sucrose production

3.298 kJ for sucrose 3.6 × 10 3 kJ total solar × 100 = 0.092 % sunlight used to produce sucrose


5.117

  1. 6 CO 2 (g) + 6 H 2 O(l) → C 6 H 12 O 6 (s) + 6 O 2 (g), ΔH = 2803 kJ

    This is the reverse of the combustion of glucose (Section 5.8 and Exercise 5.89), so ΔH = –(–2803) kJ = +2803 kJ.

    5.5 × 10 16 g CO 2 yr × 1 mol CO 2 44.01 g CO 2 × 2803 kJ 6 mol CO 2 = 5.838 × 10 17 = 5.8 × 10 17 kJ
  2. 1 W = 1 J/s; 1 W-s = 1 J

    5 . 8 3 8 × 1 0 1 7 kJ yr × 1 0 0 0 J k J × 1 y r 3 6 5 d × 1 d 2 4 h × 1 h 6 0 min × 1 min 6 0 s × 1 W-s J × 1 MW 1 × 10 6 W = 1.851 × 10 7 MW = 1.9 × 10 7 MW

    1.9 × 10 7 MW × 1 plant 10 3 M W = 1.9 × 10 4 = 19 , 000 nuclear power plants

Integrative Exercises

5.118

  1. mi/h → m/s

    1050 mi h × 1.6093 km 1 mi × 1000 m 1 km × 1 h 3600 s = 469.38 = 469.4 m/s

  2. Find the mass of one N 2 molecule in kg.

    28.0134 g N 2 1 mol × 1 mol 6 .022 × 10 23 molecules × 1 kg 1000 g = 4.6518 × 10 26 = 4.652 × 10 26 kg

    E k = 1/2 mv 2 = 1/2 × 4.6518 × 10 –26 kg × (469.38 m/s) 2

    = 5.1244 × 10 21 kg-m 2 s 2 = 5.124 × 10 21 J
  3. 5.1244 × 10 21 J molecule × 6.022 × 10 23 molecules 1 mol = 3086 J/mol = 3 .086 kJ/mol

5.119

  1. E p = mgd = 52.0 kg × 9.81 m/s 2 × 10.8 m = 5509.3 J = 5.51 kJ
  2. E k = 1 / 2 mv 2 ; v = (2E k / m) 1/2 = ( 2 × 5509.3 kg-m 2 / s 2 52.0 kg ) 1 / 2 = 14.6 m/s
  3. Yes, the diver does work on entering (pushing back) the water in the pool.

5.120

  1. CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)

    Δ H o = Δ H f CO 2 ( g ) + 2 Δ H f H 2 O(l) Δ H f CH 4 ( g ) 2 Δ H f O 2 ( g ) = 393 . 5 kJ + 2 ( 285 . 83 kJ ) ( 74 . 8 kJ ) 2 ( 0 ) = 89 0. 36 = 89 0. 4kJ / mol CH4

    The minus sign indicates that 890.4 kJ are produced per mole of CH 4 burned.

    890.36 kJ mol CH 4 × 1000 J 1 kJ × 1 mol 6 .022 × 10 23 molecules CH 4 = 1.4785 × 10 18 = 1.479 × 10 18 J/molecule


  1. 1eV = 96.485 kJ/mol

    8 keV × 1000 eV 1 keV × 96.485 kJ eV-mol × 1 mol 6 .022 × 10 23 × 1000 J kJ = 1.282 × 10 15 = 1 × 10 15 J/X - ray

    The energy produced by the combustion of 1 molecule of CH 4 (g) is much smaller than the energy of the X-ray.

5.121  (a,b) Ag + (aq) + Li(s) → Ag(s) + Li + (aq)

Δ H = Δ H f Li + ( aq ) Δ H f Ag + ( aq ) = 278 .5 kJ 105 .90 kJ = 384 .4 kJ

Fe(s) + 2 Na + (aq) → Fe 2+ (aq) + 2 Na(s)

Δ H = Δ H f Fe 2 + ( aq ) 2 Δ H f Na + ( aq ) = 87. 86 kJ 2 ( 240.1 kJ) = + 392.3 kJ

2 K(s) + 2 H 2 O(l) → 2 KOH(aq) + H 2 (g)

Δ H = 2 Δ H f KOH(aq) 2 Δ H f H 2 O(l)

= 2(–482.4 kJ) – 2(–285.83 kJ) = –393.1 kJ

  1. Exothermic reactions are more likely to be favored, so we expect the first and third reactions be favored.
  1. In the activity series of metals, Table 4.5, any metal can be oxidized by the cation of a metal below it on the table.

Ag + is below Li, so the first reaction will occur.

Na + is above Fe, so the second reaction will not occur.

H + (formally in H 2 O) is below K, so the third reaction will occur.

These predictions agree with those in part (c).

5.122

  1. Δ H = Δ H f NaNO 3 ( aq ) + Δ H f H 2 O(l) Δ H f HNO 3 ( aq ) Δ H f NaOH(aq)

    ΔH = –446.2 kJ – 285.83 kJ – (–206.6 kJ) – (–469.6 kJ) = –55.8 kJ

    Δ H = Δ H f NaCl(aq) + Δ H f H 2 O(l) Δ H f HCl(aq) Δ H f NaOH(aq)

    ΔH = –407.1 kJ – 285.83 kJ – (–167.2 kJ) – (–469.6 kJ) = –56.1 kJ

    Δ H = Δ H f NH 3 ( aq ) + Δ H f Na + ( aq ) + Δ H f H 2 O(l) Δ H f NH 4 + ( aq ) Δ H f NaOH(aq)

    = –80.29 kJ – 240.1 kJ – 285.83 kJ – (–132.5 kJ) – (–469.6 kJ) = –4.1 kJ

  2. H + (aq) + OH (aq) → H 2 O(l)
  3. The ΔH values for the first two reactions are nearly identical, –55.8 kJ and –56.1 kJ. The spectator ions by definition do not change during the course of a reaction, so ΔH is the enthalpy change for the net ionic equation. Because the first two reactions have the same net ionic equation, it is not surprising that they have the same ΔH .

  1. Strong acids are more likely than weak acids to donate H + . The neutralizations of the two strong acids are energetically favorable, whereas the neutralization of NH 4 + (aq) is significantly less favorable. NH 4 + (aq) is probably a weak acid.

5.123

  1. mol Cu = M × L = 1.00 M × 0.0500 L = 0.0500 mol

    g = mol × MM = 0.0500 × 63.546 = 3.1773 = 3.18 g Cu

  2. The precipitate is copper(II) hydroxide, Cu(OH) 2 .
  3. CuSO 4 (aq) + 2 KOH(aq) → Cu(OH) 2 (s) + K 2 SO 4 (aq), complete

    Cu 2+ (aq) + 2 OH–(aq) → Cu(OH) 2 (s), net ionic

  4. The temperature of the calorimeter rises, so the reaction is exothermic and the sign of q is negative.

    q = 6.2 o C × 100 g × 4 .184 J 1 g- o C = 2.6 × 10 3 J = 2.6 kJ

    The reaction as carried out involves only 0.050 mol of CuSO 4 and the stoichiometrically equivalent amount of KOH. On a molar basis,

    Δ H = 2.6 kJ 0 .050 mol = 52 kJ for the reaction as written in part (c)

5.124

  1. AgNO 3 (aq) + NaCl(aq) → NaNO 3 (aq) + AgCl(s)

    net ionic equation: Ag + (aq) + Cl (aq) → AgCl(s)

    Δ H o = Δ H f AgCl(s) Δ H f Ag + ( aq) Δ H f Cl ( aq )

    ΔH = –127.0 kJ – (105.90 kJ) – (–167.2 kJ) = –65.7 kJ

  2. ΔH for the complete molecular equation will be the same as ΔH? for the net ionic equation. Na + (aq) and NO 3– (aq) are spectator ions; they appear on both sides of the chemical equation. Because the overall enthalpy change is the enthalpy of the products minus the enthalpy of the reactants, the contributions of the spectator ions cancel.
  3. Δ H o = Δ H f NaNO 3 ( aq ) + Δ H f AgCl(s) Δ H f AgNO 3 ( aq ) Δ H f NaCl(aq)

    Δ H f AgNO 3 ( aq ) = Δ H f NaNO 3 ( aq ) + Δ H f AgCl(s) Δ H f NaCl(aq) Δ H o Δ H f AgNO 3 ( aq ) = 446.2 kJ + ( 127.0 kJ) ( 407.1 kJ) ( 65.7 kJ) Δ H f AgNO 3 ( aq ) = 100.4 kJ/mol

5.125

  1. 21.83 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 1 mol C 1 mol CO 2 × 12.01 g C 1 mol C = 5.9572 = 5.957 g C

    4.47 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 2 mol H 1 mol H 2 O × 1.008 g H mol H = 0.5001 = 0.500 g H

    The sample mass is (5.9572 + 0.5001) = 6.457 g


  1. 5.957 g C × 1 mol C 12 .01 g C = 0.4960 mol C;   0 .4960/0 .496 = 1

    0.500 g H × 1 mol H 1 .008 g H = 0.496 mol H;  0 .496/0 .496 = 1

    The empirical formula of the hydrocarbon is CH.

  1. Calculate " Δ H f o " for 6.457 g of the sample.

    6.457 g sample + O 2 (g) → 21.83 g CO 2 (g) + 4.47 g H 2 O(g), Δ H comb = –311 kJ

    Δ H comb = Δ H f o CO 2 ( g ) + Δ H f o H 2 O(g) Δ H f o sample Δ H f o O 2 ( g ) Δ H f o sample = Δ H f o CO 2 ( g ) + Δ H f o H 2 O(g) Δ H comb Δ H f o CO 2 ( g) = 21.83 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 393.5 kJ mol CO 2 = 195.185 = 195.2 kJ

    Δ H f o H 2 O ( g) = 4.47 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 241.82 kJ mol H 2 O = 59.985 = 60.0 kJ

    Δ H f o sample = 195.185 kJ 59.985 kJ ( 311 kJ) = 55.83 = 56 kJ

    H f o = 55.83 kJ 6 .457 g sample × 13.02 g CH unit = 112.6 = 1.1 × 10 2 kJ/CH unit

  1. The hydrocarbons in Appendix C with empirical formula CH are C 2 H 2 and C 6 H 6 .
    substance Δ H f / mo l Δ H f º / CH unit
    C 2 H 2 (g) 226.7 kJ 113.4 kJ
    C 6 H 6 (g) 82.9 kJ 13.8 kJ
    C 6 H 6 (l) 49.0 kJ 8.17 kJ
    sample 1.1 × 10 2 kJ

    The calculated value of Δ H f º / CH unit for the sample is a good match with acetylene, C 2 H 2 (g). 5.126

5.126

  1. CH 4 (g) → C(g) + 4 H(g) (i) reaction given
    CH 4 (g) → C(s) + 2 H 2 (g) (ii) reverse of formation

    The differences are: the state of C in the products; the chemical form, atoms, or diatomic molecules, of H in the products.

    1. ΔH = ΔH f C(g) + 4 ΔH f H(g) ΔH f CH 4 ( g ) = 718 .4 kJ + 4(217 .94) kJ ( 74 .8) kJ = 1665 .0 kJ
    2. Δ H = Δ H f CH 4 = ( 74.8 ) kJ = 74. 8 kJ

  1. The rather large difference in ΔH? values is due to the enthalpy difference between isolated gaseous C atoms and the orderly, bonded array of C atoms in graphite, C(s), as well as the enthalpy difference between isolated H atoms and H 2 molecules. In other words, it is due to the difference in the enthalpy stored in chemical bonds in C(s) and H 2 (g) versus the corresponding isolated atoms.

  1. CH 4 (g) + 4 F 2 (g) → CF 4 (g) + 4 HF(g) ΔH = –1679.5 kJ

    The ΔH value for this reaction was calculated in Solution 5.96.

    3.45 g CH 4 × 1 mol CH 4 16.04 g CH 4 × 0.21509 = 0.215 mol CH 4

    1.22 g F 2 × 1 mol F 2 38.00 g F 2 = 0.03211 = 0.0321 mol F 2

    There are fewer mol F 2 than CH 4 , but 4 mol F 2 are required for every 1 mol of CH 4 reacted, so clearly F 2 is the limiting reactant.

    0.03211 mol F 2 × 1679.5 kJ 4 mol F 2 = 13.48 = 13.5 kJ heat evolved

5.127

  1. C 2 H 5 OH(l) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O(l)
  2. Δ H rxn = 2 Δ H f CO 2 ( g ) + 3 Δ H f H 2 O(l) Δ H f C 2 H 5 OH(l) 3 Δ H f O 2 ( g ) = 2 ( 393 . 5 ) kJ + 3 ( 285 . 83 kJ ) ( 277 . 7 kJ ) 3 ( 0 ) = 1366 . 79 = 1367 kJ
  3. oz beer → mL beer → mL ethanol → g ethanol

    12 oz beer × 29 .5735 mL oz × 0 .042 mL ethanol mL beer × 0 .789 g ethanol mL = 11 .76 = 12 g ethanol

  4. The metabolism reaction in part (a) produces 1367 kJ/mol ethanol, part (b).

    1 1 . 7 6 g ethanol × 1 mol ethanol 4 6 . 0 7 g ethanol × 1 3 6 7 kJ mol ethanol × 1 Cal 4 . 1 8 4 kJ = 8 3 . 4 0 = 8 3 Cal

  5. Percent Cal from ethanol

    83 .40 Cal from ethanol 110 Cal total × 100 = 75 .82 = 76%


 

6 Electronic Structure of Atoms

Visualizing Concepts

6.1

  1. Analyze/Plan .  We are given the speed of sound in dry air and the frequency of the lowest audible wave. We must find the wavelength of the sound waves. For any wave, speed equals wavelength times frequency. Solve.

    Speed = λ × ν;     λ = speed/ν;     20 Hz = 20 s −1 = 20/s

    λ = speed/ν; 343 m s × s 20 = 17.15 = 17 m (to 1 sig fig, 20 m)

  2. Analyze/Plan .  We are asked to find the frequency of electromagnetic radiation with this wavelength. All electromagnetic radiation in a vacuum has the speed 2.998 × 10 8 m/s; the symbol for the speed of light is “c”. ν = c/λ. Solve .

    ν = c/ λ ; 2 . 9 9 8 × 1 0 8 m s × 1 1 7 .15 m = 1.7 × 10 7 s 1 = 1.7 × 10 7 Hz

    1.7 × 10 7 Hz × 1 MHz 1 × 1 0 6 Hz = 17 MHz  (20 MHz, to 1 sig fig)

  3. According to Figure 6.4, this would be radio frequency radiation.

6.2   Given: 2450 MHz radiation. Hz = s −1 , unit of frequency. M = 1 × 10 6 ; 2450 × 10 6 Hz = 2.45 × 10 9 Hz = 2.45 × 10 9 s −1 .

  1. Find 2.45 × 10 9 s −1 on the frequency axis of Figure 6.4. The wavelength that corresponds to this frequency is approximately 1 × 10 −1 = 0.1 m or 10 cm.
  2. No, visible radiation has wavelengths of 4 × 10 −7 to 7 × 10 −7 m, much shorter than 0.1 m.
  3. Energy and wavelength are inversely proportional. Photons of the longer 0.1 m radiation have less energy than visible photons.
  4. Radiation of 0.1 m is in the low energy portion of the microwave region. The appliance is probably a microwave oven. (Appliances with heating elements that glow red or orange give off wavelengths in the visible or the near visible portion of the infrared. The 0.1 m wavelength is too long to belong to these appliances.)

6.3

  1. By inspection, wave (a) has the longer wavelength.
  2. Wave (b) has the higher frequency because it has the shorter wavelength.
  3. Wave (b) has a higher energy because it has a higher frequency (and shorter wavelength).

6.4

  1. (iii) Betelgeuse < (i) Sun < (ii) Rigel .
  2. Black body radiation

6.5

  1. Increase.  The rainbow has shorter wavelength blue light on the inside and longer wavelength red light on the outside. (See Figure 6.4.)
  2. Decrease.  Wavelength and frequency are inversely related. Wavelength increases so frequency decreases going from the inside to the outside of the rainbow.

6.6

  1. n = 1 to n = 4
  2. n = 1 to n = 2
  3. Wavelength and energy are inversely proportional; the smaller the energy, the longer the wavelength. In order of increasing wavelength (and decreasing energy): (iii) n = 2 to n = 4 < (iv) n = 1 to n = 3 < (ii) n = 2 to n = 3 < (i) n = 1 to n = 2

6.7

  1. Wave (iii) corresponds to transition C. Transition C represents the smallest energy change, which will emit a photon with the longest wavelength.
  2. Analyze/Plan . Use Equation 6.6, which describes energy changes in the hydrogen atom, to calculate the energy of the photon emitted for each transition. Solve .

    A: n i = 2 , n f = 1 ; Δ E = 2.18 × 10 18 J [ 1 n f 2 1 n i 2 ] = 2.18 × 10 18 J (1 1 / 4 ) = 1.6 35 × 10 18 J 1.64 × 10 18 J

    B: n i = 3 , n f = 2 ; Δ E = 2.18 × 10 18 J (1/4 1/9 ) = 3.028 × 10 19 = 3.03 × 10 19 J

    C: n i = 4 , n f = 3 ; Δ E = 2.18 × 10 18 J (1/9 1 /16 ) = 1.0597 × 10 19 = 1.06 × 10 19 J

    The negative signs for ΔE indicate that the photons are emitted.

  3. A: λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 1.635 × 10 18 J = 1.22 × 10 7 m

    B: λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 3.028 × 10 19 J = 6.56 × 10 7 m

    C: λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 1.0597 × 10 19 J = 1.87 × 10 6 m

    According to Figure 6.4, visible light has wavelengths from 4 × 10 −7 m to 7.5 × 10 −7 m (400–750 nm). Transition B emits photons of visible light.

6.8

  1. ψ 2 (x) will be positive or zero at all values of x and have two maxima with larger magnitudes than the maximum in ψ(x).
    image

  1. The greatest probability of finding the electron is at the two maxima in ψ 2 (x) at x = π/2 and 3π/2.
  1. There is zero probability of finding the electron at x = π. This value is called a node.

6.9

  1. l = 1
  2. 3p y ( n = 3 shell, dumbbell shape, node at the nucleus, oriented along the y-axis)
  3. (iii) ( m l indicates the orientation of an orbital)

6.10

  1. The smallest possible n value for a d orbital is 3.
  2. For a d orbital, l = 2.
  3. For a d orbital, the largest possible value of m l is 2 (possible values are 2, 1, 0, –1, or –2)
  4. The probability density goes to zero along the xy and xz planes.

6.11  Configuration B is the correct one for a nitrogen atom in its ground state.

Configuration C violates the Pauli exclusion principle, that no two electrons can have the same set of quantum numbers. That is, two electrons in the same orbital must have opposite spins.

Configurations A, C, and D violate the Hund’s rule, that electron spin is maximized. That is, electrons will occupy degenerate orbitals singly when possible and they will have the same spin.

6.12

  1. Group 7A or 17, the halogens, the column second from the right
  2. Group 5A or 15
  3. Gallium, atomic number 31, at the intersection of row 4 and group 3A or 13
  4. All of the B groups, groups 3–12, in the middle of the major part of the table, not including the two rows of f -block elements

The Wave Nature of Light (Section 6.1)

6.13

  1. Meters (m)
  2. 1/seconds (s −1 )
  3. meters/second (m-s −1 or m/s)

6.14

  1. Wavelength (λ) and frequency (ν) are inversely proportional; the proportionality constant is the speed of light (c). ν = c/λ.
  2. Light in the 210–230 nm range is in the ultraviolet region of the spectrum. These wavelengths are slightly shorter than the 400 nm short-wavelength boundary of the visible region.

6.15

  1. True.
  2. False.  Ultraviolet light has shorter wavelengths than visible light. [See Solution 6.14(b).]
  3. False.  X-rays travel at the same speed as microwaves. (X-rays and microwaves are both electromagnetic radiation.)
  4. False.  Electromagnetic radiation and sound waves travel at different speeds. (Sound is not a form of electromagnetic radiation.)

6.16

  1. False.  The frequency of radiation decreases as the wavelength increases.
  2. True.
  3. False.  Infrared light has lower frequencies than visible light.
  4. False.  The glow from a fireplace and the energy within a microwave oven are both forms of electromagnetic radiation. (A foghorn blast is a form of sound waves, which are not accompanied by oscillating electric and magnetic fields.)

6.17 Analyze/Plan .  Use the electromagnetic spectrum in Figure 6.4 to determine the wavelength of each type of radiation; put them in order from shortest to longest wavelength. Solve .

Wavelength of X-rays < ultraviolet < green light < red light < infrared < radio waves

Check .  These types of radiation should read from left to right on Figure 6.4.

6.18   Wavelength of (a) gamma rays < (d) yellow (visible) light < (e) red (visible) light < (b) 93.1 MHz FM (radio) waves < (c) 680 kHz or 0.680 MHz AM (radio) waves

6.19 Analyze/Plan .  These questions involve relationships between wavelength, frequency, and the speed of light. Manipulate the equation ν = c/λ to obtain the desired quantities, paying attention to units. Solve .

  1. ν = c/ λ ; 2.998 × 10 8 m s × 1 10 μ m × 1 μ m 1 × 10 6 m = 3.0 × 10 13 s 1
  2. λ = c/ ν ; 2.998 × 10 8 m s × 1 s 5.50 × 10 14 = 5.45 × 10 7 m (545 nm)
  3. The radiation in (b) is in the visible region and is “visible” to humans.

    The 10 µm (1 × 10 −5 m) radiation in (a) is in the infrared region and is not visible.

  4. 50.0 μ s × 1 s 1 × 10 6 μ s × 2.998 × 10 8 m s = 1.50 × 10 4 m

Check .  Confirm that powers of 10 make sense and units are correct.

6.20

  1. ν = c/ λ ; 2.998 × 10 8 m s × 1 0.86 nm × 1 1 .0 × 10 9 m = 3.5 × 10 17 s 1
  2. λ = c/ ν ; 2 .998 × 10 8 m s × 1 s 6 .4 × 10 11 = 4 .7 × 10 4 m
  3. The radiation in (a) can be observed by an X-ray detector. The 8.6 × 10 −10 m wavelength is within the range of 10 −8 to 10 −11 m for X-rays.
  4. 0.38 ps × 1 × 10 12 s 1 ps × 2.998 × 10 8 m s = 1.1 × 10 4 m (0 .11 mm)

6.21 Analyze/Plan .  ν = c/λ; change nm → m.

Solve . ν = c/ λ ; 2.998 × 10 8 m 1 s × 1 650 nm × 1 nm 1 × 10 9 m = 4.6 × 10 14 s 1

The color is red.

Check .  (3000 × 10 5 /650 × 10 −9 ) ≈ 4.5 × 10 14 s −1 ; units are correct.


6.22   According to Figure 6.4, ultraviolet radiation has both higher frequency and shorter wavelength than infrared radiation. Looking forward to section 6.2, the energy of a photon is directly proportional to frequency (E = hν), so ultraviolet radiation yields more energy from a photovoltaic device.

Quantized Energy and Photons (Section 6.2)

6.23   (iii) Quantization means that energy changes can only happen in certain allowed increments. If the human growth quantum is one-foot, growth occurs instantaneously in one-foot increments.

6.24   Planck’s original hypothesis was that energy could only be gained or lost in discrete amounts (quanta) with a certain minimum size. The size of the minimum energy change is related to the frequency of the radiation absorbed or emitted, ΔE = hν, and energy changes occur only in multiples of hν.

Einstein postulated that light itself is quantized, that the minimum energy of a photon (a quantum of light) is directly proportional to its frequency, E = hν. If a photon that strikes a metal surface has less than the threshold energy, no electron is emitted from the surface. If the photon has energy equal to or greater than the threshold energy, an electron is emitted and any excess energy becomes the kinetic energy of the electron.

6.25 Analyze/Plan .  These questions deal with the relationships between energy, wavelength, and frequency. Use the relationships E = hν = hc/λ to calculate the desired quantities. Pay attention to units. Solve .

  1. E = h ν = 6.626 × 10 34 J-s × 2.94 × 10 14 s 1 = 1 .95 × 10 19 J
  2. E = h ν = hc λ = 6 .626 × 10 34 J-s × 2 .998 × 10 8 m/s 413 nm × 1 nm 1 × 10 9 m = 4.81 × 10 19 J
  3. λ = hc/ Δ E = 6.626 × 10 34 J-s 6 .06 × 10 19 J × 2 .998 × 10 8 m 1 s = 3.28 × 10 7 m = 328 nm

6.26 Analyze/Plan .  These questions deal with the relationships between energy, wavelength, and frequency. Use the relationships E = hν = hc/λ to calculate the desired quantities. Pay attention to units. Solve .

  1. ν = c/ λ ; 2 .998 × 10 8 m s × 1 532 nm × 1 nm 1 × 10 9 m = 5.64 × 10 14 s 1
  2. E = h ν = 6.626 × 10 34 J-s × 5.64 × 10 14 s 1 = 3 .73 × 10 19 J
  3. The energy gap between the ground and excited states is the energy of a single 532 nm photon emitted when one electron relaxes from the excited to the ground state. Δ E = 3 .73 × 10 19 J

6.27 Analyze/Plan .  Use E = hc/λ; pay close attention to units. Solve .

  1. E = hc/ λ = 6.626 × 10 34 J-s × 2 .998 × 10 8 m 1 s × 1 3.3 μ m × 1 μ m 1 × 10 6 m = 6.0 × 10 20 J

    E = hc/ λ = 6.626 × 10 34 J-s × 2 .998 × 10 8 m 1 s × 1 0.154 nm × 1 nm 1 × 10 9 m = 1.29 × 10 15 J

    Check .  (6.6 × 3/3.3) × (10 −34 × 10 8 /10 −6 ) ≈ 6 × 10 −20 J

    (6.6 × 3/0.15) × (10 −34 × 10 8 /10 −9 ) ≈ 120 × 10 −17 ≈ 1.2 × 10 −15 J

    The results are reasonable. We expect the longer wavelength 3.3 µm radiation to have the lower energy.

  1. The 3.3 µm photon is in the infrared and the 0.154 nm (1.54 × 10 −10 m) photon is in the X-ray region; the X-ray photon has the greater energy.

6.28   E = hν

AM: 6.626 × 10 34 J-s × 1010 × 10 3 1 s = 6.69 × 10 28 J

FM: 6.626 × 10 34 J-s × 98 .3 × 10 6 1 s = 6.51 × 10 26 J

The FM photon has about 100 times more energy than the AM photon.

6.29 Analyze/Plan .  Use E = hc/λ to calculate J/photon; Avogadro’s number to calculate J/mol; photon/J [the result from part (a)] to calculate photons in 1.00 mJ. Pay attention to units. Solve .

  1. E photon = hc/ λ = 6.626 × 10 34 J-s 325 × 10 9 m × 2 .998 × 10 8 m s = 6.1122 × 10 19 = 6.11 × 10 19 J/photon
  2. 6.1122 × 10 19 J 1 photon × 6.022 × 10 23 photons 1 mol = 3.68 × 10 5 J/mol = 368 kJ/mol
  3. 1 photon 6.1122 × 10 19 J × 1.00 mJ × 1 × 10 3 J 1 mJ = 1.64 × 10 15 photons

    Check .  Powers of 10 (orders of magnitude) and units are correct.

  4. If the energy of one 325 nm photon breaks exactly one bond, 1 mol of photons break 1 mol of bonds. The average bond energy in kJ/mol is the energy of 1 mol of photons (from part b) 368 kJ/mol.

6.30 242 × 10 3 J mol Cl 2 × 1 mol 6 .022 × 10 23 photons = 4.0186 × 10 19 = 4.02 × 10 19 J/photon

λ = hc/E = 6.626 × 10 34 J-s 4 .0186 × 10 19 J × 2 .998 × 10 8 m 1 s = 4.94 × 10 7 m = 494 nm

According to Figure 6.4, this is visible radiation.

6.31 Analyze/Plan .  E = hc/λ gives J/photon. Use this result with J/s (given) to calculate photons/s. Solve.

  1. The ~1 × 10 −6 m radiation is infrared but very near the visible edge.

  1. E photon = hc/ λ = 6.626 × 10 34 J-s 987 × 10 9 m × 2 .998 × 10 8 m 1 s = 2.0126 × 10 19 = 2.01 × 10 19 J/photon

    0.52 J 32 s × 1 photon 2.0126 × 10 19 J = 8.1 × 10 16 photons/s

Check .  (7 × 3/1000) × (10 −34 × 10 8 /10 −9 ) ≈ 21 × 10 −20 ≈ 2.1 × 10 −19 J/photon

(0.5/30/2) × (1/10 −19 ) = 0.008 × 10 19 = 8 × 10 16 photons/s

Units are correct; powers of 10 are reasonable.

6.32

  1. 3.55 mm = 3.55 × 10 −3 m; the radiation is microwave.
  2. E photon = hc/ λ = 6.626 × 10 34 J-s 3.55 × 10 3 m × 2 .998 × 10 8 m 1 s = 5.5957 × 10 23 = 5.60 × 10 23 J/photon

    5.5957 × 10 23 J 1 photon × 3.2 × 10 8 photons 1 s × 60 s 1 min × 60 min 1 hr = 6.4463 × 10 11 = 6.4 × 10 11 J/hr

6.33 Analyze/Plan .  Use E = hν and ν = c/λ. Calculate the desired characteristics of the photons. Assume 1 photon interacts with 1 electron. Compare E min and E 120 to calculate maximum kinetic energy of the emitted electron. Solve .

  1. E = h ν = 6.626 × 10 34 J-s × 1 .09 × 10 15 s 1 = 7.22 × 10 19 J
  2. λ = c/ ν = 2.998 × 10 8 m 1 s × 1 s 1 .09 × 10 15 = 2.75 × 10 7 m = 275 nm
  3. E 120 = hc/ λ = 6.626 × 10 34 J-s × 2 .998 × 10 8 m 1 s × 1 120 nm × 1 nm 1 × 10 9 m = 1.655 × 10 18 = 1.66 × 10 18 J

    The excess energy of the 120 nm photon is converted into the kinetic energy of the emitted electron.

    E k = E 120 – E min = 16.55 × 10 −19 J – 7.22 × 10 −19 J = 9.3 × 10 −19 J/electron

Check .  E 120 must be greater than E min in order for the photon to impart kinetic energy to the emitted electron. Our calculations are consistent with this requirement.

6.34

  1. ν = E / h = 6.94 × 10 19 J 6.626 × 10 34 J-s = 1.04739 × 10 15 = 1.05 × 10 15 s 1
  2. λ = hc/E = 6.626 × 10 34 J-s 6.94 × 10 19 J × 2.998 × 10 8 m s = 2.86 × 10 7 m = 286 nm
  3. No. The maximum 286 nm wavelength required to eject an electron from titanium is in the ultraviolet region (Figure 6.4).
  4. E 233 = hc/ λ = 6.626 × 10 34 J-s 233 × 10 9 m × 2.998 × 10 8 m s = 8.5256 × 10 19 = 8.53 × 10 19 J

    E K = E 233 – E min = 8.5256 × 10 −19 J – 6.94 × 10 −19 J = 1.5856 × 10 −19 = 1.59 × 10 −19 J


Bohr’s Model; Matter Waves (Sections 6.3 and 6.4)

6.35   When the electron in a hydrogen atom transitions from n = 1 to n = 3, the atom ”expands.” The average distance from the nucleus of an n = 3 electron is greater than that of an n = 1 electron. The volume where there is a significant probability of finding an electron increases.

6.36

  1. True.
  2. False. As the value of n increases, energy becomes less negative. The energy of the n = 2 state is higher (less negative) than that of the n = 1 state.
  3. True.

6.37 Analyze/Plan .  An isolated electron is assigned an energy of zero; the closer the electron comes to the nucleus, the more negative its energy. Thus, as an electron moves closer to the nucleus, the energy of the electron decreases and the excess energy is emitted. Conversely, as an electron moves further from the nucleus, the energy of the electron increases and energy must be absorbed. Solve .

  1. As the principle quantum number decreases, the electron moves toward the nucleus and energy is emitted.
  2. An increase in the radius of the orbit means the electron moves away from the nucleus; energy is absorbed.
  3. An isolated electron is assigned an energy of zero. As the electron moves to the n = 3 state closer to the H + nucleus, its energy becomes more negative (decreases) and energy is emitted.

6.38

  1. Absorbed.
  2. Emitted.
  3. Absorbed.

6.39 Analyze/Plan .  Equation 6.5: E = (–2.18 × 10 −18 J)(1/ n 2 ). Solve .

  1. E 2 = –2.18 × 10 −18 J/(2) 2 = –5.45 × 10 −19 J

    E 6 = –2.18 × 10 −18 J/(6) 2 = –6.0556 × 10 −20 = –0.606 × 10 −19 J

    ΔE = E 2 – E 6 = (–5.45 × 10 −19 J) – (–0.606 × 10 −19 J)

    = –4.844 × 10 −19 J = –4.84 × 10 −19 J

    λ = hc/ Δ E = 6.626 × 10 34 J-s 4.844 × 10 19 J × 2.998 × 10 8 m s = 4.10 × 10 7 m = 410 nm

  2. The visible range is 400–700 nm, so this line is visible; the observed color is violet.

    Check .  We expect E 6 to be a more positive (or less negative) than E 2 , and it is. ΔE is negative, which indicates emission. The orders of magnitude make sense and units are correct.


6.40

  1. The value of n for the electron increases, so ΔE is positive.
  2. Analyze/Plan .  Use Equation 6.6 to calculate ΔE, then λ = hc/ΔE. Solve .

    n i = 4 , n f = 9 ; Δ E = 2.18 × 10 18 J [ 1 n f 2 1 n i 2 ] = 2.18 × 10 18 J (1/81 1 / 16 )

    λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/81 1 / 16 ) = 1.82 × 10 6 m

    This wavelength will be absorbed. Energy is required to move the electron to the higher energy n = 9 state. [Note that the denominator of the preceding equation has a positive sign, because (1/81 – 1/16) is negative.]

  3. The light in (b) is in the infrared.

6.41

  1. Statement (ii) is the best explanation. Only lines with n f = 2 represent ΔE values and wavelengths that lie in the visible portion of the spectrum. Lines with n f = 3 have smaller ΔE values and lie in the lower energy, longer wavelength infrared portion of the electromagnetic spectrum.
  2. Analyze/Plan .  Use Equation 6.6 to calculate ΔE, then λ = hc/ΔE. Solve .

    n i = 3 , n f = 2 ; Δ E = 2.18 × 10 18 J [ 1 n f 2 1 n i 2 ] = 2.18 × 10 18 J (1/4 1 / 9 )

    λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/4 1 / 9 ) = 6.56 × 10 7 m

    This is the red line at 656 nm.

    n i = 4 , n f = 2 ; λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/4 1 / 16 ) = 4.86 × 10 7 m

    This is the blue-green line at 486 nm.

    n i = 5 , n f = 2 ; λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/4 1 / 25 ) = 4.34 × 10 7 m

    This is the blue-violet line at 434 nm.

    Check .  The calculated wavelengths correspond well to three lines in the H emission spectrum in Figure 6.11, so the results are sensible.

6.42

  1. From Exercise 6.41, we know that transitions with n f = 2 emit light in the visible region of the electromagnetic spectrum. Transitions with n f = 1 have larger ΔE values and shorter wavelengths than those with n f = 2. These transitions will lie in the ultraviolet region.
  2. n i = 2 , n f = 1 ; λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/1 1 / 4 ) × 1 nm 1 × 10 9 m = 121 nm

    n i = 3 , n f = 1 ; λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/1 1 / 9 ) × 1 nm 1 × 10 9 m = 103 nm


    n i = 4 , n f = 1 ; λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/1 1 / 16 ) × 1 nm 1 × 10 9 m = 97 .2 nm

    Note that all three wavelengths are in the ultraviolet region of the electromagnetic spectrum.

6.43

  1. 93.07 nm × 1 × 10 9 m 1 nm = 9.307 × 10 8 m; this line is in the ultraviolet region .
  2. Analyze/Plan .  Only lines with n f = 1 have a large enough ΔE to lie in the ultraviolet region (see Solutions 6.41 and 6.42). Solve Equation 6.6 for n i , recalling that ΔE is negative for emission. Solve .

    hc λ = 2.18 × 10 18 J [ 1 n f 2 1 n i 2 ] ; hc λ ( 2 . 1 8 × 1 0 18 J) = [ 1 1 n i 2 ]

    1 n i 2 = [ hc λ ( 2 . 1 8 × 1 0 18 J) 1 ] ; 1 n i 2 = [ 1 hc λ ( 2 . 1 8 × 1 0 18 J) ]

    n i 2 = [ 1 hc λ ( 2 . 1 8 × 1 0 18 J) ] 1 ; n i = [ 1 hc λ ( 2 . 1 8 × 1 0 18 J) ] 1 / 2

    n i = ( 1 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 9.307 × 10 8 m × 2 .18 × 10 18 J ) 1 / 2 = 7 ( n values must be integers)

    n i = 7, n f = 1

    Check .  From Solution 6.42, we know that n i > 4 for λ = 93.07 nm. The calculated result is close to 7, so the answer is reasonable.

6.44

  1. 1094 nm × 1 × 10 9 m 1 nm = 1.094 × 10 6 m; this line is in the infrared .
  2. Absorption lines with n i = 1 are in the ultraviolet and with n i = 2 are in the visible. Thus, n i ≥ 3, but we do not know the exact value of n i . Calculate the longest wavelength with n i = 3 ( n f = 4). If this is less than 1094 nm, n i > 3.

    λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 J (1/16 1 / 9 ) = 1.875 × 10 6 m

    This wavelength is longer than 1.094 × 10 −6 m, but we are in the ballpark. Use

    n i = 3 and solve for n f as in Solution 6.43. Note that ΔE is positive because we are dealing with absorption.

    n f = ( 1 n i 2 h c λ ( 2 . 1 8 × 1 0 1 8 J ) ) 1 / 2 = ( 1 / 9 6 . 6 2 6 × 1 0 3 4 J - s × 2 . 9 9 8 × 1 0 8 m / s 1 . 0 9 4 × 1 0 6 m × 2 . 1 8 × 1 0 1 8 J ) 1 / 2 = 6

    n f = 6, n i = 3

6.45 Analyze/Plan .  According to Equation 6.6 and several preceding solutions, the greater the energy of light absorbed, the lower the value of n i . The greater the energy of light, the greater its frequency, ΔE = hν.

Solve .  The order of increasing frequency (and energy) of light absorbed is:

n = 4 to n = 9 < n = 3 to n = 6 < n = 2 to n = 3 < n = 1 to n = 2


6.46   According to Equation 6.6 and several preceding solutions, the greater the energy of light emitted, the lower the value of n f . The greater the energy of light, the shorter its wavelength, ΔE = hc/λ. The order of increasing wavelength (and decreasing energy) of light emitted is:

n = 4 to n = 2 < n = 3 to n = 2 < n = 5 to n = 3 < n = 7 to n = 4

6.47 Analyze/Plan . λ = h mv ; 1 J = 1 kg-m 2 s 2 ;

Change mass to kg and velocity to m/s in each case. Solve .

  1. 50 km 1 hr × 1000 m 1 km × 1 hr 60 min × 1 min 60 s = 13.89 = 14 m/s

    λ = 6.626 × 10 34 kg-m 2 -s 1 s 2 × 1 85 kg × 1 s 13 .89 m = 5.6 × 10 37 m

  2. 10.0 g × 1 kg 1000 g = 0.0100 kg

    λ = 6.626 × 10 34 kg-m 2 -s 1 s 2 × 1 0 .0100 kg × 1 s 250 m = 2.65 × 10 34 m

  3. We need to calculate the mass of a single Li atom in kg.

    6.94 g Li 1 mol Li × 1 kg 1000 g × 1 mol 6 .022 × 10 23 Li atoms = 1.152 × 10 26 = 1.15 × 10 26 kg

    λ = 6.626 × 10 34 kg-m 2 -s 1 s 2 × 1 1 .152 × 10 26 kg × 1 s 2 .5 × 10 5 m = 2.3 × 10 13 m

  4. Calculate the mass of a single O 3 molecule in kg.

    48 .00 g O 3 1 mol O 3 × 1 kg 1000 g × 1 mol 6 .022 × 10 23 O 3 molecules = 7 .971 × 10 26 = 7 .97 × 10 26 kg

    λ = 6.626 × 10 34 kg-m 2 -s 1 s 2 × 1 7 .971 × 10 26 kg × 1 s 550 m = 1 .51 × 10 11 m (15 pm)

6.48   Find the mass of an electron on the inside back cover of the text. λ = h/mv; change velocity to m/s.

mass of muon = 206.8 × 9.1094 × 10 31 g = 1.8838 × 10 28 = 1.88 × 10 28 kg

λ = 6.626 × 10 34 kg-m 2 -s 1 s 2 × 1 1 .8838 × 10 28 kg × 1 s 8 .85 × 10 3 m/s = 3.97 × 10 10 m = 3.97 Å


6.49 Analyze/Plan .  Use v = h/mλ; change wavelength to meters and mass of neutron (back inside cover) to kg. Solve .

λ = 1.25 Å × 1 × 10 10 m 1 A = 1.25 × 10 10 m ; mass = 1.6749 × 10 27 kg

v = 6 . 6 2 6 × 1 0 3 4 k g - m 2 - s 1 s 2 × 1 1 . 6 7 4 9 × 1 0 2 7 k g × 1 1 . 2 5 × 1 0 1 0 m = 3 . 1 6 × 1 0 3 m / s

Check .  (6.6/1.6/1.25) × (10 −34 /10 −27 /10 −10 ) ≈ 3 × 10 3 m/s

6.50   m e = 9.1094 × 10 −31 kg (back inside cover of textbook)

λ = 6.626 × 10 34 kg-m 2 -s 1 s 2 × 1 9.1094 × 10 31 kg × 1 s 9 .47 × 10 6 m = 7.68 × 10 11 m

7.68 × 10 11 m × 1 Å 1 × 10 10 m = 0.768 Å

Because atomic radii and interatomic distances are on the order of 1–5 Å (Section 2.3), the wavelength of this electron is comparable to the size of atoms.

6.51 Analyze/Plan .  Use Δx ≥ h/4 π m Δv, paying attention to appropriate units. Note that the uncertainty in speed of the particle (Δv) is important, rather than the speed itself. Solve .

  1. m = 1.50 mg × 1 g 1000 mg × 1 kg 1000 g = 1.50 × 10 6 kg; Δ v = 0.01 m/s

    Δ x = 6.626 × 10 34 J-s 4 π ( 1 . 5 0 × 1 0 6 kg) (0 .01  m/s) 3.52 × 10 27 = 4 × 10 27 m

  2. m = 1.673 × 10 24 g = 1.673 × 10 27 kg; Δ v = 0.01 × 10 4 m/s

    Δ x = 6.626 × 10 34 J-s 4 π ( 1.673 × 10 27 kg ) (0 .01 × 10 4 m/s ) 3 × 10 10 m

Check .  The more massive particle in (a) has a much smaller uncertainty in position.

6.52   Δx ≥ = h/4πmΔv; use masses in kg, Δv in m/s.

  1. 6.626 × 10 34 J-s 4 π ( 9.109 × 10 31 kg ) (0 .01 × 10 5 m/s) = 6 × 10 8 m
  2. 6.626 × 10 34 J-s 4 π ( 1.675 × 10 27 kg ) (0 .01 × 10 5 m/s) = 3 × 10 11 m
  3. For particles moving with the same uncertainty in velocity, the more massive neutron has a much smaller uncertainty in position than the lighter electron. We know the position of the neutron with much greater precision.

Quantum Mechanics and Atomic Orbitals (Sections 6.5 and 6.6)

6.53

  1. False. The contour representation shows a volume where there is significant electron density. The electron can be moving anywhere within this volume.
  2. False. The probability of finding an electron at a given distance from the nucleus is the radial probability function. This is the probability density summed over all points that lie a distance r from the nucleus.

6.54

  1. False. Each electron behaves as a wave and is allowed to spread out over regions that are separated by a node.
  2. True. The radial probability function is the sum of the probability density at all points a distance r from the nucleus. The radial probability goes to zero at the distance of the node because the probability density goes to zero at this distance.
  3. False. For an s orbital, the number of nodes is n –  1.

6.55

  1. The possible values of l are ( n – 1) to 0. n = 4, l = 3, 2, 1, 0
  2. The possible values of m l are – l to + l . l = 2, m l = –2, –1, 0, 1, 2
  3. Because the value of m l is less than or equal to the value of l , m l = 2 must have an l -value greater than or equal to 2. In terms of elements that have been observed, the possibilities are 2, 3, and 4.

6.56

  1. For n = 3, there are nine uniques combinations of l and m l (values): ( l = 2;

    m l = –2, –1, 0, 1, 2; l = 1, m l = –1, 0, 1; l = 0, m l = 0).

  2. For n = 4, there are sixteen unique combinations of l and m l (values):

    ( l = 3, m l = –3 to +3; l = 2, m l = –2 to +2; l = 1, m l = –1 to +1; l = 0, = 0).

    In general, for each principal quantum number n , there are n values for l and n 2 values for m l . For each shell, there are n kinds of orbitals and n 2 total orbitals.

6.57

  1. 3p: n = 3, l = 1
  2. 2s: n = 2, l = 0
  3. 4f: n = 4, l = 3
  4. 5d: n = 5, l = 2

6.58

  1. 2, 1, 1;   2, 1, 0;   2, 1, –1
  2. 5, 2, 2;   5, 2, 1;   5, 2, 0;   5, 2, –1;   5, 2, –2

6.59

  1. 2, 1 ,0, –1, –2
  2. ½, –½

6.60

  1. 2, 3
  2. ½, –½

6.61   Impossible: (a)  1p, only l = 0 is possible for n = 1; (d)  2d, for n = 2, l = 1 or 0, but not 2

6.62

n l m l orbital
2 1 –1 2p (example)
1 0 0 1s
3 –3 2 not allowed ( l < n and + only)
3 2 –2 3d
2 0 –1 not allowed ( m l = – l to + l )
0 0 0 not allowed ( n ≠ 0)
4 2 1 4d
5 3 0 5f

6.63

image

Note that the lobes of the d xy orbital lie in the xy-plane and point between the x-axes and y-axes.

6.64

image

6.65

  1. The 1s and 2s orbitals of an H atom have the same overall spherical shape. The 2s orbital has a larger radial extension and one node, whereas the 1s orbital has continuous electron density. Because the 2s orbital is “larger,” there is greater probability of finding an electron farther from the nucleus in the 2s orbital.
  2. A single 2p orbital is directional in that its electron density is concentrated along one of the three Cartesian axes of the atom. The d x 2 y 2 orbital has electron density along both the x- and y-axes, whereas the p x orbital has density only along the x-axis.
  3. The average distance of an electron from the nucleus in a 3s orbital is greater than for an electron in a 2s orbital. In general, for the same kind of orbital, the larger the n value, the greater the average distance of an electron from the nucleus of the atom.
  4. 1s < 2p < 3d < 4f < 6s. In the H atom, orbitals with the same n value are degenerate and energy increases with increasing n value. Thus, the order of increasing energy is given above.

6.66

  1. In an s orbital, there are ( n – 1) nodes.
  2. The 2p x orbital has one node (the yz plane passing through the nucleus of the atom). The 3s orbital has two nodes.
  3. Probability density, ψ 2 (r), is the probability of finding an electron at a single point, r. The radial probability function, P(r), is the probability of finding an electron at any point that is distance r from the nucleus. Figure 6.19 contains plots of P(r) vs. r for 1s, 2s, and 3s orbitals. The most obvious features of these plots are the radii of maximum probability for the three orbitals, and the number and location of nodes for the three orbitals.

    By comparing plots for the three orbitals, we see that as n increases, the number of nodes increases and the radius of maximum probability (orbital size) increases.

  4. 2s = 2p < 3s < 4d < 5s. In the hydrogen atom, orbitals with the same n value are degenerate and energy increases with increasing n value.

Many-Electron Atoms and Electron Configurations (Sections 6.7 – 6.9)

6.67

  1. Yes. An He + ion has only one electron. In a one-electron particle, the energy of an orbital depends only on the value of n . Therefore, the 2s and 2p orbitals of He + have the same energy.
  2. Yes. In an He atom, the 2s orbital is lower in energy than the 2p orbitals. A helium atom is a two-electron particle. In multi-electron particles, electron-electron repulsions cause orbitals with the same n -value but different l -values to have different energies.

6.68

  1. The electron with the greater average distance from the nucleus feels a smaller attraction for the nucleus and is higher in energy. Thus the 3p is higher in energy than 3s.
  2. Because it has a larger n value, a 3s electron has a greater average distance from the chlorine nucleus than a 2p electron. The 3s electron experiences a smaller attraction for the nucleus and requires less energy to remove from the chlorine atom.

6.69

  1. No, both configurations follow the Pauli exclusion principle.
  2. No, both configurations obey Hund's rule.
  3. No. In the absence of a magnetic field, we cannot say which configuration has lower energy. The difference in the two configurations is the m s value of the electron in the 2s orbital. In the absence of an external magnetic field, electrons that only differ in m s have the same energy.

6.70

  1. Ag: [Kr]5s 1 4d 10 (1 unpaired electron in the 5s orbital)
  2. No. The experiment requires that the atoms in the beam each have an unpaired electron. The electron configuration of cadmium is: [Kr]5s 2 4d 10 . There are no unpaired electrons in the configuration of cadmium, so a beam of Cd atoms would not be deflected in a magnetic field.
  3. Yes. The electron configuration of fluorine is: [He]2s 2 2p 5 . Each fluorine atom has one unpaired electron in a 2p orbital. A beam of F atoms will be deflected in a magnetic field.

6.71 Analyze/Plan .  Each subshell has an l -value associated with it. For a particular l -value, permissible m l -values are – l to + l . Each m l -value represents an orbital, which can hold two electrons. Solve .

  1. 6
  2. 10
  3. 2
  4. 14

6.72

  1. 2
  2. 14
  3. 2
  4. 2

6.73

  1. ”Valence electrons” are those involved in chemical bonding. They are part (or all) of the outer-shell electrons listed after core electrons in a condensed electron configuration.
  2. ”Core electrons” are inner shell electrons that have the electron configuration of the nearest noble-gas element.
  3. Each box represents an orbital.
  4. Each half-arrow in an orbital diagram represents an electron. The direction of the half-arrow represents electron spin.

6.74

Element (a)  N (b)  Si (c)  Cl
Electron Configuration [He]2s 2 2p 3 [Ne]3s 2 3p 2 [Ne]3s 2 3p 5
Core electrons 2 10 10
Valence electrons 5 4 7
Unpaired electrons 3 2 1

6.75 Analyze/Plan .  Follow the logic in Sample Exercise 6.9. Solve .

  1. Cs: [Xe]6s 1
  2. Ni: [Ar]4s 2 3d 8
  3. Se: [Ar]4s 2 3d 10 4p 4
  4. Cd: [Kr]5s 2 4d 10
  5. U: [Rn]5f 3 6d 1 7s 2 . (Note the U and several other f- block elements have irregular d - and f -electron orders.)
  6. Pb: [Xe]6s 2 4f 14 5d 10 6p 2

6.76

  1. Mg: [Ne]3s 2 , 0 unpaired electrons
  2. Ge: [Ar]4s 2 3d 10 4p 2 , 2 unpaired electrons
  3. Br: [Ar]4s 2 3d 10 4p 5 , 1 unpaired electron
  4. V: [Ar]4s 2 3d 3 , 3 unpaired electrons
  5. Y: [Kr]5s 2 4d 1 , 1 unpaired electron
  6. Lu: [Xe]6s 2 4f 14 5d 1 , 1 unpaired electron

6.77

  1. Be, 0 unpaired electrons
  2. O, 2 unpaired electrons
  3. Cr, 6 unpaired electrons
  4. Te , 2 unpaired electrons

6.78

  1. 7A or 17 (halogens), 1 unpaired electron
  2. 4B or 4, 2 unpaired electrons
  3. 3A or 13 (row 4 and below), 1 unpaired electron
  4. the f-block elements Sm and Pu, 6 unpaired electrons

6.79

  1. The orbitals are not filled in order of increasing energy. The fifth electron would fill the 2p subshell (same n -value as 2s) before the 3s.
  2. The orbitals are not filled in order of increasing energy. After 6s, electrons fill the 4f subshell.
  3. The orbitals are not filled in order of increasing energy. The 3p subshell would fill before the 3d because it has the lower l -value and the same n -value. (If there were more electrons, 4s would also fill before 3d.)

6.80  Count the total number of electrons to assign the element.

  1. F: [He]2s 2 2p 5
  2. Ge: [Ar]4s 2 3d 10 4p 2
  3. Nb: [Kr]5s 2 4d 3

Additional Exercises

6.81

  1. λ A = 1.6 × 10 −7 m / 4.5 = 3.56 × 10 −8 = 3.6 × 10 −8 m

    λ B = 1.6 × 10 −7 m / 2 = 8.0 × 10 −8 m

  2. ν = c/λ; ν A = 2.998 × 10 8 m 1 s × 1 3.56 × 10 8 m = 8.4 × 10 15 s 1

    ν B = 2.998 × 10 8 m 1 s × 1 8.0 × 10 8 m = 3.7 × 10 15 s 1

  3. A: ultraviolet, B: ultraviolet

6.82

  1. ν = c / λ = 2 . 9 9 8 × 1 0 8 m s × 1 5 8 9 n m × 1 n m 1 × 1 0 9 m = 5 . 0 9 0 0 × 1 0 1 4 = 5 . 0 9 × 1 0 1 4 s 1
  2. E = h ν = 6 . 6 2 6 × 1 0 3 4 J - s × 5 . 0 9 0 0 × 1 0 1 4 s 1 × 6 . 0 2 2 × 1 0 2 3 p h o t o n s m o l × 0.1 mol = 2.03 × 10 4 J = 20.3 kJ
  3. Δ E = h ν = hc λ = 6 .626 × 10 34 J-s × 2 .998 × 10 8 m/s 589 nm × 1 nm 1 × 10 9 m = 3.37 × 10 19 J
  4. The 589 nm light emission is characteristic of Na + . If the pickle is soaked in a different salt long enough to remove all Na + , the 589 nm light would not be observed. Emission at a different wavelength, characteristic of the new salt, would be observed.

6.83

  1. Elements that emit in the visible: Ba (blue), Ca (violet-blue), K (violet), Na (yellow/orange). (The other wavelengths are in the ultraviolet.)
  2. Au: shortest wavelength, highest energy

    Na: longest wavelength, lowest energy

  3. λ = c/ ν = 2.998 × 10 8 m/s 9 .23 × 10 14 / s × 1 nm 1 × 10 9 m = 325 nm, Cu

6.84  All electromagnetic radiation travels at the same speed, 2.998 × 10 8 m/s. Change miles to meters and seconds to some appropriate unit of time.

391 × 10 6 mi × 1 .6093 km 1 mi × 1000 m 1 km × 1 s 2 .998 × 10 8 m × 1 min 60 s = 35.0 min

6.85

  1. ν = c/ λ = 2.998 × 10 8 m/s 320 nm × 1 nm 1 × 10 9 m = 9.4 × 10 14 s 1
  2. E = hc/ λ = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 3.2 × 10 7 m × 1 kJ 1000 J × 6.022 × 10 23 photons mole = 373 .8 = 370 kJ/mol
  3. UV-B photons have shorter wavelength and higher energy.
  4. Yes. The higher energy UV-B photons would be more likely to cause sunburn.

6.86  E = hc/λ → J/photon; total energy = power × time; photons = total energy / J / photon

E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 780 × 10 9 m = 2.5468 × 10 19 = 2.55 × 10 19 J/photon

0.10 mW = 0.10 × 10 3 J 1 s × 69 min × 60 s 1 min = 0.4140 = 0.41 J

0.4140 J × 1 photon 2.5468 × 10 19 J = 1.626 × 10 18 = 1.6 × 10 18 photons

6.87

  1. If a plant appears orange, it absorbs the complementary (opposite) color on the color wheel. The plant most strongly absorbs blue light in the range 430-490 nm.
  2. E = h c / λ = 6 . 6 2 6 × 1 0 3 4 J - s 4 5 5 n m × 2 . 9 9 8 × 1 0 8 m 1 s × 1 n m 1 × 1 0 9 m = 4 . 3 7 × 1 0 1 9 J

6.88

  1. ν = c / λ ; 2 . 9 9 8 × 1 0 8 m s × 1 5 4 2 n m × 1 n m 1 × 1 0 - 9 m = 5 . 5 3 1 4 × 1 0 1 4 = 5 . 5 3 × 1 0 1 4 s 1
  2. Calculate J/photon using E = hc/λ; change to kJ/mol.

    E photon = 6.626 × 10 34 J-s 542 × 10 9 m × 2.998 × 10 8 m s = 3.6651 × 10 19 = 3.67 × 10 19 J/photon

    3.6651 × 10 19 J photon × 6.022 × 10 23 photons mol × 1 kJ 1000 J = 220.71 = 221 kJ/mol

  3. Let E total be the total energy of an incident photon, E min be the minimum energy required to eject an electron, and E k be the “extra” energy that becomes the kinetic energy of the ejected electron.

    E total = E min + E k , E k = E total – E min = hν – hν 0 , E k = h(ν – ν 0 ). The slope of the line is the value of h, Planck’s constant.

6.89

  1. When an electron is excited to n = ∞, it is completely removed from the atom. The end result of this process is ionization, the production of H + .
  2. n i = 1 , n f = ; Δ E = 2.18 × 10 18 J [ 1 n f 2 1 n i 2 ] = 2.18 × 10 18 J (1/ 1 / 1 ) = 2.18 × 10 18 J

    λ = hc/E = 6.626 × 10 34 J-s 2.18 × 10 18 J × 2.998 × 10 8 m 1 s = 9.11 × 10 8 m = 91 .1 nm

  3. If light with a wavelength shorter than 91.1 nm is used to excite the H atom, the excess energy will become the kinetic energy of the ejected electron. (The potential energy of the ejected electron is, by definition, zero. It no longer experiences electrostatic interactions with the H atom.)
  4. The frequency associated with the wavelength calculated in part (b) is analogous to ν 0 on the plot in Exercise 6.88. Any excess kinetic energy imparted to the ejected electron corresponds to the sloped line to the right of ν 0 .

6.90

  1. “blue” cone, λ max = 450 nm = 450 × 10 −9 m

    E = hc/ λ = 6.626 × 10 34 J-s 450 × 10 9 m × 2.998 × 10 8 m 1 s = 4.41 × 10 19 J

    “green” cone, λ max = 545 nm = 545 × 10 −9 m

    E = hc/ λ = 6.626 × 10 34 J-s 545 × 10 9 m × 2.998 × 10 8 m 1 s = 3.64 × 10 19 J

    “red” cone, λ max = 585 nm = 585 × 10 −9 m

    E = hc/ λ = 6.626 × 10 34 J-s 5 85 × 10 9 m × 2.998 × 10 8 m 1 s = 3.40 × 10 19 J

  2. “blue” s c a t t e r i n g e f f i c i e n c y = ( 1 4 5 0 ) 4 ; “green” s c a t t e r i n g e f f i c i e n c y = ( 1 5 4 5 ) 4

    ratio of “blue” to “green” = ( 1 4 5 0 n m ) 4 ( 1 5 4 5 n m ) 4 = ( 5 4 5 4 5 0 ) 4 = 2 . 1 5

  3. Mainly, the shorter wavelengths perceived by the “blue” cone are scattered more efficiently, so there is more of the blue light to see. Also, the amplitude of the absorption curve for the “blue” cone is greater than the amplitudes of the other two curves. This indicates that our eyes are more sensitive to blue light than the other wavelengths. (It is also true that the intensities of the different wavelengths reaching Earth are not the same, but this information is not conveyed in the exercise.)

6.91

  1. Lines with n f = 1 lie in the ultraviolet (see Solution 6.42) and with n f = 2 lie in the visible (see Solution 6.41). Lines with n f = 3 will have smaller ΔE and longer wavelengths and lie in the infrared.
  2. Use Equation 6.6 to calculate ΔE, then λ = hc/ΔE.

    n i = 4 , n f = 3 ; Δ E = 2.18 × 10 18 J [ 1 n f 2 1 n i 2 ] = 2.18 × 10 18 J (1/9 1 / 16 )

    λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 ( 1 / 9 1 / 16 ) = 1.87 × 10 6 m

    n i = 5 , n f = 3 ; λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 ( 1 / 9 1 / 25 ) = 1.28 × 10 6 m

    n i = 6 , n f = 3 ; λ = hc/E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 2.18 × 10 18 ( 1 / 9 1 / 36 ) = 1.09 × 10 6 m

    These three wavelengths are all greater than 1 µm or 1 × 10 −6 m. They are in the infrared, close to the visible edge (0.7 × 10 −6 m).


6.92

  1. Gaseous atoms of various elements in the sun’s atmosphere typically have ground state electron configurations. When these atoms are exposed to radiation from the sun, the electrons change from the ground state to one of several allowed excited states. Atoms absorb the wavelengths of light that correspond to these allowed energy changes. All other wavelengths of solar radiation pass through the atmosphere unchanged. Thus, the dark lines are the wavelengths that correspond to allowed energy changes in atoms of the solar atmosphere. The continuous background is all other wavelengths of solar radiation.
  2. The scientist should record the absorption spectrum of pure neon or other elements of interest. The Fraunhofer lines that belong to a particular element will appear at the same wavelength as the lines in the absorption spectrum of that element.

6.93

  1. not valid, m l cannot be greater than l .
  2. valid
  3. valid
  4. not valid, the only possible values for m s are + ½ and – ½ .
  5. not valid, the maximum value for l is ( n – 1).

6.94

  1. He + is hydrogen-like because it is a one-electron particle. An He atom has two electrons. The Bohr model is based on the interaction of a single electron with the nucleus but does not accurately account for additional interactions when two or more electrons are present.
  2. Divide each energy by the smallest value to find the integer relationship.

    H: –2.18 × 10 −18 /–2.18 × 10 −18 = 1;  Z = 1

    He + : –8.72 × 10 −18 /–2.18 × 10 −18 = 4;  Z = 2

    Li 2+ : –1.96 × 10 −17 /–2.18 × 10 −18 = 9;  Z = 3

    The ground-state energies are in the ratio of 1:4:9, which is also the ratio Z 2 , the square of the nuclear charge for each particle.

    The ground state energy for hydrogen-like particles is:

    E = R H Z 2 . (By definition, n = 1 for the ground state of a one-electron particle.)

  3. Z = 6 for C 5+ . E = –2.18 × 10 −18 J (6) 2 = –7.85 × 10 −17 J

6.95 Plan .   Calculate v from kinetic energy. λ = h/mv. Solve .

E k = mv 2 / 2 ; v 2 = 2 E k /m; v = 2 E k /m

v = ( 2 × 2.147 × 10 15 kg-m 2 / s 2 9.1094 × 10 31 kg ) 1 / 2 = 6.866 × 10 7 = 6.87 × 10 7 m/s

λ = h / m v = 6 . 6 2 6 × 1 0 3 4 J - s 9 . 1 0 9 4 × 1 0 3 1 kg × 6 . 8 6 6 × 1 0 7 m / s × 1 kg - m 2 / s 2 1 J = 1 . 0 6 × 1 0 1 1 m = 1 0 . 6 p m

6.96  Heisenberg postulated that the dual nature of matter places a limitation on how precisely we can know both the position and momentum of an object. This limitation is significant at the subatomic particle level. The Star Trek transporter (presumably) disassembles humans into their protons, neutrons, and electrons, moves the particles at high speed (possibly the speed of light) to a new location, and reassembles the particles into the human. Heisenberg’s uncertainty principle indicates that if we know the momentum (mv) of the moving particles, we can’t precisely know their position (x). If a few of the subatomic particles don’t arrive in exactly the correct location, the human would not be reassembled in their original form. So, the “Heisenberg compensator” is necessary to make sure that the transported human arrives at the new location intact.


6.97

  1. A subatomic particle is so small that we must use light to measure its position. The interacting photons will impart some momentum to the subatomic particle, thus disturbing it. The shorter the wavelength of the photon, the more accurate the measurement but the greater the momentum imparted to the particle and the bigger the disturbance.
  2. An ongoing discussion in quantum theory is whether we can know the quantum states of a system without observing and thus disturbing the system. One interpretation of quantum theory indicated that a system could have multiple acceptable states before it was observed in a single state. That is, the act of observation defined the state. Schrodinger articulated this question on a macroscopic level with his “cat” paradox. Recently, physicists have devised clever ways to observe “cat states,” where the act of observing does not destroy the simultaneous states.

6.98

  1. Probability density, [ψ(r)] 2 , is the probability of finding an electron at a single point at distance r from the nucleus. The radial probability function, 4πr 2 , is the probability of finding an electron at any point on the sphere defined by radius r.

    P(r) = 4πr 2 [ψ(r)] 2 .

  2. The term 4πr 2 explains the differences in plots of the two functions. Plots of the probability density, [ψ(r) 2 ], for s orbitals shown in Figure 6.22 each have their maximum value at r = 0, with ( n – 1) smaller maxima at greater values of r. The plots of radial probability, P(r), for the same s orbitals shown in Figure 6.19 have values of zero at r = 0 and the size of the maxima increases. P(r) is the product of [ψ(r)] 2 and 4πr 2 . At r = 0, the value of [ψ(r)] 2 is finite and large, but the value of 4πr 2 is zero, so the value of P(r) is zero. As r increases, the values of [ψ(r)] 2 vary as shown in Figure 6.22, but the values of 4πr 2 increase continuously, leading to the increasing size of P(r) maxima as r increases.
  3. image

6.99

  1. The p z orbital has a nodal plane where z = 0. This is the xy plane.
  2. The d xy orbital has 4 lobes and 2 nodal planes, the 2 planes where x = 0 and y = 0. These are the yz and xz planes.
  3. The d x 2 y 2 has 4 lobes and 2 nodal planes, the planes where x 2 – y 2 = 0. These are the planes that bisect the x and y axes and contain the z axis.

6.100

  1. This frequency is in the radiowave portion of the electromagnetic spectrum. One megahertz is 1 × 10 6 Hz; 600 MHz is 6 × 10 8 Hz. According to Figure 6.4, this frequency is in the radiowave region.
  2. Δ E = h ν = 6.626 × 10 34 J-s × 450 × 10 6 s = 2.98 × 10 25 J
  3. The proton. When radiowaves are absorbed during an NMR or MRI experiment, it is the spin of a nucleus that changes. The nucleus of a hydrogen atom is a single proton.

6.101  If m s had three allowed values instead of two, each orbital would hold three electrons instead of two. Assuming that the same orbitals are available (that there is no change in the n , l , and m l values), the number of elements in each of the first four rows would be:

1 st row:  1 orbital  × 3 =   3 elements

2 nd row:  4 orbitals × 3 = 12 elements

3 rd row:  4 orbitals × 3 = 12 elements

4 th row:  9 orbitals × 3 = 27 elements

The s-block would be 3 columns wide, the p-block 9 columns wide and the d-block 15 columns wide.

6.102

  1. Br: [Ar]4s 2 3d 10 4p 5 , 1 unpaired electron
  2. Ga: [Ar]4s 2 3d 10 4p 1 , 1 unpaired electron
  3. Hf: [Xe]6s 2 4f 14 5d 2 , 2 unpaired electrons
  4. Sb: [Kr]5s 2 4d 10 5p 3 , 3 unpaired electrons
  5. Bi: [Xe]6s 2 4f 14 5d 10 6p 3 , 3 unpaired electrons
  6. Sg: [Rn]7s 2 5f 14 6d 4 , 4 unpaired electrons

6.103  The core would be the electron configuration of element 118. If no new subshell begins to fill, the condensed electron configuration of element 126 would be similar to those of elements vertically above it on the periodic chart, Pu and Sm. The condensed configuration would be [118]8s 2 6f 6 . On the other hand, the 5g subshell could begin to fill after 8s, resulting in the condensed configuration [118]8s 2 5g 6 . Exceptions are also possible (likely).

6.104

  1. Neutral H atoms have a single unpaired electron. For a beam of H atoms to be deflected by a magnetic field, the unpaired electrons must interact with the magnetic field. For a single beam of atoms to be split into two beams that are deflected in opposite directions, the unpaired electron on each atom must have some characteristic that can interact in two different ways with a magnetic field. We call that characteristic “electron spin.” The significance of this observation is experimental evidence for electron spin.

  1. If the strength of the magnetic field were increased, the magnitude of the deflection would increase.
  1. If H atoms were replaced by He atoms, no deflection would occur. Helium has no unpaired electrons to interact with the magnetic field.
  1. The electron configuration of Ag is [Kr]5s 1 4d 10 . Neutral Ag atoms each have a single unpaired electron. Each unpaired electron has two possible m s values and a beam of Ag atoms will be split in two by the magnetic field.

Integrative Exercises

6.105

  1. We know the wavelength of microwave radiation, the volume of coffee to be heated, and the desired temperature change. Assume the density and heat capacity of coffee are the same as pure water. We need to calculate: (i) the total energy required to heat the coffee and (ii) the energy of a single photon to find (iii) the number of photons required.
    1. From Chapter 5 , the heat capacity of liquid water is 4.184 J/g-°C.

      To find the mass of 200 mL of coffee at 23 °C, use the density of water given in Appendix B.

      200 mL × 0 .997 g 1 mL = 199.4 = 199 g coffee

      4.184 J 1 g- ° C × 199.4 g × (60 ° C 23 ° C) = 3.087 × 10 4 J = 31 kJ

    2. E = hc/ λ = 6.626 × 10 34 J-s × 2 .998 × 10 8 m 1 s × 1 0.112 m = 1 .77 × 10 24 J 1 photon
    3. 3.087 × 10 4 J × 1 photon 1.774 × 10 24 J = 1.7 × 10 28 photons

    (The answer has 2 sig figs because the temperature change, 37 °C, has 2 sig figs.)

  2. 1 W = 1 J/s. 900 W = 900 J/s. From part (a), 31 kJ are required to heat the coffee.

    3.087 × 10 4 J × 1 s 900 J = 34.30 = 34 s

6.106 Δ H rxn o = Δ H f o O 2 ( g ) + Δ H f o O(g) Δ H f o O 3 ( g )

Δ H rxn o = 0 + 247.5 kJ 142.3 kJ = + 105.2 kJ

105.2 kJ mol O 3 × 1 mol O 3 6.022 × 10 23 molecules × 1000 J 1 kJ = 1.747 × 10 19 J O 3 molecule

Δ E = hc/ λ ; λ = hc Δ E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 1.747 × 10 19 J = 1.137 × 10 6 m

Radiation with this wavelength is in the infrared portion of the spectrum. (Clearly, processes other than simple photodissociation cause O 3 to absorb ultraviolet radiation.)


6.107

  1. The electron configuration of Zr is [Kr]5s 2 4d 2 and that of Hf is [Xe]6s 2 4f 14 5d 2 . Although Hf has electrons in f orbitals as the rare earth elements do, the 4f subshell in Hf is filled, and the 5d electrons primarily determine the chemical properties of the element. Thus, Hf should be chemically similar to Zr rather than the rare earth elements.
  2. ZrCl 4 (s) + 4 Na(l) → Zr(s) + 4 NaCl(s)

    This is an oxidation–reduction reaction; Na is oxidized and Zr is reduced.

  3. 2 ZrO 2 (s) + 4 Cl 2 (g) + 3 C(s) → 2 ZrCl 4 (s) + CO 2 (g) + 2 CO(g)

    55.4 g ZrO 2 × 1 mol ZrO 2 123.2 g ZrO 2 × 2 mol ZrCl 4 2 mol ZrO 2 × 233.0 g ZrCl 4 1 mol ZrCl 4 = 105 g ZrCl 4

  4. In ionic compounds of the type MCl 4 and MO 2 , the metal ions have a 4+ charge, indicating that the neutral atoms have lost 4 electrons. Zr, [Kr]5s 2 4d 2 , loses the 4 electrons beyond its Kr core configuration. Hf, [Xe]6s 2 4f 14 5d 2 , similarly loses its four 6s and 5d electrons, but not electrons from the “complete” 4f subshell.

6.108

  1. Each oxide ion, O 2– , carries a 2- charge. Each metal oxide is a neutral compound, so the metal ion or ions must adopt a total positive charge equal to the total negative charge of the oxide ions in the compound. The table below lists the electron configuration of the neutral metal atom, the positive charge of each metal ion in the oxide, and the corresponding electron configuration of the metal ion.
    1. K:    [Ar] 4s 1 1+     [Ar]
    2. Ca:  [Ar] 4s 2 2+     [Ar]
    3. Sc:   [Ar] 4s 2 3d 1 3+      [Ar]
    4. Ti:   [Ar] 4s 2 3d 2 4+      [Ar]
    5. V:    [Ar] 4s 2 3d 3 5+      [Ar]
    6. Cr:   [Ar] 4s 1 3d 5 6+      [Ar]

    Each metal atom loses all (valence) electrons beyond the Ar core configuration. In K 2 O, Sc 2 O 3 , and V 2 O 5 , where the metal ions have odd charges, 2 metal ions are required to produce a neutral oxide.

    1. potassium oxide
    2. calcium oxide
    3. scandium (III) oxide
    4. titanium (IV) oxide
    5. vanadium (V) oxide
    6. chromium (VI) oxide

    (Roman numerals are required to specify the charges on the transition metal ions, because more than one stable ion may exist.)


  1. Recall that Δ H f ° = 0 for elements in their standard states. In these reactions, M(s) and H 2 (g) are elements in their standard states.
    1. K 2 O(s) + H 2 (g) → 2 K(s) + H 2 O(g)

      Δ H ° = Δ H f ° H 2 O(g) + 2 Δ H f ° K(s) Δ H K 2 O(s) Δ H f ° H 2 ( g )

      Δ H ° = 241.82 kJ + 2(0) ( 363.2 kJ) 0 = 121.4 kJ

    2. CaO(s) + H 2 (g) → Ca(s) + H 2 O(g)

      Δ H ° = Δ H f ° H 2 O(g) + Δ H f ° Ca(s) Δ H f ° CaO(s) Δ H f ° H 2 ( g )

      Δ H ° = 241.82 kJ + 0 ( 635.1 kJ) 0 = 393.3 kJ

    3. TiO 2 (s) + 2 H 2 (g) → Ti(s) + 2 H 2 O(g)

      Δ H ° = 2 Δ H f ° H 2 O(g) + Δ H f ° Ti(s) Δ H f ° TiO 2 (s) 2 Δ H f ° H 2 ( g )

      = 2(–241.82) + 0 – (–938.7) – 2(0) = 455.1 kJ

    4. V 2 O 5 (s) + 5 H 2 (g) → 2 V(s) + 5 H 2 O(g)

      Δ H ° = 5 Δ H f ° H 2 O(g) + 2 Δ H f ° V(s) Δ H f ° V 2 O 5 (s) 5 Δ H f ° H 2 ( g )

      = 5(–241.82) + 2(0) – (–1550.6) – 5(0) = 341.5 kJ

  1. Δ H f ° becomes more negative moving from left to right across this row of the periodic chart. Because Sc lies between Ca and Ti, the median of the two Δ H f ° values is approximately –785 kJ/mol. However, the trend is clearly not linear. Dividing the Δ H f ° values by the positive charge on the pertinent metal ion produces the values –363, –318, –235, and –310. The value between Ca 2+ (–318) and Ti 4+ (–235) is Sc 3+ (–277). Multiplying (–277) by 3, a value of approximately –830 kJ results. A reasonable range of values for Δ H f ° of Sc 2 O 3 (s) is then –785 to –830 kJ/mol.

6.109

  1. Bohr’s theory was based on the Rutherford “nuclear” model of the atom. That is, Bohr theory assumed a dense positive charge at the center of the atom and a diffuse negative charge (electrons) surrounding it. Bohr’s theory then specified the nature of the diffuse negative charge. The prevailing theory before the nuclear model was Thomson’s plum pudding or watermelon model, with discrete electrons scattered about a diffuse positive charge cloud. Bohr’s theory could not have been based on the Thomson model of the atom.
  2. De Broglie’s hypothesis is that electrons exhibit both particle and wave properties. Thomson’s conclusion that electrons have mass is a particle property, whereas the nature of cathode rays is a wave property. De Broglie’s hypothesis actually rationalizes these two seemingly contradictory observations about the properties of electrons.

6.110

  1. 238 U: 92 p, 146 n, 92 e; 235 U: 92 p, 143 n, 92 e

    In keeping with the definition isotopes, only the number of neutrons is different in the two nuclides. Because the two isotopes have the same number of electrons, they will have the same electron configuration.

  2. U: [Rn]7s 2 5f 4
  3. From Figure 6.31, the actual electron configuration is [Rn]7s 2 5f 3 6d 1 . The energies of the 6d and 5f orbitals are very close, and electron configurations of many actinides include 6d electrons.
  4. 92 238 U 90 234 Th + 2 4 He 234 Th has 90 p, 144 n, 90 e . 238 U has lost 2 p, 2 n, 2 e .

    These are organized into 2 4 He shown in the nuclear reaction above.

  5. From Figure 6.31, the electron configuration of Th is [Rn]7s 2 6d 2 . This is not really surprising because there are so many rare earth electron configurations that are exceptions to the expected orbital filling order. However, Th is the only rare earth that has two d valence electrons. Furthermore, the configuration of Th is different than that of Ce, the element above it on the periodic chart, so the electron configuration is at least interesting.

 

7 Periodic Properties of the Elements

Visualizing Concepts

7.1 Analyze/Plan. Consider Equation 7.1 in relation to the illustration (and Figure 7.2). The intensity of the bulb represents the nuclear charge, Z. The thickness of the frosting represents the shielding, S. Solve.

  1. Moving from boron to carbon, the intensity of the bulb increases because Z increases from 5 to 6. The thickness of the frosting stays the same because the core electron configuration is the same for both atoms (The added electron occupies its own p orbital, so electron repulsion doesn’t change much.)
  2. Moving from boron to aluminum, the intensity of the bulb increases because Z increases from 5 to 13. The thickness of the frosting also increases because Al has the core configuration of Ne, while B has the core of He. (We know from the chapter that the increase in Z dominates and Z eff increases slightly going down a column; the 3p valence electron of Al "sees” a brighter light than the 2p valence electron of B.)

7.2   The order of radii is Br > Br > F, so the largest brown sphere is Br , the intermediate blue one is Br, and the smallest red one is F.

7.3

  1. Mg 2+ is isoelectronic with Ne, K + , and Cl are isoelectronic with Ar, and Se 2– is isoelectronic with Kr. The atomic radii of the noble gases increase moving down the column, so this gives the rough order of size for the corresponding isoelectronic ions. The Cl ion is larger than K + because it has a smaller positive nuclear charge holding the same number and configuration of electrons. The order of ionic radii is then Mg 2+ < K + < Cl < Se 2– ; these ions match the spheres moving from left to right.
  2. Ca 2+ and S 2– are both isoelectronic with Ar, as are K + and Cl . For ions in an isoelectronic series, the larger the nuclear charge, the smaller the ionic radius. Ca 2+ is smaller than K + and fits between the two leftmost spheres. S 2– is larger than Cl and fits between the two rightmost spheres.

7.4   The red sphere represents a metal and the blue sphere represents a nonmetal. The size of the red sphere decreases on reaction, so it loses one or more electrons and becomes a cation. Metals lose electrons when reacting with nonmetals, so the red sphere represents a metal. The size of the blue sphere increases on reaction, so it gains one or more electrons and becomes an anion. Nonmetals gain electrons when reacting with metals, so the blue sphere represents a nonmetal.


7.5

  1. The bonding atomic radius of A, r A , is d 1 /2. The distance d 2 is the sum of the bonding atomic radii of A and X, r A + r X . We know that r A = d 1 /2, so d 2 = r X + d 1 /2, r X = d 2 – d 1 /2.
  2. The length of the X–X bond is 2r X .

    2r X = 2 (d 2 – d 1 /2) = 2d 2 – d 1 .

7.6

  1. The 3 d subshell is missing.
  2. Statement (ii) is the best description of why the 2 s and 2 p subshells have different energies in Na.
  3. In a Na atom, the highest energy electron is in the 3 s subshell.
  4. In a Na vapor lamp, the highest energy 3 s electron is excited into the empty 3 p subshell.

7.7   The trend for bonding atomic radius (1) is shown in chart (iii).

The trend for first ionization energy (2) is shown in chart (ii).

The trend for effective nuclear charge is shown in chart (i).

7.8

  1. X + 2 F 2 → XF 4
  2. If X is a nonmetal, XF 4 is a molecular compound. If X is a metal, XF 4 is ionic. For an ionic compound with this formula, X would have a charge of 4+, and a much smaller bonding atomic radius than F . X in the diagram has about the same bonding radius as F, so it is likely to be a nonmetal.

Periodic Table; Effective Nuclear Charge (Sections 7.1 and 7.2)

7.9

  1. The results are 2, 8, 18, 32.
  2. The atomic numbers of the noble gases are 2, 10, 18, 36, 54, and 86. The differences between sequential pairs of these atomic numbers is 8, 8, 18, 18, and 32. These differences correspond to the results in (a). They represent the filling of new subshells when moving across the next row of the periodic chart.
  3. The Pauli exclusion principle is the source of the “2” in the expressions in part (a). The Pauli principle states that no two electrons can have the same four quantum numbers. Because m s has only two possible values, the consequence is that an atomic orbital can hold a maximum of “2” electrons.

7.10   Assuming eka - means one place below or under, eka-manganese on Figure 7.1 is technetium, Tc.

7.11

  1. According to Figure 7.1, of the elements listed, only Fe was known before 1700.
  2. The seven metals known in ancient times, Fe, Cu, Ag, Sn, Au, Hg, and Pb, are mostly near the bottom of the activity series, Table 4.5. These less active metals are present in nature in elemental form; they can be observed directly and their isolation does not require chemical processing.

7.12

  1. In order of increasing atomic mass, the element following chlorine is potassium, K.
  2. Potassium is a reactive metal that is solid at room temperature and pressure. Elements in group 8A are the noble gases. All are unreactive nonmetals that exist as gases at ambient conditions.

7.13 Analyze/Plan . Z eff values for elements 3–18 are shown as the green line in Figure 7.5. Use Equation 7.1 to calculate Z eff values for elements 1 and 2, H and He. Compare the values. Solve .

Z eff = Z – S, where Z is the atomic number and S is the number of core electrons. For H and He, the valence electrons have n = 1, and there are no core electrons; S = 0.

H: 1 – 0 = 1; He: 2 – 0 = 2.

The minimum value of Z eff from Figure 7.5 is 1 for Li and Na. The maximum value is 8 for Ne and Ar.

For elements 1–18, H, Li and Na have minimum values of Z eff ; Ne and Ar have maximum values.

7.14  Statement (iii) is incorrect. Because of the nearly uniform spherical distribution of the core electrons, they screen much more effectively than valence electrons.

7.15

  1. Analyze/Plan . Z eff = Z – S. Find the atomic number, Z, of Na and K. Write their electron configurations and count the number of core electrons. Assume S = number of core electrons.

    Solve . Na: Z = 11; [Ne]3s 1 . In the Ne core there are 10 electrons. Z eff = 11 − 10 = 1. K: Z = 19; [Ar]4s 1 . In the Ar core there are 18 electrons. Z eff = 19 − 18 = 1.

  2. Analyze/Plan . Z eff = Z – S. Write the complete electron configuration for each element to show counting for Slater’s rules. S = 0.35 [# of electrons with same n ] + 0.85 [# of electrons with ( n −1)] + 1[# of electrons with ( n −2)].

    Solve . Na: 1s 2 2s 2 2p 6 3s 1 . S = 0.35(0) + 0.85(8) + 1(2) = 8.8. Z eff = 11 – 8.8 = 2.2

    K: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 . S = 0.35(0) + 0.85(8) + 1(10) = 16.8. Z eff = 19 – 16.8 = 2.2

  3. For both Na and K, the two values of Z eff are 1.0 and 2.2. The Slater value of 2.2 is closer to the values of 2.51 (Na) and 3.49 (K) obtained from detailed calculations.
  4. Both approximations, “core electrons 100% effective” and Slater, yield the same value of Z eff for Na and K. Neither approximation accounts for the gradual increase in Z eff moving down a group.
  5. Following the trend from detailed calculations, we predict a Z eff value of approximately 4.5 for Rb.

7.16  Follow the method in the preceding question to calculate Z eff values.

  1. Si: Z = 14; [Ne]3s 2 3p 2 . 10 electrons in the Ne core. Z eff = 14 – 10 = 4

    Cl: Z = 17; [Ne]3s 2 3p 5 . 10 electrons in the Ne core. Z eff = 17 – 10 = 7

  2. Si: 1s 2 2s 2 2p 6 3s 2 3p 2 . S = 0.35(3) + 0.85(8) + 1(2) = 9.85. Z eff = 14 – 9.85 = 4.15

    Cl: 1s 2 2s 2 2p 6 3s 2 3p 5 . S = 0.35(6) + 0.85(8) + 1(2) = 10.90. Z eff = 17 – 10.90 = 6.10

  3. The Slater values of 4.15 (Si) and 6.10 (Cl) are closer to the results of detailed calculations, 4.29 (Si) and 6.12 (Cl).

  1. The Slater method of approximation more closely approximates the gradual increase in Z eff moving across a row. The “core 100%-effective” approximation underestimates Z eff for Si but overestimates it for Cl. Slater values are closer to detailed calculations and a better indication of the change in Z eff moving from Si to Cl.
  1. Relative to Si, P has one more proton (Z + 1) and one more 3p electron (S + 0.35). It is reasonable to predict that the difference in Z eff will be +0.65. That is, Z eff for P will be (4.15 + 0.65) = 4.80.

7.17  Krypton has a larger nuclear charge (Z = 36) than argon (Z = 18). The shielding of electrons in the n = 3 shell by the 1s, 2s, and 2p core electrons in the two atoms is approximately equal, so the n = 3 electrons in Kr experience a greater effective nuclear charge and are thus situated closer to the nucleus.

7.18  Mg < P < K < Ti < Rh. The shielding of electrons in the n = 3 shell by 1s, 2s, and 2p core electrons in these elements is approximately equal, so the effective nuclear charge increases as Z increases.

Atomic and Ionic Radii (Section 7.3)

7.19  The quantity described in (b) must be measured experimentally in order to determine the bonding atomic radius of an atom. Bonding atomic radius is a property of a bonded atom. The measurement must be done on an atom participating in a chemical bond.

7.20

  1. In the figure, the relevant distance is 3.72 Å. This is the distance between two Ar atoms that are touching. One-half this distance is the effective radius of an Ar atom in the close packed structure, 1.86 Å.
  2. According to Figure 7.7, the bonding atomic radius of Ar is 1.06 Å. This value is significantly smaller than 1.86 Å, the effective radius of Ar in the close packed solid.
  3. No, the Ar atoms are not held together by chemical bonds in the close packed solid. According to Figure 7.7, an Ar–Ar chemical bond would have a distance of approximately 2.12 Å, a much closer approach than the 3.72 Å in the close packed solid.

7.21

  1. The atomic ( metallic ) radius of W is the interatomic W–W distance divided by two, 2.74 Å/2= 1.37 Å.
  2. Under high pressure, we expect atoms in a pure substance to move closer together. That is, the distance between W atoms will decrease.

7.22  Statement (iv) is incorrect. Moving left to right in a particular period, the significant nuclear buildup while adding electrons into the same d subshell causes Z to increase and radii to decrease.

7.23  From bonding atomic radii in Figure 7.7, As–I = 1.19 Å + 1.39 Å = 2.58 Å. This is very close to the experimental value of 2.55 Å in AsI 3 .

7.24  Bi−I = 2.81 Å = r Bi + r I . From Figure 7.7, r I = 1.39 Å.

r Bi = [Bi−I] – r I = 2.81 Å – 1.39 Å = 1.42 Å.


7.25 Plan .  Locate each element on the periodic charge and use trends in radii to predict their order. Solve .

  1. Cs > K > Li
  2. Pb > Sn > Si
  3. N > O > F

7.26

  1. Na < Ca < Ba
  2. As < Sn < In
  3. Be < Si < Al. This order assumes the increase in radius from the second to the third row is greater than the decrease moving right in the third row. Radii in Figure 7.7 confirm this assumption.

7.27

  1. False. Cations are smaller than their corresponding neutral atoms. Electrostatic repulsions are reduced by removing an electron from a neutral atom, Z eff increases, and the cation is smaller.
  2. True. [See (a) above.]
  3. False. I is bigger than Cl . Going down a column, the n value of the valence electrons increases and they are farther from the nucleus. Valence electrons also experience greater shielding by core electrons. The greater radial extent of the valence electrons outweighs the increase in Z, and the size of particles with like charge increases.

7.28

  1. As Z stays constant and the number of electrons increases, the electron-electron repulsions increase, the electrons spread apart, and the anion becomes larger. The reverse is true for the cation, which becomes smaller than the neutral atom.

    I > I > I +

  2. For cations with the same charge, ionic radii increase going down a column because there is an increase in the principle quantum number and the average distance from the nucleus of the outer electrons.

    Ca 2+ > Mg 2+ > Be 2+

  3. Fe: [Ar]4s 2 3d 6 ; Fe 2+ : [Ar]3d 6 ; Fe 3+ : [Ar]3d 5 . The 4s valence electrons in Fe are on average farther from the nucleus than the 3d electrons, so Fe is larger than Fe 2+ . Because there are five 3d orbitals, in Fe 2+ at least one orbital must contain a pair of electrons. Removing one electron to form Fe 3+ significantly reduces repulsion, increasing the nuclear charge experienced by each of the other d electrons and decreasing the size of the ion. Fe > Fe 2+ > Fe 3+

7.29   Ga 3+ : none; Zr 4+ : Kr; Mn 7+ : Ar; I : Xe; Pb 2+ : Hg

7.30

  1. Cl : Ar
  2. Sc 3+ : Ar
  3. Fe 2+ : [Ar]3d 6 . There is no neutral atom with the same electron configuration. Fe 2+ has 24 electrons. Neutral Cr has 24 electrons, [Ar]4s 1 3d 5 . Because transition metals fill the s subshell first but also lose s electrons first when they form ions, many transition metal ions do not have neutral atoms with the same electron configuration.
  4. Zn 2+ : [Ar]3d 10 . There is no neutral atom with same electron configuration.
  5. Sn 4+ : [Kr]4d 10 ; a neutral Pd atom has 46 electrons and an anomalous electron configuration which is the same as the electron configuration of Sn 4+ .

7.31

  1. Analyze/Plan . Follow the logic in Sample Exercise 7.4.

    Solve . Na + is smaller. Because F and Na + are isoelectronic, the ion with the larger nuclear charge, Na + , has the smaller radius.

  2. Analyze/Plan . The electron configuration of the ions is [Ne] or [He]2s 2 2p 6 . The ions have either 10 core electrons or 2 core electrons. Apply Equation 7.1 to both cases and check the result.

    Solve . F : Z = 9. For 10 core electrons, Z eff = 9 – 10 = −1. Although we might be able to interpret a negative value for Z eff , positive values will be easier to compare; we will assume a He core of 2 electrons.

    F , Z = 9. Z eff = 9 − 2 = 7.   Na + : Z eff = 11 – 2= 9

  3. Analyze/Plan . The electron of interest has n = 2. There are seven other n = 2 electrons, and two n = 1 electrons.

    Solve . S = 0.35(7) + 0.85(2) + 1(0) = 4.15

    F : Z eff = 9 – 4.15 = 4.85.   Na + : Z eff = 11 – 4.15 = 6.85

  4. For isoelectronic ions (without d electrons), the electron configurations and therefore shielding values (S) are the same. Only the nuclear charge changes. So, as nuclear charge (Z) increases, effective nuclear charge (Z eff ) increases and ionic radius decreases.

7.32

  1. K + (larger Z) is smaller.
  2. Cl and K + : [Ne]3s 2 3p 6 . 10 core electrons

    Cl , Z = 17. Z eff = 17 – 10 = 7

    K + , Z = 19. Z eff = 19 – 10 = 9

  3. Valence electron, n = 3; seven other n = 3 electrons; eight n = 2 electrons; two n = 1 electrons. S = 0.35(7) + 0.85(8) + 1(2) = 11.25

    Cl : Z eff = 17 – 11.25 = 5.75. K + : Z eff = 19 – 11.25 = 7.75

  4. For isoelectronic ions (without d electrons), the electron configurations and therefore shielding values (S) are the same. Only the nuclear charge changes. So, as nuclear charge (Z) increases, effective nuclear charge (Z eff ) increases and ionic radius decreases.

7.33 Analyze / Plan .  Use relative location on periodic chart and trends in atomic and ionic radii to establish the order.

  1. Cl < S < K
  2. K + < Cl < S 2–
  3. Even though K has the largest Z value, the n -value of the outer electron is larger than the n -value of valence electrons in S and Cl so K atoms are largest. When the 4s electron is removed, K + is isoelectronic with Cl and S 2– . The larger Z value causes the 3p electrons in K + to experience the largest effective nuclear charge and K + is the smallest ion.

7.34

  1. Se < Se 2– < Te 2–
  2. Co 3+ < Fe 3+ < Fe 2+
  3. Ti 4+ < Sc 3+ < Ca
  4. Be 2+ < Na + < Ne

7.35

  1. O 2– is larger than O because the increase in electron repulsions that accompany addition of an electron causes the electron cloud to expand.
  2. S 2– is larger than O 2– because for particles with like charges, size increases going down a family.
  3. S 2– is larger than K + because the two ions are isoelectronic and K + has the larger Z and Z eff .
  4. K + is larger than Ca 2+ because the two ions are isoelectronic and Ca 2+ has the larger Z and Z eff .

7.36  Make a table of d(measured), d(ionic radii), d(covalent radii), as well as differences between measured and estimated values. The estimated distances are just the sum of the various ionic radii from Figure 7.8 and covalent radii from Figure 7.7. All distances and differences are given in Å. Use these values to judge accuracy in part (c).

(a,b)

d(meas) (a) d(ion) (b) Δ(ion – meas) (c) d(cov) (c) Δ(cov – meas)
Li−F 2.01 2.09 0.08 1.85 –0.16
Na−Cl 2.82 2.83 0.01 2.68 –0.14
K−Br 3.30 3.34 0.04 3.23 –0.11
Rb−I 3.67 3.72 0.05 3.59 –0.08
(c) Distance estimates from bonding atomic radii are not as accurate as those from ionic radii. This indicates that bonding in the series LiF, NaCl, KBr, and RbI is more accurately described as ionic, rather than covalent. The details of these two bonding models will be discussed in Chapter 8 .

Ionization Energies; Electron Affinities (Sections 7.4 and 7.5)

7.37  Al(g) → Al + (g) + e ; Al + (g) → Al 2+ (g) + e ; Al 2+ (g) → Al 3+ (g) + e

The process for the first ionization energy requires the least amount of energy. When an electron is removed from an atom or ion, electrostatic repulsions are reduced, Z eff increases, and the energy required to remove the next electron increases. This rationale is confirmed by the ionization energies listed in Table 7.2.

7.38

  1. Pb(g) → Pb + (g) + e ; Pb + (g) → Pb 2+ (g) + e
  2. Zr 3+ (g) → Zr 4+ (g) + e

7.39 Analyze/Plan . We are asked about the second ionization energy of three elements. This involves removing an electron from the 1+ ion of each element. Write the electron configurations of each ion and consider the attraction for the nucleus of the valence electron to be lost. Solve .

The electron configurations are: Li + , 1s 2 or [He]; Be + , [He]2s 1 ; K + , [Ne] 3s 2 3p 6 or [Ar]. Be has one more valence electron to lose whereas Li + has the stable noble gas configuration of He. It requires much more energy to remove a 1s core electron close to the nucleus of Li + than a 2s valence electron farther from the nucleus of Be + . K + also has a stable noble gas configuration. The electron to be lost is a core 2p electron. This electron is farther from the nucleus than the 1s electron in Li + and will require less energy to remove. Of these three elements, Li has the highest second ionization energy.


7.40

  1. False. Ionization energies are always positive quantities.
  2. False. F has a greater first ionization energy than O, because Z eff for F is greater than Z eff for O.
  3. True.
  4. False. Ionization energies are not cumulative. The third ionization energy is the energy needed to remove an electron from a gas phase ion with a 2+ charge.

7.41

  1. In general, the smaller the atom, the larger its first ionization energy.
  2. According to Figure 7.10, He has the largest and Cs has the smallest first ionization energy of the nonradioactive elements.

7.42

  1. Moving from F to I in group 7A, first ionization energies decrease and atomic radii increase. The greater the atomic radius, the smaller the electrostatic attraction of an outer electron for the nucleus and the smaller the ionization energy of the element.
  2. First ionization energies increase slightly going from K to Kr and atomic sizes decrease. As valence electrons are drawn closer to the nucleus (atom size decreases), it requires more energy to completely remove them from the atom (first ionization energy increases). Each trend has a discontinuity at Ga, owing to the increased shielding of the 4p electrons by the filled 3d subshell.

7.43 Plan .  Use periodic trends in first ionization energy. Solve .

  1. Cl
  2. Ca
  3. K
  4. Ge
  5. Sn

7.44  Greater distance of valence electrons from the nucleus predicts lower first ionization energy in all the pairs of elements below. Z eff decreases moving left along a row but increases slightly moving down column. These trends are not (solely) predictive of first ionization energy for the pairs of elements in this exercise.

  1. Ba. Recall that transition metals like Ti lose n s electrons first when forming ions. The 6s valence electrons in Ba are farther from the nucleus and have a smaller first ionization energy than the 4s valence electrons of Ti.
  2. Ag. Recall that transition elements lose n s electrons first when forming ions. The 5s valence electron of Ag is farther from the nucleus and has a lower first ionization energy than the 4s valence electron of Cu.
  3. Ge. The 4p valence electrons in Ge have a smaller first ionization energy than the 3p valence electrons in Cl. Going from Cl to Ge, the decrease in Z eff moving four places to the left may more than compensate for the small increase moving one place down. If so, the trends in Z eff and distance of valence electrons from the nucleus cooperate to produce the (significantly) lower first ionization energy for Ge.
  4. Pb. The 6p valence electrons in Pb are farther from the nucleus and have a smaller first ionization energy than the 5p valence electrons in Sb, despite the buildup in nuclear charge (Z) associated with filling the 4f subshell between Sb and Pb.

7.45 Plan .  Follow the logic of Sample Exercise 7.7. Solve .

  1. Co 2+ : [Ar]3d 7
  2. Sn 2+ : [Kr]5s 2 4d 10
  3. Zr 4+ : [Kr], noble-gas configuration
  4. Ag + : [Kr]4d 10
  5. S 2– : [Ne]3s 2 3p 6 , noble-gas configuration

7.46

  1. Ru 3+ : [Kr]3d 5
  2. As 3– : [Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas configuration
  3. Y 3+ : [Kr], noble-gas configuration
  4. Pd 2+ : [Kr]3d 8
  5. Pb 2+ : [Xe]6s 2 4f 14 5d 10
  6. Au 3+ : [Xe]4f 14 5d 8

7.47 Plan .  Focus on transition metals, which have d electrons in their outer shell. Use Figure 7.15 to find representative oxidations states for transition metals. Note that, by definition, metals lose electrons to form positive ions.

Solve . Elements in group 10 and beyond have at least eight d electrons. Of these, the group 10 metals all form 2+ ions with the electron configuration n d 8 . Of the elements in groups 11 and 12, only Au adopts a sufficiently high positive charge to form an ion with the configuration n d 8 . (Other possibilities not listed on Figure 7.15 exist.)

Ni 2+ : [Ar]3d 8 ;  Pd 2+ : [Kr]4d 8 ;  Pt 2+ : [Xe]4f 14 5d 8 ;  Au 3+ : [Xe]4f 14 5d 8

7.48 The 2+ ions of group 8 metals and the 3+ ions of group 9 metals have the electrons configuration nd 6 . (Other possibilities not listed on Figure 7.15 exist.)

Fe 2+ : [Ar]3d 6 ;  Ru 2+ : [Kr]4d 6 ;  Os 2+ : [Xe] 4f 14 5d 6

Co 3+ : [Ar]3d 6 ;  Rh 3+ : [Kr]4d 6 ;  Ir 3+ : [Xe] 4f 14 5d 6

7.49 Analyze/Plan . The second electron affinity of Cl is addition of an electron to Cl . Write the chemical equation and electron configurations for the relevant ions. Consider the attraction of the added electron for the Cl nucleus. Solve .

Second Electron affinity: Cl (g) + e Cl 2 ( g ) [ Ne ] 3 s 2 3 p 6 [Ne]3s 2 3 p 6 4 s 1

The Cl ion has the stable noble gas electron configuration of Ar. According to Figure 7.12 the first electron affinity of Ar is positive. Because the Cl ion has the electron configuration of Ar and a smaller nuclear charge, we predict that the second electron affinity of Cl will be positive as well. A positive value means that Cl 2– ion is unstable and will not form. It is probably not possible to directly measure the second electron affinity of Cl. [Sometimes unstable species have a finite lifetime before decomposing, so a very fast measurement may be possible.]

7.50  No. The process described by electron affinity can be written as: A + e A

If ΔE for this process is negative, it means that the energy of A is lower than the total energy of A plus the energy of a free electron. If electron affinity is negative, the entity that is lower in energy, or more stable, is the added electron. An electron in an atom or ion is stabilized by its attraction for the atomic nucleus and is lower in energy than a free electron.


7.51 Analyze/Plan . Write chemical equations for the electron affinity of K + and K. Consider the effective nuclear charge, Z – S, experienced by the added electron. Solve .

K + (g) + e K ( g ) ; [ Ne ] 3 s 2 3 p 6 or [Ar] [Ne]3s 2 3 p 6 4 s 1 or [Ar] 4 s 1 K(g) + e K ( g ) [ Ar ] 4 s 1 [Ar] 4 s 2

The electron affinity of K + is more negative. The full nuclear charge, Z, is the same for all K atoms and ions. In both cases, the added electron occupies a 4s subshell; the screening by the Ar core of electrons is the same. The single difference is the electron-electron repulsion experienced by the electron added to neutral K(g). This repulsion raises the energy of K relative to K and the added electron. According to Figure 7.12, the electron affinity of K is negative. The electron affinity of K + will also be negative and have a greater magnitude.

7.52  Ionization energy of F :  F (g) → F(g) + e

Electron affinity of F:   F(g) + e → F (g)

The two processes are the reverse of each other. The energies are equal in magnitude but opposite in sign. I 1 (F ) = –E (F)

7.53 Analyze/Plan . Consider the definitions of ionization energy and electron affinity, along with the appropriate electron configurations. Solve .

  1. Ionization energy of Ne: Ne(g) Ne + (g) + e
    [He]2s 2 2p 6 [He]2s 2 2p 5
    Electron affinity of F: F(g) + e F (g)
    [He]2s 2 2p 5 [He]2s 2 2p 6
  2. The I 1 of Ne is positive, whereas E 1 of F is negative. All ionization energies are positive.
  3. One process is apparently the reverse of the other, with one important difference. The Z (and Z eff ) for Ne is greater than Z (and Z eff ) for F . We expect the magnitude of I 1 (Ne) to be somewhat greater than the magnitude of E 1 (F). [Repulsion effects approximately cancel; repulsion decrease upon I 1 causes smaller positive value; repulsion increase upon E 1 causes smaller negative value.]

7.54 Ca + ( g) + e Ca(g) [ Ar ] 4 s 1 [Ar] 4s 2

Statements (i) and (ii) are true.

Properties of Metals and Nonmetals (Section 7.6)

7.55

  1. Decrease
  2. Increase
  3. The smaller the first ionization energy of an element, the greater the metallic character of that element. The trends in (a) and (b) are the opposite of the trends in ionization energy.

7.56  Element Y has the greater metallic character. Metallic character increases as ionization energy decreases.

7.57 Analyze/Plan . Use Figure 7.13, “Metals, metalloids, and nonmetals,” and Figure 7.15, “Representative oxidation states of the elements,” to inform our discussion.

Solve . Agree. An element that commonly forms a cation is a metal. The only exception to this statement shown on Figure 7.15 is antimony, Sb, a metalloid that commonly forms cations. Although Sb is a metalloid, it is far down (in the fifth row) on the chart and likely to have significant metallic character.

7.58  Disagree. According to Figure 7.15, both Sb and Te are metalloids and commonly form ions. Sb forms cations and Te forms anions.

7.59 Analyze/Plan .  Ionic compounds are formed by combining a metal and a nonmetal; molecular compounds are formed by two or more nonmetals. Solve .

Ionic: SnO 2 , Al 2 O 3 , Li 2 O, Fe 2 O 3 ; molecular: CO 2 , H 2 O

7.60  Follow the logic in Sample Exercise 7.8. Scandium is a metal, so we expect Sc 2 O 3 to be ionic. Metal oxides are usually basic and react with acid to form a salt and water. We choose HNO 3 (aq) as the acid for our equation.

Sc 2 O 3 (s) + 6 HNO 3 (aq) → 2 Sc(NO 3 ) 3 (aq) + 3 H 2 O(l)

The net ionic equation is:

Sc 2 O 3 (s) + 6 H + (aq) → 2 Sc 3+ (aq) + 3 H 2 O(l)

7.61 Analyze/Plan . Decide whether MnO is an acidic or basic oxide.

Solve . MnO will react more readily with HCl(aq). Manganese is a metal. Oxides of metals usually act like bases, which react readily with the strong acid HCl(aq).

7.62  The more nonmetallic the central atom, the more acidic the oxide. In order of increasing acidity: CaO < Al 2 O 3 < SiO 2 < CO 2 < P 2 O 5 < SO 3

7.63 Analyze/Plan .  Cl 2 O 7 is a molecular compound formed by two nonmetallic elements. More specifically, it is a nonmetallic oxide and acidic. Solve .

  1. Dichlorine heptoxide
  2. Elemental chlorine and oxygen are diatomic gases.

    2 Cl 2 (g) + 7 O 2 (g) → 2 Cl 2 O 7 (l)

  3. Cl 2 O 7 is an acidic oxide, so it will be more reactive to base, OH .

    Cl 2 O 7 (l) + 2 OH (aq) → 2 ClO 4 (aq) + H 2 O(l)

  4. The oxidation state of Cl in Cl 2 O 7 is +7. In this oxidation state, the electron configuration of Cl is [He]2s 2 2p 6 or [Ne].

7.64

  1. XCl 4 (l) + 2 H 2 O(l) → XO 2 (s) + 4 HCl(g)

    The second product is HCl(g).


  1. If X were a metal, both the oxide and the chloride would be high melting solids. If X were a nonmetal, XO 2 would be a nonmetallic, molecular oxide and probably gaseous, like CO 2 , NO 2 , and SO 2 . Neither of these statements describes the properties of XO 2 and XCl 4 , so X is probably a metalloid.
  1. Use the Handbook of Chemistry to find formulas and melting points of oxides, and formulas and boiling points of chlorides of selected metalloids.
    metalloid formula of oxide m.p. of oxide formula of chloride b.p. of chloride
    boron B 2 O 3 460 °C BCl 3 12 °C
    silicon SiO 2 ~1700 °C SiCl 4 58 °C
    germanium

    GeO

    GeO 2

    710 °C

    ~1100 °C

    GeCl 2

    GeCl 4

    decomposes

    84 °C

    arsenic

    As 2 O 3

    As 2 O 5

    315 °C

    315 °C

    AsCl 3 132 °C

    Boron, arsenic, and, by analogy, antimony, do not fit the description of X because the formulas of their oxides and chlorides are wrong. Silicon and germanium, in the same family, have oxides and chlorides with appropriate formulas. Both SiO 2 and GeO 2 melt above 1000 °C, but the boiling point of SiCl 4 is much closer to that of XCl 4 . Element X is silicon.

7.65

  1. BaO(s) + H 2 O(l) → Ba(OH) 2 (aq)
  2. FeO(s) + 2 HClO 4 (aq) → Fe(ClO 4 ) 2 (aq) + H 2 O(l)
  3. SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq)
  4. CO 2 (g) + 2 NaOH(aq) → Na 2 CO 3 (aq) + H 2 O(l)

7.66

  1. K 2 O(s) + H 2 O(l) → 2 KOH(aq)
  2. P 2 O 3 (l) + 3 H 2 O(l) → 2 H 3 PO 3 (aq)
  3. Cr 2 O 3 (s) + 6 HCl(aq) → 2 CrCl 3 (aq) + 3 H 2 O(l)
  4. SeO 2 (s) + 2 KOH(aq) → K 2 SeO 3 (aq) + H 2 O(l)

Group Trends in Metals and Nonmetals (Sections 7.7 and 7.8)

7.67

  1. Ca and Mg are both metals; they tend to lose electrons and form cations when they react. Ca is more reactive because it has a lower ionization energy than Mg. The Ca valence electrons in the 4s orbital are less tightly held because they are farther from the nucleus than the 3s valence electrons of Mg.
  2. K and Ca are both metals; they tend to lose electrons and form cations when they react. K is more reactive because it has the lower first ionization energy. The 4s valence electron in K is less tightly held because it has the same n value as the valence electrons of Ca and experiences smaller Z and Z eff .

7.68  Rb: [Kr]5s 1 , r = 2.11 Å  Ag: [Kr]5s 1 4d 10 , r = 1.53 Å

The electron configurations both have a [Kr] core and a single 5s electron; Ag has a completed 4d subshell as well. The smaller radius of Ag indicates that the 5s electron in Ag experiences a much greater effective nuclear charge than the 5s electron in Rb. Ag has a much larger Z (47 vs. 37), and although the 4d electrons in Ag shield the 5s electron somewhat, the increased shielding does not compensate for the large increase in Z. Ag is much less reactive (less likely to lose an electron) because its 5s electron experiences a much larger effective nuclear charge and is more difficult to remove.

7.69

  1. 2 K(s)  + Cl 2 (g) → 2 KCl(s)
  2. SrO(s) + H 2 O(l) → Sr(OH) 2 (aq)
  3. 4 Li(s)  + O 2 (g) → 2 Li 2 O(s)
  4. 2 Na(s) + S(l) → Na 2 S(s)

7.70

  1. 2 Cs(s) + 2 H 2 O(l) → 2 CsOH(aq) + H 2 (g)
  2. Sr(s) + 2 H 2 O(l) → Sr(OH) 2 (aq) + H 2 (g)
  3. 2 Na(s) + O 2 (g) → Na 2 O 2 (s) (See Equation 7.20.)
  4. Ca(s) + I 2 (s) → CaI 2 (s)

7.71

  1. The reactions of the alkali metals with hydrogen and with a halogen are redox reactions. In both classes of reaction, the alkali metal loses electrons and is oxidized. Both hydrogen and the halogen gain electrons and are reduced. Hydrogen or the halogen act as oxidizing agents in these reactions.

    Ca(s) + F 2 (g) → CaF 2 (s)    Ca(s) + H 2 (g) → CaH 2 (s)

  2. The oxidation number of Ca in both products is +2. The electron configuration is that of Ar, [Ne]3s 2 3p 6 .

7.72

  1. 2 K(s) + H 2 (g) → 2 KH(s)
  2. K(g)      → K + (g)       419 kJ   (I 1 of K)

    H(g) + e → H (g)      –73 kJ   (E 1 of H)

    K(g) + H(g) → K + (g) + H (g) 346 kJ

    H(g)      → H + (g)      1312 kJ   (I 1 of H)

    K(g) + e → K (g)      –48 kJ   (E 1 of K)

    K(g) + H(g) → K (g) + H + (g) 1264 kJ

  3. Both reactions are endothermic; the first reaction is less unfavorable and therefore more favorable than the second.
  4. The more energetically favorable reaction in part (c) produces hydride ions (H–) and potassium ions (K + ), so it is reasonable to describe potassium hydride as containing hydride ions.

7.73

Br Cl
(a) [Ar]4s 2 4p 5 [Ne]3s 2 3p 5
(b) –1 –1
(c) 1140 kJ/mol 1251 kJ/mol
(d) reacts slowly to form HBr+HOBr reacts slowly to form HCl+HOCl
(e) –325 kJ/mol –349 kJ/mol
(f) 1.20 Å 1.02 Å

The n = 4 valence electrons in Br are farther from the nucleus and less tightly held than the n = 3 valence electrons in Cl. Therefore, the ionization energy of Cl is greater, the electron affinity is more negative and the atomic radius is smaller.

7.74 Plan .  Predict the physical and chemical properties of At based on the trends in properties in the halogen (7A) family. Solve .

  1. F, at the top of the column, is a diatomic gas; I, immediately above At, is a diatomic solid; the melting points of the halogens increase going down the column. At is likely to be a diatomic solid at room temperature.
  2. Like the other halogens, we expect it to be a nonmetal. According to Figure 7.13, there are no metalloids in row 6 of the periodic table, and At is a nonmetal. (Looking forward to Chapter 8 , the most likely way for At to satisfy the octet rule is for it to gain an electron to form At , which makes it a nonmetal.)
  3. All halogens form ionic compounds with Na; they have the generic formula NaX. The compound formed by At will have the formula NaAt.

7.75

  1. The term “inert” was dropped because it no longer described all the group 8A elements.
  2. In the 1960s, scientists discovered that Xe would react with substances such as F 2 and PtF 6 that have a strong tendency to remove electrons. Thus, Xe could not be categorized as an “inert” gas.
  3. The group is now called the noble gases.

7.76

  1. Xe has a lower ionization energy than Ne. The valence electrons in Xe are much farther from the nucleus than those of Ne ( n = 5 vs n = 2) and much less tightly held by the nucleus; they are more “willing” to be shared than those in Ne. Also, Xe has empty 5d orbitals that can help to accommodate the bonding pairs of electrons, whereas Ne has all its valence orbitals filled.
  2. In the CRC Handbook of Chemistry and Physics , 79th edition, Xe–F bond distances in gas phase molecules are listed as: XeF 2 , 1.977 Å; XeF 4 , 1.94 Å; XeF 6 , 1.89 Å. From Figure 7.7, the sum of atomic radii for Xe and F is (1.40 Å + 0.57 Å) = 1.97 Å. This number represents an “average” or “typical” distance and agrees well with the bond distance in XeF 2 . Bond lengths in specific compounds are not exactly equal to the sum of covalent radii. Physical state, electronic, and steric factors affect bond lengths in specific compounds.

7.77

  1. 2 O 3 (g) → 3 O 2 (g)
  2. Xe(g) + F 2 (g) → XeF 2 (g)

    Xe(g) + 2 F 2 (g) → XeF 4 (s)

    Xe(g) + 3 F 2 (g) → XeF 6 (s)

  3. S(s) + H 2 (g) → H 2 S(g)
  4. 2 F 2 (g) + 2 H 2 O(l) → 4 HF(aq) + O 2 (g)

7.78

  1. Cl 2 (g) + H 2 O(l) → HCl(aq) + HOCl(aq)
  2. Ba(s) + H 2 (g) → BaH 2 (s)
  3. 2 Li(s) + S(s) → Li 2 S(s)
  4. Mg(s) + F 2 (g) → MgF 2 (s)

Additional Exercises

7.79   Up to Z = 82, there are three instances where atomic weights are reversed relative to atomic numbers: Ar and K; Co and Ni; Te and I.

7.80

  1. 2s
  2. Slater’s rules provide a method for calculating the shielding, S, and Z eff experienced by a particular electron in an atom. Slater assigns a shielding value of 0.35 to electrons with the same n -value, assuming that s and p electrons shield each other to the same extent. However, because s electrons have a finite probability of being very close to the nucleus (Figure 7.4), they shield p electrons more than p electrons shield them. To account for this difference, assign a slightly larger shielding value to s electrons and a slightly smaller shielding value to the p electrons. This will produce a slightly greater S and smaller Z eff for p electrons than for s electrons with the same n -value.

7.81

  1. P: [Ne]3s 2 3p 3 . Z eff = Z − S = 15 – 10 = 5.
  2. Four other n = 3 electrons, eight n = 2 electrons, two n = 1 electron. S = 0.35(4) + 0.85(8) + 1(2) = 10.2. Z eff = Z − S = 15 – 10.2 = 4.8.
  3. The 3s electrons penetrate the [Ne] core electrons (by analogy to Figure 7.4) and experience less shielding than the 3p electrons. That is, S is greater for 3p electrons, owing to the penetration of the 3s electrons, so Z – S (3p) is less than

    Z – S (3s).

  4. The 3p electrons are the outermost electrons; they experience a smaller Z eff than 3s electrons and thus a smaller attraction for the nucleus, given equal n -values. The first electron lost is a 3p electron. Each 3p orbital holds one electron, so there is no preference as to which 3p electron will be lost.

7.82  Atomic size (bonding atomic radius) is strongly correlated to Z eff , which is determined by Z and S. Moving across the representative elements, electrons added to ns or np valence orbitals do not effectively screen each other. The increase in Z is not accompanied by a similar increase in S; Z eff increases and atomic size decreases. Moving across the transition elements, electrons are added to ( n– 1)d orbitals and become part of the core electrons, which do significantly screen the ns valence electrons. The increase in Z is accompanied by a larger increase in S for the ns valence electrons; Z eff increases more slowly and atomic size decreases more slowly.

7.83

  1. The estimated distances in the table below are the sum of the radii of the group 5A elements and H from Figure 7.7.
    bonded atoms estimated distance measured distance
    P–H 1.38 1.419
    As–H 1.50 1.519
    Sb–H 1.70 1.707

    In general, the estimated distances are very slightly shorter than the measured distances. (Recall that the radii in Figure 7.7 come from measuring many different molecules for each element, not just the bonds listed in this exercise.)

  2. The principal quantum number of the outer electrons and thus the average distance of these electrons from the nucleus increases from P ( n = 3) to As ( n = 4) to Sb ( n = 5). This causes the systematic increase in M–H distance.

7.84  She is correct. Xenon bonds with certain elements to form compounds, so its bonding atomic radius is an average of experimentally determined values. To date, no compound containing Ne has been observed, so its “atomic radius” is an estimate. Measured values are always more realistic than estimates.

7.85

  1. Assume that the bonding atomic radius of the element will be one-half of the bond distance in the element.

    r As = (2.48 Å)/2 = 1.24 Å;    r Cl = (1.99 Å)/2 = 0.995 Å

    The As–Cl distance is the sum of these radii: 1.24 Å + 0.995 Å = 2.235 = 2.24 Å.

  2. From Figure 7.7, the predicted As–Cl distance = 1.19 Å + 1.02 Å = 2.21 Å

7.86  The estimated A–B distance is (r A + r B ) = (A–A)/2 + (B–B)/2. Because the AB 2 molecule is linear, the distance between the two terminal B atoms is twice the A–B distance, 2[(A–A)/2 + (B–B)/2] = (A–A) + (B–B). This is just the sum of the bond lengths of the two diatomic molecules. The separation between the two B nuclei in AB 2 is 2.36 Å + 1.94 Å = 4.30 Å.

7.87

  1. The most common oxidation state of the chalcogens is –2, whereas that of the halogens is –1.
  2. The family listed has the larger value of the stated property.

    atomic radii, chalcogens

    ionic radii of the most common oxidation state, chalcogens

    first ionization energy, halogens

    second ionization energy, halogens


7.88

Y:  [Kr]5s 2 4d 1 , Z = 39 Zr: [Kr] 5s 2 4d 2 , Z = 40
La: [Xe]6s 2 5d 1 , Z = 57 Hf: [Xe] 6s 2 4f 14 5d 2 , Z = 72

The completed 4f subshell in Hf leads to a much larger change in Z going from Zr to Hf (72 – 40 = 32) than in going from Y to La (57 – 39 = 18). The 4f electrons in Hf do not completely shield the valence electrons, so there is also a larger increase in Z eff . This significant increase in Z eff going from Zr to Hf causes the two elements to have the same radii, even though the valence electrons of Hf have a larger n value than those of Zr. (This phenomenon is called the “lantanide contraction.”)

7.89

  1. Co 4+ is smaller.
  2. Co 4+ , 0.67 Å < Co 3+ , 0.75 Å < Li + , 0.90 Å

    Values from WebElements©, CN 6, high spin (for comparing equivalent ion environments)

  3. As Li + ions are inserted, smaller Co 4+ ions are reduced to larger Co 3+ ions and the lithium colbalt oxide will expand.
  4. “Sodium colbalt oxide” will probably not work as an electrode material, because Na + ions are much larger than Li + ions, which are larger than Co 4+ and Co 3+ ions. Na + ions would be too large to insert into the electrode without disrupting the structure of the material.
  5. An alternative metal for a sodium version of the electrode would have redox-active ions with larger ionic radii than the Co 4+ and Co 3+ ions. Moving left along the fourth row of the periodic table, Fe 3+ /Fe 2+ and Mn 3+ /Mn 2+ ion couples are possibilities. Both have radii larger than Co 4+ /Co 3+ ions. Mn 3+ is more redox-active than Fe 3+ and may be a more effective electrode material.

7.90

  1. 2 Sr(s) + O 2 (g) → 2 SrO(s)
  2. Assume that the corners of the cube are at the centers of the outermost O 2– ions, and that the edges each pass through the center of one Sr 2+ ion. The length of an edge is then r(O 2– ) + 2r(Sr 2+ ) + r(O 2– ) = 2r(O 2– ) + 2r(Sr 2+ ) = 2(1.32 Å) + 2(1.26 Å) = 5.16 Å.
  3. Density is the ratio of mass to volume.

    d = mass SrO in cube vol cube = # SrO units × mass of SrO vol cube

    Calculate the mass of 1 SrO unit in grams and the volume of the cube in cm 3 ; solve for number of SrO units.

    103.62 g SrO mol × 1 mol SrO 6.022 × 10 23 SrO units = 1.7207 × 10 22 = 1.721 × 10 22 g/SrO unit

    V = ( 5.16 ) 3 Å 3 × ( 1 × 10 8 ) 3 cm 3 Å 3 = 1.3739 × 10 22 = 1.37 × 10 22 cm 3

    d = number of SrO units × 1 .7207 × 10 22 g/SrO unit 1.3739 × 10 22 cm 3 = 5.10 g/cm 3


    number of SrO units = 5.10 g/cm 3 × 1.3739 × 10 22 cm 3 1.7207 × 10 22 g/SrO unit = 4.07 units

    Because the number of formula units must be an integer, there are four SrO formula units in the cube. Using average values for ionic radii to estimate the edge length probably leads to the small discrepancy.

7.91  C: 1s 2 2s 2 2p 2 . I 1 through I 4 represent loss of the 2p and 2s electrons in the outer shell of the atom. The values of I 1– I 4 increase as expected. The nuclear charge is constant, but removing each electron reduces repulsive interactions between the remaining electrons, so effective nuclear charge increases and ionization energy increases. I 5 and I 6 represent loss of the 1s core electrons. These 1s electrons are much closer to the nucleus and experience the full nuclear charge (they are not shielded), so the values of I 5 and I 6 are significantly greater than I 1– I 4 . I 6 is larger than I 5 because all repulsive interactions have been eliminated.

7.92  Only statement (ii) is true.

We expect electron affinities for Group 4A to be more negative than Group 3A, based on increasing effective nuclear charge moving from left to right across a row. Electron-electron repulsion causes the electron affinities of Groups 5A and 6A to be less negative than expected from effective nuclear charge trends.

7.93

A(g) → A + (g) + e ionization energy of A
A(g) + e → A–(g) electron affinity of A
A(g) + A(g) → A + (g) + A–(g) ionization energy of A + electron affinity of A

The energy change for the reaction is the ionization energy of A plus the electron affinity of A.

This process is endothermic for both nonmetals and metals. Considering data for Cl and Na from Figures 7.10 and 7.12, the endothermic ionization energy term dominates the exothermic electron affinity term, even for Cl, which has the most exothermic electron affinity listed.

7.94

  1. O: [He]2s 2 2p 4

    image

    O 2– : [He]2s 2 2p 6 = [Ne]

    image

  2. O 3– : [Ne]3s 1 The third electron would be added to the 3s orbital, which is farther from the nucleus and more strongly shielded by the [Ne] core. The overall attraction of this 3s electron for the O nucleus is not large enough for O 3– to be a stable particle.

7.95

  1. P: [Ne] 3s 2 3p 3 ; S: [Ne] 3s 2 3p 4 . In P, each 3p orbital contains a single electron, whereas in S one 3p orbital contains a pair of electrons. Removing an electron from S eliminates the need for electron pairing and reduces electrostatic repulsion, so the overall energy required to remove the electron is smaller than in P, even though Z is greater.

  1. C: [He] 2s 2 2p 2 ; N: [He] 2s 2 2p 3 ; O: [He] 2s 2 2p 4 . An electron added to an N atom must be paired in a relatively small 2p orbital, so the additional electron-electron repulsion more than compensates for the increase in Z and the electron affinity is smaller (less exothermic) than that of C. In an O atom, one 2p orbital already contains a pair of electrons, so the additional repulsion from an extra electron is offset by the increase in Z and the electron affinity is greater (more exothermic). Note from Figure 7.12 that the electron affinity of O is only slightly more exothermic than that of C, although the value of Z has increased by 2.
  1. O + : [He] 2s 2 2p 3 ; O 2+ : [He] 2s 2 2p 2 ; F: [He] 2s 2 2p 5 ; F + : [He] 2s 2 2p 4 . Both ‘core-only’ [Z eff (F) = 7; Z eff (O + ) = 6] and Slater [Z eff (F) = 5.2; Z eff (O + ) = 4.9] predict that F has a greater Z eff than O + . Variation in Z eff does not offer a satisfactory explanation. The decrease in electron-electron repulsion going from F to F + energetically favors ionization and causes it to be less endothermic than the second ionization of O, where there is no significant decrease in repulsion.
  1. Mn 2+ : [Ar]3d 5 ; Mn 3+ ; [Ar] 3d 4 ; Cr 2+ : [Ar] 3d 4 ; Cr 3+ : [Ar] 3d 3 ; Fe 2+ : [Ar] 3d 6 ; Fe 3+ : [Ar] 3d 5 . The third ionization energy of Mn is expected to be larger than that of Cr because of the larger Z value of Mn. The third ionization energy of Fe is less than that of Mn because going from 3d 6 to 3d 5 reduces electron repulsions, making the process less endothermic than predicted by nuclear charge arguments.

7.96

  1. Cl , K +
  2. Mn 2+ , Fe 3+
  3. Sn 2+ , Sb 3+

7.97

  1. (ii)
  2. (v)
  3. (i)

7.98

  1. The group 2B metals have complete ( n– 1)d subshells. An additional electron would occupy an n p subshell and be substantially shielded by both n s and ( n– 1)d electrons. Overall this is not a lower energy state than the neutral atom and a free electron.
  2. Valence electrons in Group 1B elements experience a relatively large effective nuclear charge because of the buildup in Z with the filling of the ( n– 1)d subshell (and for Au, the 4f subshell.) Thus, the electron affinities are large and negative. Group 1B elements are exceptions to the usual electron filling order and have the generic electron configuration n s 1 ( n– 1)d 10 . The additional electron would complete the n s subshell and experience repulsion with the other n s electron. Going down the group, size of the n s subshell increases and repulsion effects decrease. That is, effective nuclear charge is greater going down the group because it is less diminished by repulsion, and electron affinities become more negative.

7.99

  1. For both H and the alkali metals, the added electron will complete an n s subshell (1s for H and n s for the alkali metals) so shielding and repulsion effects will be similar. For the halogens, the electron is added to an n p subshell, so the energy change is likely to be quite different.

  1. True. Only He has a smaller estimated “bonding” atomic radius, and no known compounds of He exist. The electron configuration of H is 1s 1 . The single 1s electron experiences no repulsion from other electrons and feels the full unshielded nuclear charge. It is held very close to the nucleus. The outer electrons of all other elements that form compounds are shielded by a spherical inner core of electrons and are less strongly attracted to the nucleus, resulting in larger bonding atomic radii.
  1. Ionization is the process of removing an electron from an atom. For the alkali metals, the n s electron being removed is effectively shielded by the core electrons, so ionization energies are low. For the halogens, a significant increase in nuclear charge occurs as the n p orbitals fill, and this is not offset by an increase in shielding. The relatively large effective nuclear charge experienced by n p electrons of the halogens is similar to the unshielded nuclear charge experienced by the H 1s electron. Both H and the halogens have large ionization energies.
  1. Ionization energy of hydride:  H (g) → H(g) + e
  1. Electron affinity of hydrogen:  H(g) + e → H (g)

    The two processes in parts (d) and (e) are the exact reverse of one another. The value for the ionization energy of hydride is equal in magnitude but opposite in sign to the electron affinity of hydrogen.

7.100  Because Xe reacts with F 2 , and O 2 has approximately the same ionization energy as Xe, O 2 will probably react with F 2 . Possible products would be O 2 F 2 , analogous to XeF 2 , or OF 2 .

O 2 (g) + F 2 (g) → O 2 F 2 (g)

O 2 (g) + 2 F 2 (g) → 2 OF 2 (g)

7.101  The first ionization energies are: Ag, 731 kJ/mol; Mn, 717 kJ/mol. According to Figure 7.13, we define metallic character as showing the opposite trend as ionization energy. That is, the smaller the ionization energy, the greater the metallic character. Because Mn has the smaller ionization energy, it should have the greater metallic character. (It is difficult to predict the relative metallic character of these two elements from trends. Ag is one row lower but four columns further right than Mn; these are opposing trend directions.)

7.102  The most likely product is (i). The ionization energy of K is less than that of H, so K will lose an electron and H will gain one; the (mostly) ionic solid KH is the product. This is one instance when H acts like a nonmetal, even though it appears with the metals on the periodic chart.

7.103  Statement (ii) is the best explanation.

Statements (ii) and (iii) are both true, assuming “smaller” in statement (iii) means more negative. In this reaction, Na or Cs loses an electron and acts like a metal. Statement (ii) about ionization energy is more directly applicable. Electron affinity is about an atom gaining an electron.

7.104

  1. All alkali metals except Li form metal peroxides when they react with oxygen; the formation of a peroxide (or a superoxide) eliminates Li. The lilac-purple flame indicates that the metal is potassium (see Figure 7.22).

  1. K 2 O 2 ( s) + potassium peroxide 2 H 2 O(l) H 2 O 2 (aq) + 2 KOH (aq) hydrogen peroxide

    2 KO 2 ( s) + potassium superoxide 2 H 2 O(l) H 2 O 2 (aq) + 2 KOH (aq) + O 2 ( g ) hydrogen peroxide

    Both potassium peroxide and potassium superoxide react with water to form hydrogen peroxide. The white solid could be either potassium salt.

7.105

  1. The pros are that Zn and Cd are in the same family, have the same electron configuration and thus similar chemical properties. The same can be said for Zn 2+ and Cd 2+ ions. Because of their chemical similarity, we expect Cd 2+ to easily substitute for Zn 2+ in flexible molecules. The main difference is that Zn 2+ , with an ionic radius of 0.88 Å, is much smaller than Cd 2+ , with an ionic radius of 1.09 Å. Although Zn 2+ is beneficial in living systems, Cd 2+ is toxic. This difference in biological function could be related to the size difference and is a definite con.
  2. Cu + is isoelectronic with Zn 2+ . That is, the two ions have the same number of electrons and the same electron configurations. The ionic radius of Cu + is 0.91 Å, very similar to that of Zn 2+ . We expect Cu + to be a reasonable substitute for Zn 2+ in terms of chemical properties and size. Electrostatic interactions may vary, because of the difference in charges of the two ions. (All ionic radii are taken from WebElements©.)

7.106

  1. Plan .  Use qualitative physical (bulk) properties to narrow the range of choices, then match melting point and density to identify the specific element. Solve .

    Hardness varies widely in metals and nonmetals, so this information is not too useful. The relatively high density, appearance, and ductility indicate that the element is probably less metallic than copper. Focus on the block of nine main group elements centered around Sn. Pb is not a possibility because it was used as a comparison standard. The melting point of the five elements closest to Pb are:

    Tl, 303.5 °C; In, 156.1 °C; Sn, 232 °C; Sb, 630.5 °C; Bi, 271.3 °C

    The best match is In. To confirm this identification, the density of In is 7.3 g/cm 3 , also a good match to properties of the unknown element.

  2. To write the correct balanced equation, determine the formula of the oxide product from the mass data, assuming the unknown is In.

    5.08 g oxide – 4.20 g In = 0.88 g O

    4.20 g In/114.82 g/mol = 0.0366 mol In; 0.0366/0.0366 = 1

    0.88 g O/16.00 g/mol = 0.0550 mol O; 0.0550/0.0366 = 1.5

    Multiplying by two produces an integer ratio of 2 In: 3 O and a formula of In 2 O 3 . The balanced equation is: 4 In(s) + 3 O 2 (g) → 2 In 2 O 3 (s)

  3. According to Figure 7.1, the element In was discovered between 1843 and 1886. The investigator who first recorded this data in 1822 could have been the first to discover In.

7.107 Plan. According to the periodic table on the inside cover of your text, element 116 is in group 6A, so element 117 will be in group 7A, the halogens. Write the electron configuration and use information from Figures 7.7, 7.10, 7.12, 7.15, and Table 7.7 along with periodic trends to estimate values for properties. Remember that element 117 is two rows below iodine, and that the increase in Z and Z eff that accompanies filling of the f orbitals will decrease the size of the changes in ionization energy, electron affinity and atomic size. Solve.

Electron configuration:    [Rn]7s 2 5f 14 6d 10 7p 5

First ionization energy:    805 kJ/mol

Electron affinity:       –235 kJ/mol

Atomic size:         1.65 Å

Common oxidation state:   –1

7.108

  1. Si and Ge are in group 4A and have 4 valence electrons. GaAs and GaP have their first element in group 3A with 3 valence electrons and their second element in group 5A with 5 valence electrons. Cd in CdS and CdSe is in group 2B and has 2 valence electrons, whereas S and Se are in group 6A with 6 valence electrons. In each case, the two elements in the compound semiconductor have an average of 4 valence electrons.
  2. The roman numerals represent the number of valence electrons in the component elements of the compound semiconductor. CdS and CdSe are II-VI materials, whereas GaAs and GaP are III-V materials.
  3. Replace Ga with In: InP, InAs, InSb; replace Se with Te: CdTe. It is problematic to replace Cd with Hg, because Hg is toxic. ZnS is ionic and an insulator, so Zn may not be a good substitute for Cd.

Integrative Exercises

7.109

  1. ν = c/λ; 1 Hz = 1 s −1

    Ne : ν = 2.998 × 10 8 m/s 14.610 Å × 1 Å 1 × 10 10 m = 2.052 × 10 17 s 1 = 2.052 × 10 17 Hz

    Ca: ν = 2.998 × 10 8 m/s 3.358 × 10 10 m = 8.928 × 10 17 Hz

    Zn: ν = 2.998 × 10 8 m/s 1.435 × 10 10 m = 20.89 × 10 17 Hz

    Zr: ν = 2.998 × 10 8 m/s 0.786 × 10 10 m = 38.14 × 10 17 = 38.1 × 10 17 Hz

    Sn: ν = 2.998 × 10 8 m/s 0.491 × 10 10 m = 61.06 × 10 17 = 61.1 × 10 17 Hz


  1. Element Z ν ν 1/2
    Ne 10 2.052 × 10 17 4.530 × 10 8
    Ca 20 8.928 × 10 17 9.449 × 10 8
    Zn 30 20.89 × 10 17 14.45 × 10 8
    Zr 40 38.14 × 10 17 19.5 × 10 8
    Sn 50 61.06 × 10 17 24.7 × 10 8
    image
  2. The plot in part (b) indicates that there is a linear relationship between atomic number and the square root of the frequency of the X-rays emitted by an element. Thus, elements with each integer atomic number should exist. This relationship allowed Moseley to predict the existence of elements that filled “holes” or gaps in the periodic table.
  3. For Fe, Z = 26. From the graph, ν 1/2 = 12.5 × 10 8 , ν = 1.56 × 10 18 Hz.

    λ = c/ν = 2.998 × 10 8 m/s 1.56 × 10 18 s 1 × 1 Å 1 × 10 10 m = 1.92 Å

  4. λ = 0.980 Å = 0.980 × 10 −10 m

    ν = c/λ = 2.998 × 10 8 m/s 0.980 × 10 10 m = 30.6 × 10 17 Hz; ν 1 / 2 = 17.5 × 10 8

    From the graph, ν 1/2 = 17.5 × 10 8 , Z = 36. The element is krypton, Kr.

7.110

  1. Li: [He]2s 1 . Assume that the [He] core is 100% effective at shielding the 2s valence electron Z eff = Z – S ≈ 3 – 2 = 1+.
  2. The first ionization energy represents loss of the 2s electron.

    ΔE = energy of free electron (n = ∞) – energy of electron in ground state ( n = 2)

    ΔE = I 1 = [–2.18 × 10 −18 J (Z 2 /∞ 2 )] – [–2/18 × 10 −18 J (Z 2 /2 2 )]

    ΔE = I 1 = 0 + 2.18 × 10 −18 J (Z 2 /2 2 )

    For Li, which is not a one-electron particle, let Z = Z eff .

    ΔE ≈ 2.18 × 10 −18 J (+1 2 /4) ≈ 5.45 × 10 −19 J/atom


  1. Change the result from part (b) to kJ/mol so it can be compared to the value in Table 7.4.

    5 .45 × 10 19 J atom × 6.022 × 10 23 atom mol × 1 kJ 1000 J = 328 kJ/mol

    The value in Table 7.4 is 520 kJ/mol. This means that our estimate for Z eff was a lower limit, that the [He] core electrons do not perfectly shield the 2s electron from the nuclear charge.

  1. From Table 7.4, I 1 = 520 kJ/mol.

    520 kJ mol × 1000 J kJ × 1 mol 6 .022 × 10 23 atoms = 8.6350 × 10 19 J/atom

    Use the relationship for I 1 and Z eff developed in part (b).

    Z eff 2 = 4 ( 8.6350 × 10 19 J) 2.18 × 10 18 J = 1.5844 = 1.58 ; Z eff = 1.26

    This value, Z eff = 1.26, based on the experimental ionization energy, is greater than our estimate from part (a), which is consistent with the explanation in part (c).

7.111

  1. E = hc/λ; 1 nm = 1 × 10 −9 m; 58.4 nm = 58.4 × 10 −9 m;

    1 eV = 96.485 kJ/mol, 1 eV - mol = 96.485 kJ

    E = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 58.4 × 10 9 m = 3.4015 × 10 18 = 3.40 × 10 18 J/photon

  2. Hg(g) → Hg + (g) + 1e
  3. I 1 = E 58.4 – E K = 3.4015 × 10 −18 J – 1.72 × 10 −18 J = 1.6815 × 10 −18 = 1.68 × 10 −18 J/atom

    1.6815 × 10 18 J atom × 1 kJ 1000 J × 6.022 × 10 23 atoms mol = 1012.6 = 1.01 × 10 3 kJ/mol

  4. From Figure 7.10, iodine (I) appears to have the ionization energy closest to that of Hg, approximately 1000 kJ/mol.

7.112

  1. The X-ray source had an energy of 1253.6 eV. Change eV to J/photon and use the relationship λ = hc/E to find wavelength.

    1253.6 eV × 1.602 × 10 19 J eV = 2.0083 × 10 16 = 2.008 × 10 16 J/photon

    λ = 6.6261 × 10 34 J-s × 2 .9979 × 10 8 m/s 2. 0083 × 10 16 J = 9.8911 × 10 10 m = 0.98911 nm = 9.891 Å

  2. Express energies of Hg 4f and O 1s electrons in terms of kJ/mol for comparison with data from Figure 7.10 of the text.

    Hg 4f: 105 eV × 1.602 × 10 19 J eV × 6.022 × 10 23 atoms mol = 10 , 130 = 1.01 × 10 4 kJ/mol

    O 1s: 531 eV × 1.602 × 10 19 J eV × 6.022 × 10 23 atoms mol = 51 , 227 = 5.12 × 10 4 kJ/mol


    By definition, the first ionization energy is the minimum energy required to remove the first electron from an atom. This first electron is the highest energy valence electron in the neutral atom. We expect the energies of valence electrons to be higher than those of core electrons, and first ionization energies to be less than the energy required to remove a lower energy core electron.

    For Hg, the first ionization energy is 1007 kJ/mol, whereas the XPS energy of the 4f electron is 10,100 kJ/mol. The energy required to remove a 4f core electron is 10 times the energy required to remove a 6s valence electron.

    For O, the first ionization energy is 1314 kJ/mol, whereas the XPS energy of a 1s electron is 51,200 kJ/mol. The energy required to remove a 1s core electron is 50 times that required to remove a 2p valence electron.

  1. Hg 2+ : [Xe]4f 14 5d 10 ; valence electrons are 5d

    O 2– : [He]2s 2 2p 6 or [Ne]; valence electrons are 2p

7.113

  1. Mg 3 N 2
  2. Mg 3 N 2 (s) + 3 H 2 O(l) → 3 MgO(s) + 2 NH 3 (g)

    The driving force is the production of NH 3 (g).

  3. After the second heating, all the Mg is converted to MgO.

    Calculate the initial mass Mg.

    0.486 g MgO × 24 .305 g Mg 40.305 g MgO = 0.293 g Mg

    x = g Mg converted to MgO; y = g Mg converted to Mg 3 N 2 ; x = 0.293 – y

    g MgO = x ( 40 .305 g MgO 24.305 g Mg ) ; g Mg 3 N 2 = y ( 100 .929 g Mg 3 N 2 72.915 g Mg )

    g MgO + g Mg 3 N 2 = 0.470

    ( 0.293 y) ( 40 .305 24 .305 ) + y ( 100 .929 72 .915 ) = 0.470

    (0.293 – y)(1.6583) + y(1.3842) = 0.470

    –1.6583 y + 1.3842 y = 0.470 – 0.48588

    –0.2741 y = –0.01588 = –0.016

    y = 0.05794 = 0.058 g Mg in Mg 3 N 2

    g Mg 3 N 2 = 0.05794 g Mg × 100 .929 g Mg 3 N 2 72.915 g Mg = 0.0802 = 0.080 g Mg 3 N 2

    mass % Mg 3 N 2 = 0.0802 g Mg 3 N 2 0.470 g (MgO + Mg 3 N 2 ) × 100 = 17 %

    (The final mass % has 2 sig figs because the mass of Mg obtained from solving simultaneous equations has 2 sig figs.)


  1. 3 Mg(s) + 2 NH 3 (g) → Mg 3 N 2 (s) + 3 H 2 (g)

    6.3 g Mg × 1 mol Mg 24.305 g Mg = 0.2592 = 0.26 mol Mg

    2.57 g NH 3 × 1 mol NH 3 17.031 g NH 3 = 0.1509 = 0.15 mol NH 3

    0.2592 mol Mg × 2 mol NH 3 3 mol Mg = 0.1728 = 0.17 mol NH 3

    0.26 mol Mg requires more than the available NH 3 so NH 3 is the limiting reactant.

    0.1509 mol NH 3 × 3 mol H 2 2 mol NH 3 × 2.016 g H 2 mol H 2 = 0.4563 = 0.46 g H 2

  1. Δ H rxn ° = Δ H f ° Mg 3 N 2 (s) + 3 Δ H f ° H 2 ( g ) 3 Δ H f ° Mg(s) 2 Δ H f ° NH 3 ( g )

    = –461.08 kJ + 3(0) – 3(0) – 2(–46.19) = –368.70 kJ

7.114

  1. r Bi = r BiBr 3 r Br = 2.63 Å 1.20 Å = 1.43 Å
  2. Bi 2 O 3 (s) + 6 HBr(aq) → 2 BiBr 3 (aq) + 3 H 2 O(l)
  3. Bi 2 O 3 is soluble in acid solutions because it acts as a base and undergoes acid-base reactions like the one in part (b). It is insoluble in base because it cannot act as an acid. Thus, Bi 2 O 3 is a basic oxide, the oxide of a metal. Based on the properties of its oxide, Bi is characterized as a metal.
  4. Bi: [Xe]6s 2 4f 14 5d 10 6p 3 . Bi has five outer electrons in the 6p and 6s subshells. If all five electrons participate in bonding, compounds such as BiF 5 are possible. Also, Bi has a large enough atomic radius (1.43 Å) and low-energy orbitals available to accommodate more than four pairs of bonding electrons.
  5. The high ionization energy and relatively large negative electron affinity of F, coupled with its small atomic radius, make it the most electron withdrawing of the halogens. BiF 5 forms because F has the greatest tendency to attract electrons from Bi. Also, the small atomic radius of F reduces repulsions between neighboring bonded F atoms. The strong electron withdrawing properties of F are also the reason that only F compounds of Xe are known.

7.115

  1. 4 KO 2 (s) + 2 CO 2 (g) → 2 K 2 CO 3 (s) + 3 O 2 (g)
  2. K, +1; O, –1/2 (O 2– is superoxide ion); C, +4; O, –2 → K, +1; C, +4; O, –2; O, 0

    Oxygen (in the form of superoxide) is oxidized (to O 2 ) and reduced (to O 2– ).

  3. 18.0 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 4 mol KO 2 2 mol CO 2 × 71.10 g KO 2 1 mol KO 2 = 58.2 g KO 2

    18.0 g CO 2 × 1 mol CO 2 44.01 g CO 2 × 3 mol O 2 2 mol CO 2 × 32.00 g O 2 1 mol O 2 = 19.6 g O 2


 

8 Basic Concepts of Chemical Bonding

Visualizing Concepts

8.1 Analyze/Plan . Count the number of electrons in the Lewis symbol. This corresponds to the ‘A’-group number of the family. Solve .

  1. Group 4A or 14
  2. Group 2A or 2
  3. Group 5A or 15

    (These are the appropriate groups in the s and p blocks, where Lewis symbols are most useful.)

8.2 Analyze . Given the size and charge of four different ions, determine their ionic bonding characteristics.

Plan . The magnitude of lattice energy is directly proportional to the charges of the two ions and inversely proportional to their separation. E el = −Q 1 Q 2 /d. Apply these concepts to A, B, X, and Y.

  1. AY and BX have a 1:1 ratio of cations and anions. In an ionic compound, the total positive and negative charges must be equal. To form a 1:1 compound, the magnitude of positive charge on the cation must equal the magnitude of negative charge on the anion. A 2+ combines with Y 2– and B + combines with X– to form 1:1 compounds.
  2. AY has the larger lattice energy. The A−Y and B−X separations are nearly equal. (A is smaller than B, but X is smaller than Y, so the differences in cation and anion radii approximately cancel.) In AY, Q 1 Q 2 = (2)(2) = 4, whereas in BX, Q 1 Q 2 = (1)(1) = 1.

8.3 Analyze . Given a schematic “slab” of NaCl(s), answer questions regarding the various ions and the electrostatic interactions among them. Plan. E el = −Q 1 Q 2 /d. Use geometry to estimate or calculate distances when needed. Solve.

  1. The smaller purple balls represent Na + cations. Na + has a completed n = 2 shell, whereas Cl has a completed n = 3 shell.
  2. The larger green balls represent Cl anions.
  3. Four. Green-purple interactions are attractive; these are electrostatic attractions between two oppositely charged ions. The sign of E el for these interactions is negative (−).
  4. Four. Green-green (and purple-purple) interactions are repulsive; these are electrostatic attractions between two ions with the same charge. The sign of E el for these interactions is positive (+).

  1. Larger. Because the anions and cations have the same magnitude of charge (1− and 1+), the magnitude of their interactions depends on the distance between the ions; the shorter the distance, the larger the magnitude of the interaction. The distances between any green and any purple ion are the same, d . The magnitude between any two like-colored ions is the hypotenuse of a right triangle with distance 2 d . The shorter attractive interactions have the greater magnitude. Because there are equal numbers of attractive and repulsive interactions, the sum of the attractive interactions is larger.
  1. Positive. If this pattern of ions was extended indefinitely in two dimensions, the magnitude of the total attractive interactions would be greater than the magnitude of the total repulsive interactions. Lattice energy is the energy required to overcome attractive interactions and separate the particles into gas phase ions. The lattice energy would be positive.

8.4 Analyze/Plan . Count the valence electrons in the orbital diagram, take ion charge into account, and find the element with this orbital electron count on the periodic table. Write the complete electron configuration for the ion. Solve .

  1. This ion has six 4d-electrons. Transition metals, or d-block elements, have valence electrons in d-orbitals. Transition metal ions first lose electrons from the 5s orbital, then from 4d if required by the charge. This 2+ ion has lost two electrons from 5s, none from 4d. The transition metal with six 4d-electrons is ruthenium, Ru.
  2. The electron configuration of Ru is [Kr]5s 2 4d 6 . (The configuration of Ru 2+ is [Kr]4d 6 ).

8.5 Analyze/Plan . This question is a “reverse” Lewis structure. Count the valence electrons shown in the Lewis structure. For each atom, assume zero formal charge and determine the number of valence electrons an unbound atom has. Name the element. Solve .

A: 1 shared e pair = 1 valence electron + 3 unshared pairs = 7 valence electrons, F

E: 2 shared pairs = 2 valence electrons + 2 unshared pairs = 6 valence electrons, O

D: 4 shared pairs = 4 valence electrons, C

Q: 3 shared pairs = 3 valence electrons + 1 unshared pair = 5 valence electrons, N

X: 1 shared pair = 1 valence electron, no unshared pairs, H

Z: same as X, H

Check . Count the valence electrons in the Lewis structure. Does the number correspond to the molecular formula CH 2 ONF? 12 e pair in the Lewis structure. CH 2 ONF = 4 + 2 + 6 + 5 + 7 = 24 e , 12 e pair. The molecular formula we derived matches the Lewis structure.

8.6

(a) HNO 2 , 18 valence e , 9 e pairs NO 2 , 18 valence e , 9 e pairs
image
image
  1. The formal charge on N is zero, in both species.

  1. NO 2 is expected to exhibit resonance; the double bond can be drawn to either oxygen atom. An alternate resonance structure for HNO 2 can be drawn, but it has nonzero formal charges on the oxygen atoms. This structure is less likely than the one shown above.
  1. The N=O bond length in HNO 2 will be shorter than the N–O lengths in NO 2 , assuming that the structure shown above is the main contributor to the structure of HNO 2 . This is a reasonable assumption because the Lewis structure in part (a) minimizes formal charges. Because there are two equivalent resonance structures for NO 2 , the N–O lengths are approximately an average of N–O single and double bond lengths. These are longer than the full N=O double bond in HNO 2 .

8.7 Analyze/Plan . Because there are no unshared pairs in the molecule, we use single bonds to H to complete the octet of each C atom. For the same pair of bonded atoms, the greater the bond order, the shorter and stronger the bond. Solve .

  1. Four. Moving from left to right along the molecule, the first C needs two H atoms, the second needs one, the third needs none, and the fourth needs one. The complete molecule is:
    image
  2. In order of increasing bond length: 3 < 1 < 2
  3. Bond 3 is strongest. For the same pair of bonded atoms, the shorter the bond length, the stronger the bond.

8.8 Analyze/Plan . Given an oxyanion of the type XO 4 n– , find the identity of X from elements in the third period. Use the generic Lewis structure to determine the identity of X, and to draw the ion-specific Lewis structures. Use the definition of formal charge, [# of valence electrons − # of nonbonding electrons – (# of bonding electrons/2)], to draw Lewis structures where X has a formal charge of zero. Solve.

  1. According to the generic Lewis structure, each anion has 12 nonbonding and 4 bonding electron pairs for a total of 32 electrons. Of these 32 electrons, the 4 O atoms contribute (4 × 6) = 24, and the overall negative charges contribute 1, 2, or 3. # X electrons = 32 – 24 – n.

    For n = 1−, X has (32 – 24 –1) = 7 valence electrons. X is Cl, and the ion is ClO 4 .

    For n = 2–, X has (32 – 24 – 2) = 6 valence electrons. X is S, and the ion is SO 4 2– .

    For n = 3–, X has (32 – 24 – 3) = 5 valence electrons. X is P, and the ion is PO 4 3– .

    Check . The identity of the ions is confirmed in Table 2.5.

  2. In the generic Lewis structure, X has 0 nonbonding electrons and (8/2) = 4 bonding electrons. Differences in formal charge are because of different numbers of valence electrons on X.

    For PO 4 3– , formal charge of P is (5 – 4) = +1.

    For SO 4 2– , formal charge of S is (6 – 4) = +2.

    For ClO 4 , formal charge of Cl is (7 – 4) = +3.


  1. To reduce the formal charge of X to zero, X must have more bonding electrons. This is accomplished by changing the appropriate number of lone pairs on O to multiple bonds between X and O.
    image

Lewis Symbols (Section 8.1)

8.9

  1. False. The total number of electrons in an atom is the same as its atomic number. Valence electrons are those that take part in chemical bonding, those in the outermost shell of the atom.
  2. image A nitrogen atom has 5 valence electrons.
  3. image The atom (Si) has 4 valence electrons.

8.10

  1. False. The valence shell of H is n = 1, which holds a maximum of 2 electrons.
  2. S: [Ne]3s 2 3p 4 A sulfur atom has 6 valence electrons, so it must gain 2 electrons to achieve an octet.
  3. 1s 2 2s 2 2p 3 = [He]2s 2 2p 3 The atom (N) has 5 valence electrons and must gain 3 electrons to achieve an octet.

8.11

  1. Si: 1s 2 2s 2 2p 6 3s 2 3p 2 .
  2. Four.
  3. The 3s and 3p electrons are valence electrons.

8.12

  1. Ti: [Ar]4s 2 3d 2 . Ti has 4 valence electrons. These valence electrons are available for chemical bonding, whereas core electrons do not participate in chemical bonding.
  2. Hf: [Xe]6s 2 4f 14 5d 2
  3. If Hf and Ti both behave as if they have 4 valence electrons, the 6s and 5d orbitals in Hf behave as valence orbitals and the 4f behaves as a core orbital. This is reasonable because 4f is complete and 4f electrons are, on average, closer to the nucleus than 5d or 6s electrons. The core orbitals for Hf are then [Xe]4f 14 .

8.13

  1. image
  2. image
  3. image
  4. image

8.14

  1. image
  2. image
  3. image
  4. image

Ionic Bonding (Section 8.2)

8.15

  1. image
  2. 2 electrons are transferred.
  3. Mg loses electrons.

8.16

  1. image
  2. CaF 2
  3. 2 electrons are transferred.
  4. Ca loses electrons.

8.17

  1. AlF 3
  2. K 2 S
  3. Y 2 O 3
  4. Mg 3 N 2

8.18

  1. BaF 2
  2. CsCl
  3. Li 3 N
  4. Al 2 O 3

8.19

  1. Sr 2+ : [Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas configuration
  2. Ti 2+ : [Ar]3d 2
  3. Se 2– : [Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas configuration
  4. Ni 2+ : [Ar]3d 8
  5. Br–: [Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas configuration
  6. Mn 3+ : [Ar]3d 4

8.20

  1. Cd 2+ : [Kr]4d 10
  2. P 3– : [Ne]3s 2 3p 6 = [Ar], noble-gas configuration
  3. Zr 4+ : [Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas configuration
  4. Ru 3+ : [Kr]4d 5
  5. As 3– : [Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas configuration
  6. Ag + : [Kr]4d 10

8.21

  1. Endothermic. Lattice energy is the energy required to totally separate 1 mole of solid ionic compound into its gaseous ions. E el for attractive interactions among ions is negative, so the energy required to overcome these attractions and separate the ions is positive.
  2. NaCl(s) → Na + (g) + Cl (g)
  3. Salts like NaCl that have singly charged ions will have smaller lattice energies compared with salts that have doubly charged ions. The magnitude of lattice energy depends on the magnitudes of the charges of the two ions, their radii, and the arrangement of ions in the lattice. The main factor is the charges because the radii of ions do not vary over a wide range.

8.22

  1. NaCl, 788 kJ/mol; KF, 808 kJ/mol

    Given that crystal structure and ionic charges are the same for the two compounds, the difference in lattice energy is because of the difference in ion separation (d). Lattice energy is inversely proportional to ion separation (d), so we expect the compound with the smaller lattice energy, NaCl, to have the larger ion separation. That is, the Na−Cl distance should be longer than the K−F distance.

  2. Na−Cl, 1.16 Å + 1.67 Å = 2.83 Å

    K−F, 1.52 Å + 1.19 Å = 2.71 Å

    This estimate of the relative ion separations agrees with the estimate from lattice energies. Ionic radii indicate that the Na−Cl distance is longer than the K−F distance.

8.23 Analyze/Plan . Assign ion charges by the position of the elements in the periodic table. Lattice energy is directly related to the product of ion charges and inversely related to the ion separation. The dominant factor is ion charges, because the difference between ion separations from one compound to another is not as large as the possible difference between the products of ion charges. Solve .

  1. Na + , 1+; Ca 2+ , 2+
  2. F–, 1– ; O 2– , 2–
  3. CaO will have the larger lattice energy. Lattice energy is directly related to the magnitudes of ion charges. CaO has larger cation and anion charges.
  4. Consider the relationship between the lattice energies of CaO and NaF. (Assume the lattice energy of NaF, 910 kJ, has 3 significant figures, similar to other values in Table 8.1.)

    CaO NaF = 3414 kJ 910 kJ = 3.75

    The ratio of the lattice energies is approximately 4 and the ratio of the two products of cation and anion charges is [(2)(2)/(1)(1)] = 4. If the charges in ScN are 3+ and 3–, respectively, the lattice enthalpy of ScN will be approximately (3)(3)(910) = 8190 kJ.

    Since the calculated ratio is less than the integer value of 4, we expect 8190 kJ to be slightly greater than the measured lattice energy of ScN. From Table 8.1, the measured lattice energy of ScN is 7547, slightly less than our estimate. Recall that lattice energy is also inversely related to ion separation, which is probably greater for ScN than NaF. This also predicts that the measured lattice energy of ScN will be less than our estimate, which is based only on the differences in ion charges.

8.24

  1. According to Equation 8.4, electrostatic attraction increases with increasing charges of the ions and decreases with increasing radius of the ions. Thus, lattice energy (i) increases as the charges of the ions increase and (ii) decreases as the sizes of the ions increase.
  2. KI < LiBr < MgS < GaN. Lattice energy increases as the charges on the ions increase. The ions in KI and LiBr all have 1+ and 1− charges. K + is larger than Li + , and I– is larger than Br–. The ion separation is larger in KI, so it has the smaller lattice energy.

8.25

  1. K−F, 1.52 Å + 1.19 Å = 2.71 Å

    Na−Cl, 1.16 Å + 1.67 Å = 2.83 Å

    Na−Br, 1.16 Å + 1.82 Å = 2.98 Å

    Li−Cl, 0.90 Å + 1.67 Å = 2.57 Å

  2. The order of decreasing lattice energy should be the order of increasing ion separation: LiCl > KF > NaCl > NaBr
  3. From Table 8.1: LiCl, 1030 kJ; KF, 808 kJ; NaCl, 788 kJ; NaBr, 732 kJ The predictions from ionic radii are correct.

8.26   Trend (a) is because of differences in ionic radii. The ions have 1+ and 1– charges in all three compounds. In (b), the ions in the two compounds have different charges, which dominates the lattice energy trend. In (c), ion charge is the main influence, but differences in ionic radii predict the same trend.

8.27   Statement (a) is the best explanation. Equation 8.4 predicts that as the oppositely charged ions approach each other, the energy of interaction will be large and negative. This more than compensates for the energy required to form Ca 2+ and O 2– from the neutral atoms.

8.28   Ba(s) → Ba(g); Ba(g) → Ba + (g) + e ; Ba + (g) → Ba 2+ (g) + e ;

I 2 (s) → 2 I(g); 2 I(g) + 2 e → 2 I–(g), exothermic;

Ba 2+ (g) + 2 I– (g) → BaI 2 (s), exothermic

8.29   RbCl(s) → Rb + (g) + Cl (g) ΔH (lattice energy) = ?

By analogy to NaCl, Figure 8.6, the lattice energy is

Δ H latt = Δ H f ° RbCl(s) + Δ H f ° Rb(g) + Δ H f ° Cl(g) + I 1 (Rb) + E (Cl) = ( 430.5 kJ) + 85 .8 kJ + 121 .7 kJ + 403 kJ + ( 349 kJ) = + 692 kJ/mol

8.30

  1. MgCl 2 , 2326 kJ; SrCl 2 , 2127 kJ. Because the ionic radius of Ca 2+ is greater than that of Mg 2+ , but less than that of Sr 2+ , the ion separation (d) in CaCl 2 will be intermediate as well. We expect the lattice energy of CaCl 2 to be in the range 2200−2250 kJ.
  2. By analogy to Figure 8.6:

    Δ H latt = Δ H f ° CaCl 2 + Δ H f ° Ca(g) + 2 Δ H f ° Cl(g) + I 1 (Ca) + I 2 (Ca) + 2 E (Cl) = ( 795.8 kJ) + 179.3 kJ + 2 ( 121 .7 kJ) + 590 kJ + 1145 kJ + 2 ( 349 kJ) = + 2256 kJ

    This value is near the range predicted in part (a).

Covalent Bonding, Electronegativity, and Bond Polarity (Sections 8.3 and 8.4)

8.31

  1. The bonding in (iii) water and (iv) oxygen is likely to be covalent. (i) iron and (ii) sodium chloride contain metal atoms, which are unlikely to participate in covalent bonding in simple substances. (v) argon is a monatomic gas whose atoms do not strongly interact.
  2. Substance XY is likely to be covalent because it is a gas even below room temperature.

8.32   K and Ar. K is an active metal with 1 valence electron. It is most likely to achieve an octet by losing this single electron and to participate in ionic bonding. Ar has a stable octet of valence electrons; it is not likely to form chemical bonds of any type.

8.33 Analyze/Plan . Follow the logic in Sample Exercise 8.3. Solve .

image

Check . Each pair of shared electrons in SiCl 4 is shown as a line; each atom is surrounded by an octet of electrons.

  1. 4
  2. 7
  3. 8
  4. 8
  5. 4

8.34

image
  1. 5
  2. 7
  3. 8
  4. 8
  5. 3

8.35

  1. image
  2. There are four bonding electrons (two bonding electron pairs) in the structure of O 2 .
  3. The greater the number of shared electron pairs between two atoms, the shorter the distance between the atoms. If O 2 has two bonding electron pairs, the O–O distance will be shorter than the O–O single bond distance.

8.36

  1. The H atoms must be terminal because H can form only one bond.

    14 e , 7 e pairs

    image
  2. There are two bonding electrons (one bonding electron pair) between the two O atoms.
  3. Longer. The oxygen atoms in H 2 O 2 share one pair of electrons, whereas those in O 2 share two pairs (Solution 8.35). The fewer the number of shared electron pairs between two atoms, the longer the distance between them.

8.37   Statement (b) is false. Electron affinity is a property of gas phase atoms or ions, whereas electronegativity is a property of bonded atoms in a molecule.

8.38

  1. The electronegativity of the elements increases going from left to right across a row of the periodic table.
  2. Electronegativity generally decreases going down a family of the periodic table.
  3. False. Elements with the largest ionization energies are the most electronegative.

8.39 Plan . Electronegativity increases going up and to the right in the periodic table. Solve .

  1. Mg
  2. S
  3. C
  4. As

Check . The electronegativity values in Figure 8.8 confirm these selections.

8.40   Electronegativity increases going up and to the right in the periodic table.

  1. O
  2. Al
  3. Cl
  4. F

8.41   The bonds in (a), (c), and (d) are polar because the atoms involved differ in electronegativity. The more electronegative element in each polar bond is:

  1. F
  2. O
  3. I

8.42   The more different the electronegativity values of the two elements, the more polar the bond.

  1. O–F < C–F < Be F. This order is clear from the periodic trend.
  2. S–Br < C–P < O–Cl. Refer to the electronegativity values in Figure 8.8 to confirm the order of bond polarity. The 3 pairs of elements all have the same positional relationship on the periodic table. The more electronegative element is one row above and one column to the left of the less electronegative element. This leads us to conclude that ΔEN is similar for the 3 bonds, which is confirmed by values in Figure 8.8. The most polar bond, O–Cl, involves the most electronegative element, O. Generally, the largest electronegativity differences tend to be between row 2 and row 3 elements. The 2 bonds in this exercise involving elements in row 2 and row 3 do have slightly greater ΔEN than the S–Br bond, between elements in rows 3 and 4.
  3. C–S < N–O < B–F. You might predict that N–O is least polar because the elements are adjacent on the table. However, the big decrease going from the second row to the third means that the electronegativity of S is not only less than that of O, but essentially the same as that of C. C–S is the least polar.

8.43

  1. Analyze/Plan . Q is the charge at either end of the dipole. Q = μ/r. The values for HBr are μ = 0.82 D and r = 1.41 Å. Change Å to m; use the definition of debyes and the charge of an electron to calculate effective charge in units of e. Solve
.

Q = μ r = 0.82 D 1.41 Å × 1 Å 1 × 10 10 m × 3.34 × 10 30 C-m 1 D × 1 e 1.60 × 10 19 C = 0.12 e

  • Decrease. Q = μ/r, μ = Q × r. If r decreases and Q remains the same, μ decreases.
  • 8.44

    1. The more electronegative element, Br, will have a stronger attraction for the shared electrons and adopt a partial negative charge.
    2. Q is the charge at either end of the dipole.

      Q = μ r = 1.21 D 2 .49 Å × 1 Å 1 × 10 10 m × 3.34 × 10 30 C-m 1 D × 1 e 1.60 × 10 19 C = 0.1014 = 0.101 e

      The charges on I and Br are 0.101 e .


    8.45 Analyze/Plan . Generally, compounds formed by a metal and a nonmetal are described as ionic, whereas compounds formed from two or more nonmetals are covalent. However, substances with metals in a high oxidation state often have properties of molecular compounds. In this exercise we know that one substance in each pair is molecular and one is ionic; we may need to distinguish by comparison. Solve .

    1. SiF 4 , metalloid and nonmetal, molecular, silicon tetrafluoride

      LaF 3 , metal and nonmetal, ionic, lanthanum(III) fluoride

    2. FeCl 2 , metal and nonmetal, ionic, iron(II) chloride

      ReCl 6 , metal in high oxidation state, Re(VI), molecular, rhenium hexachloride

    3. PbCl 4 , metal and nonmetal, Pb(IV) is relatively high oxidation state, molecular (by contrast with RbCl, which is definitely ionic), lead tetrachloride

      RbCl, metal and nonmetal, ionic, rubidium chloride

    8.46   Generally, compounds formed by a metal and a nonmetal are described as ionic, whereas compounds formed from two or more nonmetals are covalent. However, substances with metals in a high oxidation states often have properties of molecular compounds.

    1. TiCl 4 , metal and nonmetal, Ti(IV) is a relatively high oxidation state, molecular (by contrast with CaF 2 , which is definitely ionic), titanium tetrachloride

      CaF 2 , metal and nonmetal, ionic, calcium fluoride

    2. ClF 3 , two nonmetals, molecular, chlorine trifluoride

      VF 3 , metal and nonmetal, ionic, vanadium(III) fluoride

    3. SbCl 5 , metalloid and nonmetal, molecular, antimony pentachloride

      AlF 3 , metal and nonmetal, ionic, aluminum fluoride

    Lewis Structures; Resonance Structures (Sections 8.5 and 8.6)

    8.47 Analyze . Counting the correct number of valence electrons is the foundation of every Lewis structure. Plan/Solve .

    1. Count valence electrons: 4 + (4 × 1) = 8 e , 4 e pairs. Follow the procedure in Sample Exercise 8.6.
      image
    2. Valence electrons: 4 + 6 = 10 e , 5 e pairs
      image
    3. Valence electrons: [6 + (2 × 7)] = 20 e , 10 e pairs
      image
      1. Place the S atom in the middle and connect each F atom with a single bond; this requires 2 e pairs.

    1. ii.  Complete the octets of the F atoms with nonbonded pairs of electrons; this requires an additional 6 e pairs.

    2. iii.  The remaining 2 e pairs complete the octet of the central S atom.

    1. (Draw the structure that obeys the octet rule, for now.) 32 valence e , 16 e pairs
      image
    1. Follow Sample Exercise 8.8. 20 valence e , 10 e pairs
      image
    1. 14 valence e , 7 e pairs
      image

      Check . In each molecule, bonding e pairs are shown as lines, and each atom is surrounded by an octet of electrons (duet for H).

    8.48

    1. 12 valence e , 6 e pairs
      image
    2. 14 valence e , 7 e pairs
      image
    3. 50 valence e , 25 e pairs
      image

      (The Lewis structure that obeys the octet rule)

    4. 26 valence e , 13 e pairs
      image
    5. 26 valence e , 13 e pairs
      image

      (The Lewis structure that obeys the octet rule)

    6. NH 2 Cl 14 e , 7 e pairs
      image

    8.49   Statement (b) is most true. (The other four are clearly false.) Keep in mind that when it is necessary to place more than an octet of electrons around an atom to minimize formal charge, there may not be a “best” Lewis structure.


    8.50

    1. 26 e , 13 e pairs
      image

      The octet rule is satisfied for all atoms in the structure.

    2. F is more electronegative than P. Assuming F atoms hold all shared electrons, the oxidation number of each F is –1. The oxidation number of P is +3.
    3. Assuming perfect sharing, the formal charges on all F and P atoms are 0.

    8.51 Analyze/Plan . Draw the correct Lewis structure: count valence electrons in each atom, total valence electrons and electron pairs in the molecule or ion; connect bonded atoms with a line, place the remaining e pairs as needed, in nonbonded pairs or multiple bonds, so that each atom is surrounded by an octet (or duet for H). Calculate formal charges: assign electrons to individual atoms [nonbonding e + 1/2 (bonding e )]; formal charge = valence electrons – assigned electrons. Assign oxidation numbers, assuming that the more electronegative element holds all electrons in a bond.

    Solve . Formal charges are shown near the atoms, oxidation numbers (ox. #) are listed below the structures.

    1. 16 e , 8 e pairs
      image

      ox. #: O, –2; C, +4; S, –2

    2. 26 valence e , 13 e pairs
      image

      ox #: S, +4; Cl, –1; O, –2

    3. 26 valence e , 13 e pairs
      image

      ox. #: Br, +5; O, –2

    4. 20 valence e , 10 e pairs
      image

      ox. #: Cl, +3; H, +1; O, –2

    Check . Each atom is surrounded by an octet (or duet) and the sum of the formal charges and oxidation numbers is the charge on the particle.

    8.52   Formal charges are given near the atoms, oxidation numbers are listed below the structures.

    1. 18 e , 9 e pairs
      image

      ox. #: S, +4; O, –2

    2. 24 e , 12 e pairs
      image

      ox. #: S, +6; O, –2


    1. 26 e , 13 e pairs
      image

      ox. #: S, +4; O, –2

    1. SO 2 < SO 3 < SO 3 2–

      Double bonds are shorter than single bonds. SO 2 has two resonance structures with alternating single and double bonds, for an approximate average ‘‘one-and-a-half” bond. SO 3 has three resonance structures with one double and two single bonds, for an approximately, ‘‘one-and-a-third” bond. SO 3 2– has all single bonds. The order of increasing bond length is the order of decreasing bond type.

      SO 2 (1.5) < SO 3 (1.3) < SO 3 2– (1.0).

    8.53

    1. Plan . Count valence electrons, draw all possible correct Lewis structures, taking note of alternate placements for multiple bonds. Solve .

      18 e , 9 e pairs

      image

      Check . The octet rule is satisfied.

    2. Plan . Isoelectronic species have the same number of valence electrons and the same electron configuration. Solve .

      A single O atom has 6 valence electrons, so the neutral ozone molecule O 3 is isoelectronic with NO 2 .

      image

      Check . The octet rule is satisfied.

    3. Because each N–O bond has partial double bond character, the N–O bond length in NO 2 should be shorter than N–O single bonds but longer than N=O double bonds.

    8.54

    1. 18 e , 9 e pairs
      image
    2. Yes, resonance structures are required to describe the structure.

    1. The Lewis structure of CO 2 (16 e , 8 e pairs) is
      image

      In CO 2 , the C–O bonds are full double bonds with two shared pairs of electrons. In HCO 2 , the two resonance structures indicate that the C–O bonds have partial, but not full, double bond character. The C–O bond lengths in formate will be longer than those in CO 2 .

    8.55 Plan/Solve . The Lewis structures are as follows:

    image

    The more pairs of electrons shared by two atoms, the shorter the bond between the atoms. The average number of electron pairs shared by C and O in the three species is 3 for CO, 2 for CO 2 , and 1.33 for CO 3 2– . The order of bond lengths, from shortest to longest, is: CO < CO 2 < CO 3 2– .

    8.56   The Lewis structures are as follows:

    image

    The average number of electron pairs in the N–O bond is 3.0 for NO + , 1.5 for NO 2 , and 1.33 for NO 3 . The more electron pairs shared between two atoms, the shorter the bond. The order of N–O bond lengths from shortest to longest is: NO + < NO 2 < NO 3 .

    8.57

    1. False. Because of resonance (see Figure 8.15), the C–C bonds in benzene are the same length, but they are shorter than a typical single C–C bond and longer than a typical double C–C bond.
    2. HCCH, 10 e , 5 e pr
      image

      False. The C–C bond in acetylene is an isolated triple bond; it is shorter than an isolated double bond and therefore shorter than the average C–C bond in benzene.


    8.58

    1. image
    2. The resonance model of this molecule has bonds that are neither single nor double, but somewhere in between. This results in bond lengths that are intermediate between C–C single and double bond lengths.
    3. Four. Among the three resonance structures, there are four C–C bonds that appear twice as double bonds and once as a single bond. These are shorter than the others. (The other seven C–C bonds appear twice as single bonds and once as a double bond.)

    Exceptions to the Octet Rule (Section 8.7)

    8.59 Analyze/Plan . In order to decide whether a molecule is an exception to the octet rule, examine the Lewis structure. Does the Lewis structure have an odd number of electrons, an atom with less than eight electrons, or an atom with more than eight electrons? Solve .

    1. CO 2 , 16 e , 8 e pr
      image
      H 2 O, 8 e , 4 e pr
      image
      NH 3 , 8 e , 4 e pr
      image
      PF 3 , 26 e , 13 e pr
      image
      AsF 5 , 40 e , 20 e pr
      image

      None of the molecules have an odd number of electrons. CO 2 , H 2 O, NH 3 , and PF 3 obey the octet rule. In AsF 5 , the central As atom is bound to five F atoms, so it has 10 electrons around it. AsF 5 is an exception to the octet rule.

    2. BH 4 , 8 e , 4 e pr
      image

    1. B 3 N 3 H 6 , 30 e , 15 e pr
      image
      BCl 3 , 24 e , 12 e pr
      image

      [The structure shown minimizes formal charges and is the dominant form (see Section 8.7)]

      BCl 3 is an exception to the octet rule; the B atom has an incomplete octet.

    8.60

    1. 7, 1
    2. 6, 2
    3. 5, 3
    4. 4, 4

    8.61 Analyze/Plan. For each species, count the number of valence electrons and electron pairs. Draw the dominant Lewis structure. For the purpose of this exercise, assume that the dominant Lewis structure is the one that minimizes formal charge.

    ClO, 13 e , 6.5 e pairs

    image

    Odd number of electrons

    Does not obey the octet rule

    ClO–, 14 e , 7 e pairs

    image

    Obeys the octet rule

    ClO 2 , 20 e , 10 e pairs

    image

    Cl has expanded octet

    Does not obey the octet rule

    ClO 3 , 26 e , 13 e pairs

    image

    Cl has expanded octet

    Does not obey the octet rule

    ClO 4 , 32 e , 16 e pairs

    image

    Cl has expanded octet

    Does not obey the octet rule

    In each species, Cl has a zero formal charge and O atoms that form double bonds have zero formal charge. The O atoms that from single bonds have –1 formal charge. For ClO 2 , ClO 3 , and ClO 4 , structures that do not minimize formal charge but obey the octet rule can be drawn. The octet rule vs minimum formal charge debate is ongoing.


    8.62   The second friend is more correct. In the third row and beyond, atoms have the space and available orbitals to accommodate extra electrons. Because atomic radius increases going down a family, elements in the third period and beyond are less subject to destabilization from additional electron-electron repulsions. It is also true, but probably not as important, that elements in the third shell and beyond contain empty d orbitals that are relatively close in energy to valence orbitals (the ones that accommodate the octet).

    8.63

    1. 8 e , 4 e pairs [PH 3 ]
      image

      Obeys octet rule.

    2. 6 e , 3 e pairs
      image

      Does not obey the octete rule. Central Al has 6 electrons (impossible to satisfy octet rule with only 6 valence electrons).

    3. 16 e , 8 e pairs
      image

      3 resonance structures; all obey octet rule.

    4. 20 e , 10 e pairs
      image

      Obeys octet rule.

    5. 48 e , 24 e pairs [SnF 6 2– ]
      image

      Does not obey octet rule. Central Sn has 12 electrons.

    8.64

    1. 11 e , 5.5 e pairs
      image

      Does not obey the octet rule. N has only 7 electrons.


    1. 24 e , 12 e pairs
      image

      Does not obey the octet rule. B has only 6 electrons.

    1. 22 e , 11 e pairs
      image

      Does not obey the octet rule. Central I has 10 electrons.

    1. 32 e , 16 e pairs
      image

      The structure on the left obeys the octet rule, whereas the one on the right minimizes formal charges but does not obey the octet rule. The P in the right structure has 10 electrons.

    1. 36 e , 18 e pairs
      image

      Does not obey the octet rule. Central Xe has 12 electrons.

    8.65

    1. 16 e , 8 e pairs
      image

      This structure violates the octet rule; Be has only 4 e around it.

    2. image
    3. The formal charges on each of the atoms in the four resonance structures are:
      image

      Formal charges are minimized on the structure that violates the octet rule; this form is probably dominant.

    8.66

    1. 26 e , 13 e pairs
      image

    1. Yes, the structure with no double bonds obeys the octet rule for all atoms.
    1. The structure with one double bond has 3 resonance structures (3 possible positions for the double bond), as does the structure with 2 double and 1 single bond (3 possible positions for the single bond). The total number of resonance structures is then 8.
    1. The structure with 3 double bonds minimizes formal charges on all atoms.

    8.67

    1. Analyze/Plan . Given H 2 SO 4 with H attached to O, assume S is central and bound to the four O atoms. Draw a Lewis structure where S and O obey the octet rule and H atoms have two electrons and are terminal. Solve .

      32 e , 16 e pr

      image
    2. Analyze/Plan . Starting with the Lewis structure from part (a), rearrange electrons to minimize formal charge. Formal charge = valence e – assigned e . To have a formal charge of zero, both S and O should have 6 assigned electrons.

      Assigned e = [nonbonding e + ½(bonding e )] Solve .

      For the structure in part (a),

      S:        assigned electrons = 0 + ½(8) = 4;

      FC = 6 valence e – 4 assigned e = +2

      O (terminal):   assigned electrons = 6 + ½(2) = 7;

      FC = 6 valence e – 7 assigned e = –1

      O (bound to H): assigned electrons = 4 + ½(4) = 6;

      FC = 6 valence e – 6 assigned e = 0

      To minimize formal charges, change one nonbonding e pair on each terminal O atom into a bonding e pair between that atom and S. That is, the bonds between the terminal O atoms and S become double bonds.

      image

    8.68

    1. 32 e , 16 e pairs
      image
    2. image

    Strengths and Lengths of Covalent Bonds (Section 8.8)

    8.69 Analyze . Given: structural formulas. Find: enthalpy of reaction.

    Plan. Count the number and kinds of bonds that are broken and formed by the reaction. Use bond enthalpies from Table 8.3 and Equation 5.32 to calculate the overall enthalpy of reaction, ΔH. Solve .

    1. ΔH = 2 D(O H) + D(O C) + 4D(C H) + D(C = C) 2 D(O H) D(O C) 4 D ( C H ) D(C C ) ΔH = D ( O O ) + D ( C = C ) 2 D ( O C ) D ( C C ) = 146 + 614 2 ( 358 ) 348 = 304 kJ
    2. ΔH = 5 D(C H) + D(C N) + D(C=C) 5 D(C H) D(C N) 2 D ( C C ) = D(C = C) 2 D(C C) = 614 2 ( 348 ) = 82 kJ
    3. ΔH = 6 D(N Cl) 3 D(Cl Cl) D(N N) = 6( 200 ) 3 ( 242 ) 941 = 467 kJ

    8.70

    1. ΔH = 3 D(C Br) + D(C H)+D(Cl Cl) 3 D(C Br) D ( C Cl ) D(H Cl) = D(C H)+D(Cl Cl) D ( C Cl ) D(H Cl) ΔH = 413 + 242 328 431 = 104 kJ
    2. ΔH = 4 D(C H) + 2 D(C S) + 2D(S H) + D(C C) + 2 D ( H Br ) 4D(S H) D(C C) 2 D ( C Br ) 4 D(C H) = 2 D(C S) + 2 D ( H Br ) 2 D ( S H ) 2 D(C Br)
    3. ΔH = 4 D(N H) + D(N N) + D(Cl Cl) 4 D(N H) 2 D ( N Cl ) = D(N N) + D(Cl Cl) 2 D ( N Cl ) ΔH = 163 + 242 2 ( 200 ) = 5 kJ

    8.71

    1. False. For the same atom pair, the longer the bond the smaller the bond enthalpy.
    2. False. See average bond enthalpies in Table 8.3.
    3. False. The single bond lengths in Table 8.4 are all less than 5 Å.
    4. False. Breaking a chemical bond requires an input of energy.
    5. True.

    8.72

    1. True. See Table 8.4. In general for the same atom pair, the greater the number of bonds between atoms, the shorter the bond.
    2. False. The Lewis structure of O 2 has two pairs of bonding electrons and four pairs of nonbonding electrons.

    1. False. The C–O bond in CO is shorter, because the two atoms share three electron pairs, while in CO 2 , each C–O bond involves two shared electron pairs.
    1. False. The Lewis structure of O 2 has an isolated double bond. Ozone has resonance structures; the average O–O bond length will be shorter than a single but longer than a double bond.
    1. False. The more electronegative an atom, the more strongly it holds its own electrons. It forms the minimum number of bonds needed to satisfy the octet rule.

    8.73   Ionic bond enthalpies depend on the charge and size of the participating ions. The Ca–O bond will be stronger than the Na–Cl bond, because the ion charges are greater.

    8.74   Ionic bond enthalpies depend on the charge and size of the participating ions. The Cs–F and Li–F bonds have the same anion and cation charges but Cs + has a much larger ionic radius than Li + . The Cs–F bond, with the larger ion separation, will have the smaller ionic bond enthalpy.

    8.75

    1. CH 2 , 6 e , 3 e pr image

      The C atom in carbene is extremely electron deficient, which causes carbene to be very reactive.

    2. 2 CH 2 , → C 2 H 4
      image

      The product of this reaction contains a C–C double bond, with a typical bond length of 1.34 Å. (See Table 8.4.)

    8.76   NO, 11 e , 5.5 e pr image

    NO + , 10 e , 5 e pr image

    The bond in NO is longer than the bond in NO + . In the neutral NO molecule, the N atom has an incomplete octet and the N–O bond is formally a double bond. In NO + , the odd electron has been lost and the N–O bond is a triple bond. For the same pair of bonded atoms, a double bond is longer than a triple bond.

    Additional Exercises

    8.77   A triple C–C bond. A 1.15 Å bond length is relatively short for any atom pair. Table 8.4 indicates than an average triple C–C bond length is 1.20 Å.

    8.78   A double N–N bond. Table 8.4 indicates than an average double N–N bond length is 1.24 Å. (The bonding atomic radius of N is less than that of C, so we expect N–N bonds of all types to be shorter than C–C bonds.)

    8.79

    1. All the compounds in the series contain Group 2A metal cations with oxidation numbers of +2. The oxidation number of hydrogen is then –1. (H– is named hydride.)

    1. Be 2+ is the smallest cation, so BeH 2 has the shortest cation-anion distance. All compounds in the series have the same product of ion charges, so the trend in cation-anion distance dictates the trend in lattice energy. Lattice energy is inversely related to ion separation, so the compound with the largest lattice energy, BeH 2 , has the shortest cation-anion distance.
    1. It costs 3205 kJ of energy to break one mole of BeH 2 into its component gas phase ions. Charge attraction is a stabilizing force; energy is required to overcome it.
    1. Mg. Lattice energy depends on ionic radius and charge. The charges are the same in the series, so ionic radius is the discriminating factor. The compound with a lattice energy nearest 2870 kJ/mol has the Group 2A cation with the radius most similar to Zn 2+ .

    8.80

    1. image

      The difference in lattice energy between LiCl and LiI is 104 kJ. The difference between NaCl and NaI is 106 kJ; the difference between NaCl and NaBr is 56 kJ, or 53% of the difference between NaCl and NaI. Applying this relationship to the Li salts, 0.53(104 kJ) = 55 kJ difference between LiCl and LiBr. The approximate lattice energy of LiBr is (834 – 55) kJ = 779 kJ.

    2. image

      By analogy to the Na salts, the difference between lattice energies of CsCl and CsBr should be approximately 53% of the difference between CsCl and CsI. The lattice energy of CsBr is approximately 627 kJ.

    3. image

      By analogy to the oxides, the difference between the lattice energies of MgCl 2 and CaCl 2 should be approximately 66% of the difference between MgCl 2 and SrCl 2 . That is, 0.66(199 kJ) = 131 kJ. The lattice energy of CaCl 2 is approximately (2326 – 131) kJ = 2195 kJ.

    8.81   The charge on M is likely to be 3+. According to Table 8.1, the lattice energy for an ionic compound with the general formula MX and a charge of 2+ on the metal will be in the range of 3-4 × 10 3 kJ/mol. The charge on M must be greater than 2+. ScN, where the charge on Sc is 3+, has a lattice energy of 7547 kJ/mol. It is reasonable to conclude that the charge on M is 3+, and the M–X distance is greater than the Sc–N distance.


    8.82

    1. Six. Figure 8.4 shows a small part of the CaO (or NaCl) structure. Think of it as a cube with faces and edges. Focus on one purple Ca 2+ cation in the interior of a face (not on an edge). It is touching 4 O 2– anions also in the face. Additionally, this same Ca 2+ touches an O 2– behind and in front of it. The O 2– behind it is inside the cube, but the one “in front” of it is in the next layer of ions not shown in the figure. This is a total of six O 2– anions touching a single Ca 2+ cation.
    2. Energy would be consumed. Electrostatic attraction holds ion pairs together in a 3-dimensional structure such as the one in Figure 8.4. This energy of electrostatic attraction must be overcome in order to convert a crystal of CaO into a collection of widely separated Ca–O ion pairs.
    3. The ionic radii from Figure 7.8 are: Ca 2+ , 1.14 Å; O 2– , 1.26 Å. The lattice energy is 3414 kJ/mol.

      Ε el = κ Q 1 Q 2 d ; d = 2.40 Å; Q 1 = 2(1.60 × 10 –19 C), Q 2 = –2(1.60 × 10–19 C)

      E = 8.99 × 10 9 J-m C 2 × 4 ( 1.60 × 10 19 C) 2 2.40 Å × 1 Å 1 × 10 10 m = 3.836 × 10 18 = 3.84 × 10 18 J

    4. On a molar basis: (–3.836 × 10– 18 J)(6.022 × 10 23 ) = –2.310 × 10 6 J = –2310 kJ

      Note that the absolute value of this potential energy is less than the lattice energy of CaO, 3414 kJ/mol. The difference represents the additional energy required to separate the individual Ca 2+ O 2– ion pairs from their three-dimensional array similar to the one in Figure 8.4.

    5. The electrostatic interactions in a crystal lattice are more complicated than those in a single ion pair.

    8.83   By analogy to the Born-Haber cycle for NaCl(s), Figure 8.6, the enthalpy of formation for NaCl 2 (s) is

    Δ H f o NaCl 2 (s) = −ΔH latt NaCl 2 + Δ H f o Na(g) + 2 Δ H f o Cl(g) + I 1 (Na) + I 2 (Na) + 2 E(Cl)

    1. Δ H f o NaCl 2 (s) = −ΔH latt NaCl 2 + 107.7 kJ + 2(121.7 kJ) + 496 kJ + 4562 kJ + 2(−349 kJ)

      Δ H f o NaCl 2 (s) = −ΔH latt NaCl 2 + 4711 kJ

      The collective energy of the “other” steps in the cycle (vaporization and ioniza-tion of Na 2+ , dissociation of Cl 2 and electron affinity of Cl) is +4711 kJ. In order for the sign of Δ H f o NaCl 2 to be negative, the lattice energy would have to be greater than 4711 kJ.

    2. Δ H f o NaCl 2 (s) = −(2326 kJ) + 4711 kJ = 2385 kJ

      This value is large and positive.

    8.84

    1. Yes. If X and Y have different electronegativities, they have different attractions for the electrons in the molecule. The electron density around the more electronegative atom will be greater, producing a charge separation or dipole in the molecule.
    2. Yes. μ = Qr. The dipole moment, μ, is the product of the magnitude of the separated charges, Q, and the distance between them, r. The longer the bond between X and Y, the larger the dipole moment.

    8.85

    1. B–O. The most polar bond will be formed by the two elements with the greatest difference in electronegativity. Because electronegativity increases moving right and up on the periodic table, the possibilities are B–O and Te O. These two bonds are likely to have similar electronegativity differences (3 columns apart vs. 3 rows apart). Values from Figure 8.8 confirm the similarity, and show that B–O is slightly more polar.
    2. Te I. Both are in the fifth row of the periodic table and have the two largest covalent radii among this group of elements.
    3. TeI 2 . Te needs to participate in two covalent bonds to satisfy the octet rule, and each I atom needs to participate in one bond, so by forming a TeI 2 molecule, the octet rule can be satisfied for all three atoms.
      image
    4. B 2 O 3 . Although this is probably not a purely ionic compound, it can be understood in terms of gaining and losing electrons to achieve a noble-gas configuration. If each B atom were to lose 3 e and each O atom were to gain 2 e , charge balance and the octet rule would be satisfied.

      P 2 O 3 . Each P atom needs to share 3 e and each O atom 2 e to achieve an octet. Although the correct number of electrons seem to be available, a correct Lewis structure is difficult to imagine. In fact, phosphorus (III) oxide exists as P 4 O 6 rather than P 2 O 3 (Chapter 22).

    8.86

    1. Q = μ r = 1 . 2 4 D 1 . 6 0 Å × 1 Å 1 × 1 0 1 0 m × 3 . 3 4 × 1 0 3 0 C - m 1 D × 1 e 1 . 6 0 × 1 0 1 9 C = 0 . 1 6 1 8 = 0 . 1 6 2 e
    2. From Figure 8.8, the electronegativity of Cl is 3.0 and that of O is 3.5. Because O is the more electronegative element, we expect it to have a partial negative charge in the ClO molecule.
    3. 13 e , 6.5 e pairs
      image

      According to formal charge arguments, the Lewis structure on the right is dominant. In both structures, the less electronegative Cl atom is electron-deficient. However, the small electronegativity difference and calculated charges both point to a slightly polar covalent molecule. The true bonding situation is a blend of the two extreme Lewis structures, with the right-most structure making the larger contribution.

  • Because ClO– has an overall charge of 1–, the sum of the formal charges in any correct Lewis structure is 1–. We expect the more electronegative O atom to carry the negative formal charge. The best Lewis structure for ClO– is then
    image

    The formal charge on Cl in this structure is 0.


  • 8.87

    1. Estimate relative attraction for the bonding electron pair by calculating the relative electronegativity of the two atoms. From Figure 8.8, the electronegativity of Br is 2.8 and of Cl is 3.0.

      Br has 2.8/(3.0 + 2.8) = 0.48 of the charge of the bonding e pair.

      Cl has 3.0/(3.0 + 2.8) = 0.52 of the charge of the bonding e pair.

      This amounts to 0.52 × 2 e = 1.04 e on Cl or 0.04 e more than a neutral Cl atom. This implies a –0.04 charge on Cl and +0.04 charge on Br.

    2. From Figure 7.7, the covalent radius of Br is 1.20 Å and of Cl is 1.02 Å. The Br–Cl separation is 2.22 Å.

      μ = Qr = 0.04 e × 1.60 × 10 19 C e × 2.22 Å × 1 × 10 10 m Å × 1 D 3 .34 × 10 30 C-m = 0.43 D

    3. Q = μ r = 0.57 D 2 .22 Å × 1 Å 1 × 10 10 m × 3.34 × 10 30 C-m 1 D × 1 e 1.60 × 10 19 C = 0.054 e

      From this calculation, the partial charge on Br is +0.054 and on Cl is –0.054.

    8.88

    1. 2 NaAlH 4 (s) → 2 NaH(s) + 2 Al(s) + 3 H 2 (g)
    2. Hydrogen is the only nonmetal in NaAlH 4 , so we expect it to be most electronegative. (The position of H on the periodic table is problematic. Its electronegativity does not fit the typical trend for Gp 1A elements.) For the two metals, Na and Al, electronegativity increases moving up and to the right on the periodic table, so Al is more electronegative. The least electronegative element in the compound is Na.
    3. Covalent bonds hold polyatomic anions together; elements involved in covalent bonding have smaller electronegativity differences than those that are involved in ionic bonds. Possible covalent bonds in NaAlH 4 are Na−H and Al−H. Al and H have a smaller electronegativity difference than Na and H and are more likely to form covalent bonds. The anion has an overall 1− charge, so it can be thought of as four hydride ions and one Al 3+ ion. The formula is AlH 4– . For the purpose of counting valence electrons, assume neutral atoms.

      8 e 4 e pairs

      image
    4. The formal charge of H in AlH 4– is 0. (The formal charge of Al is −1. This brings the sum of formal charges to −1, the overall charge of the polyatomic anion.)

    8.89

    1. I 3– , 22 e , 11 e pr
      image
    2. F 2 , HF, CF 4 , SiF 4 , etc.

    1. No. I 3– violates the octet rule but is quite stable. Violating the octet rule does not prevent the existence of a substance.
    1. This classmate is at least partly correct. The presence of five bonding and nonbonding electron pairs around a central atom as small as F would generate significant electron-electron repulsion. These repulsions would destabilize F 3– .

    8.90   Formal charge (FC) = # valence e – (# nonbonding e + 1/2 # bonding e )

    1. 18 e , 9 e pairs
      image

      FC for the central O = 6 – [2 + 1/2 (6)] = +1

    2. 48 e , 24 e pairs
      image

      The three nonbonded pairs on each F have been omitted.

    3. 17 e ; 8 e pairs, 1 odd e
      image

      The odd electron is probably on N because it is less electronegative than O. Assuming the odd electron is on N, FC for N = 5 – [1 + 1/2 (6)] = +1. If the odd electron is on O, FC for N = 5 – [2 + 1/2 (6)] = 0.

    4. 28 e , 14 e pairs
      image

      FC for I = 7 – [4 + 1/2 (6)] = 0

    5. 32 e , 16 e pairs
      image

      FC for Cl = 7 – [0 + 1/2 (8)] = +3

    8.91   14e , 7 e pairs

    image

    32 e , 16 e pairs

    image
    1. FC on Cl in ClO = 7 – [6 + 1/2(2)] = 0
    2. FC on Cl in ClO 4 = 7 – [0 + 1/2(8)] = +3
    3. The oxidation number of Cl in ClO is

      [ON + (–2)] = –1; ON of Cl = +1

    4. The oxidation number of Cl in ClO 4– is

      [ON + 4(–2)] = –1; ON of Cl = +7


    1. The more positive the formal charge and the higher the oxidation number of a bonded atom, the greater the electron deficiency at that atom. The atom with the higher oxidation number is more likely to accept electrons from another compound and be reduced. Perchlorate, ClO 4– , is much more likely to be reduced.

    8.92

    1. image

      In the leftmost structure, the more electronegative O atom has the negative formal charge, so this structure is likely to be most important.

    2. No single resonance structure rationalizes both observed bond lengths. In general, the more shared pairs of electrons between two atoms, the shorter the bond, and vice versa. That the N–N bond length in N 2 O is slightly longer than the typical N≡N indicates that the middle and right resonance structures where the N atoms share less than three electron pairs are contributors to the true structure. That the N–O bond length is slightly shorter than a typical N=O indicates that the middle structure, where N and O share more than two electron pairs, does contribute to the true structure. This physical data indicates that although formal charge can be used to predict which resonance form will be more important to the observed structure, the influence of minor contributors on the true structure cannot be ignored.

    8.93

    1. 12 + 3 + 15 = 30 valence e , 15 e pairs.
      image

      Structures with H bound to N and nonbonded electron pairs on C can be drawn, but the structures above minimize formal charges on the atoms.

    2. The resonance structures indicate that triazine will have six equal C–N bond lengths, intermediate between C–N single and C–N double bond lengths. (See Solutions 8.57 and 8.58.) From Table 8.4, an average C–N length is 1.43 Å, a C=N length is 1.38 Å. The average of these two lengths is 1.405 Å. The C–N bond length in triazine should be in the range 1.40–1.41 Å.

    8.94

    1. 24 + 4 + 14 = 42 valence e , 21 e pairs.
      image
    2. image

    1. No. In benzene, the six C atoms are equivalent. In ortho-dichlorobenzene, the two C atoms bound to Cl are not equivalent to the four C atoms bound to H. In the two resonance structures above, one has a double bond between the C atoms bound to Cl, and the other has a single bond in this position. The two ortho-dichlorobenzene resonance structures are not equivalent like the resonance structures of benzene.

    Integrative Exercises

    8.95

    1. False. The B–A=B structure says nothing about the nonbonding electrons in the molecule. One possible example is NO 2 , which has an odd electron.
    2. True.

    8.96 Δ H = 8 D(C H) D(C C) 6 D(C H) D(H H) = 2 D(C H) D(C C) D(H H) = 2 ( 413 ) 348 436 = + 42 kJ Δ H = 8 D(C H)+1/2 D(O=O) D(C C) 6 D(C H) 2 D(O H) = 2 D(C H)+1/2 D(O = O) D(C C) 2 D(O H) = 2 ( 413 ) + 1 / 2 ( 495 ) 348 2 ( 463 ) = 200 kJ

    The fundamental difference in the two reactions is the formation of 1 mol of H–H bonds versus the formation of 2 mol of O–H bonds. The latter is much more exothermic, so the reaction involving oxygen is more energetically favorable.

    8.97

    1. Δ H = 5 D(C H) + D(C C) + D(C O) + D(O H) 6 D(C H) 2 D(C O) = D ( C C ) + D(O H) D(C H) D(C D) = 348 kJ + 463 kJ 413 kJ 358 kJ Δ H = + 4 0 kJ ; ethanol has the lower enthalpy
    2. Δ H = 4 D(C H) + D(C C) + 2 D(C O) 4 D(C H) D(C C) D(C=O) = 2 D ( C O ) D(C=O) = 2(358 kJ) 799 kJ Δ H = 83 kJ; acetaldehyde has the lower enthalpy
    3. Δ H = 8 D(C H) + 4 D(C C) + D(C=C) 8 D(C H) 2 D(C C) 2 D(C=C) = 2 D ( C C ) D(C=C) = 2(358 kJ) 614 kJ Δ H = 82 kJ; cyclopentene has the lower enthalpy
    4. Δ H = 3 D(C H) + 4 D(C N) + D(C N) 3 D(C H) D(C C) D(C C) = D ( C N ) D(C C) = 293 kJ 348 kJ Δ H = 55 kJ; acetonitrile has the lower enthalpy

    8.98

    1. Ti 2+ : [Ar]3d 2 ; Ca : [Ar]4s 2 .
    2. Ca has no unpaired electrons and Ti 2+ has two. The two valence electrons in Ca are paired in the 4s orbital. Each of the two valence electrons in Ti 2+ occupies its own 3d orbital (Hund’s rule).
    3. To be isoelectronic with Ca 2+ , Ti would have a 4+ charge.

    8.99

    1. H 2 O 2 , 14 e , 7 e pr
      image
    2. Referring to Table 8.3, the single O–O bond is the weakest bond in H 2 O 2 .
    3. The average bond enthalpy of one mole of single O–O bonds is 146 kJ. The enthalpy of one single O–O bond is

      1 4 6 kJ mol × 1 mol 6 . 0 2 2 × 1 0 2 3 bonds × 1 0 0 0 J k J = 2 . 4 2 4 4 × 1 0 1 9 J = 2 . 4 2 × 1 0 19 J

      λ = hc E = 6 . 6 2 6 × 1 0 3 4 J- s 2 . 4 2 4 4 × 1 0 1 9 J × 2 . 9 9 8 × 1 0 8 m s × 1 × 1 0 9 nm m = 8 1 9 . 3 7 = 8 1 9 nm

      [Recall that the wavelength range for visible light is typically 400-750 nm. All visible light has energy sufficient to break a single O–O bond.]

    8.100  The pathway to the formation of K 2 O can be written:

    2 K(s) 22 K(g) 2 Δ H f K(g) 2 K(g) 2 K + ( g ) + 2 e 2 I 1 (K) 1 / 2 O 2 ( g ) O(g) Δ H f O(g) O ( g ) + e O ( g ) E 1 ( O ) O ( g ) + e O ( g ) E 2 ( O ) 2 K + ( g ) + O 2 ( g ) K 2 O(s) Δ H latt K 2 O(s) 2 K(s) + 1 / 2 O 2 ( g ) K 2 O(s) Δ H f K 2 O(s)

    Δ H f o K 2 O(s) = 2 Δ H f o K(g) + 2 I 1 ( K) + Δ H f o O(g) + E 1 (O) + E 2 ( O) Δ H latt K 2 O(s)

    E 2 (O) = ΔH f o K 2 O(s) + ΔH latt K 2 O(s) 2 ΔH f o K(g) 2 I t ( K) ΔH f o O(g) E 1 ( O) E 2 (O) = 363 . 2 kJ + 2238 kJ 2 ( 89 . 99 ) kJ 2 ( 419 ) kJ 247 . 5 kJ ( 141 ) kJ = + 750 kJ

    8.101   To calculate empirical formulas, assume 100 g of sample.

    1. 76.0 g Ru 101.07 g/mol = 0.752 mol Ru; 0.752/0.752 = 1 Ru

      24.0 g O 15.9994 g/mol = 1.50 mol O; 1.50/0.762 = 2 O

      The empirical formula of compound 1 is RuO 2 .

    2. 61 .2 g Ru 101 .07 g/mol = 0.6055 mol Ru; 0.6055/0.6055 = 1 Ru

      38 .8 g O 15 .9994 g/mol = 2.425 mol O; 2.425/0.6055 = 4 O

      The empirical formula of compound 2 is RuO 4 .


    1. The lower melting yellow compound is molecular. Substances with metals in high oxidation states are often molecular. RuO 4 contains Ru(VIII), whereas RuO 2 contains Ru(IV), so RuO 4 is more likely to be molecular. The yellow compound is RuO 4 .
    1. The very high melting black compound is ionic. The black compound is RuO 2 .
    1. Yellow RuO 4 is molecular.
    1. Black RuO 2 is ionic.

    8.102

    1. Even though Cl has the greater (more negative) electron affinity, F has a much larger ionization energy, so the electronegativity of F is greater.

      F: k(I–EA) = k(1681 – (–328)) = k(2009)

      Cl: k(I–EA) = k(1251 – (–349)) = k(1600)

    2. Electronegativity is the ability of an atom in a molecule to attract electrons to itself. It can be thought of as the ability to hold its own electrons (as measured by ionization energy) and the capacity to attract the electrons of other atoms (as measured by electron affinity). Thus, both properties are relevant to the concept of electronegativity.
    3. EN = k(I – EA). For F: 4.0 = k(2009), k = 4.0/2009 = 2.0 × 10– 3
    4. Cl: EN = 2.0 × 10– 3 (1600) = 3.2

      O: EN = 2.0 × 10– 3 (1314 – (–141)) = 2.9

    5. F: (I+EA)/2 = (1681 – 328)/2 = 676.5 = 677

      To scale the value to 4.0 for F, 4.0 = k(677), k = 4.0/677 = 5.9 × 10– 3

      Cl: 5.9 × 10– 3 (1251 – 349)/2 = 2.7

      Br: 5.9 × 10– 3 (1140 – 325)/2 = 2.4

      I: 5.9 × 10– 3 (1008 – 295)/2 = 2.1

      On this scale, the electronegativity of Br is 2.4.

    8.103

    1. Assume 100 g.

      14.52 g C × 1 mol 12.011 g C = 1.209 mol C; 1 .209/1 .209 = 1

      1.83 g H × 1 mol 1.008 g H = 1.816 mol H; 1 .816/1 .209 = 1.5

      64.30 g Cl × 1 mol 35 .453 g Cl = 1.814 mol Cl; 1 .814/1 .209 = 1.5

      19.35 g O × 1 mol 15 .9994 g O = 1.209 mol O; 1 .209/1 .209 = 1.0

      Multiplying by 2 to obtain an integer ratio, the empirical formula is C 2 H 3 Cl 3 O 2 .

    2. The empirical formula mass is 2(12.0) + 3(1.0) + 3(35.5) + 2(16) = 165.5. The empirical formula is the molecular formula.

    1. 44 e , 22 e pairs
      image

    8.104

    1. Assume 100 g.

      62.04 g Ba × 1 mol 137.33 g Ba = 0.4518 mol Ba; 0 .4518/0 .4518 = 1.0

      37.96 g N × 1 mol 14.007 g N = 2.710 mol N; 2 .710/0 .4518 = 6.0

      The empirical formula is BaN 6 . Ba has an ionic charge of 2+, so there must be two 1– azide ions to balance the charge. The formula of each azide ion is N 3– .

    2. 16 e , 8 e pairs
      image
    3. The structure with two double bonds minimizes formal charges and is probably the main contributor.
    4. The two N–N bond lengths will be equal. The two minor contributors would individually cause unequal N–N distances, but collectively they contribute equally to the lengthening and shortening of each bond. The N–N distance will be approximately 1.24 Å, the average N=N distance.

    8.105

    1. C 2 H 2 : 10 e , 5 e pair     N 2 : 10 e , 5 e pair
      image
    2. The enthalpy of formation for N 2 is 0 kJ/mol and for C 2 H 2 is 226.77 kJ/mol. N 2 is an extremely stable, unreactive compound. Under appropriate conditions, it can be either oxidized or reduced. C 2 H 2 is a reactive gas, used in combination with O 2 for welding and as starting material for organic synthesis.
    3. 2 N 2 (g) + 5 O 2 (g) → 2 N 2 O 5 (g)

      2 C 2 H 2 (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O(g)

    4. ΔH rxn o ( N 2 ) = 2 Δ H f o N 2 O 5 (g) 2 ΔH f o N 2 (g) 5 ΔH f o O 2 (g) = 2 ( 11 . 3 0 ) 2 ( 0 ) 5 ( 0 ) = 22 . 6 0 kJ ΔH rxn o ( N 2 ) = 11.30 kJ/mol N 2

      Δ H rxn o ( C 2 H 2 ) = 4 Δ H f o CO 2 (g) + 2 Δ H f o H 2 O(g) 2 Δ H f o C 2 H 2 (g) 5 Δ H f o O 2 (g) = 4( 393 .5 kJ) + 2( 241 .82 kJ) 2(226 .77 kJ) 5(0) = 2511 .18 kJ Δ H ox o ( C 2 H 2 ) = 1255.6 kJ/mol C 2 H 2

    5. N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

      ΔH rxn o ( N 2 ) = 2 ΔH f o N H 3 (g) ΔH f o N 2 (g) 3 ΔH f o H 2 (g) = 2 ( 46 . 19 ) ( 0 ) 3 ( 0 ) = 92 . 38kJ ΔH rxn o ( N 2 ) = 46 .19 kJ/mol N 2


    1. C 2 H 2 (g) + 3 H 2 (g) → 2 CH 4 (g)

      Δ H rxn o ( C 2 H 2 ) = 2 ΔH f o CH 4 (g) 2 ΔH f o C 2 H 2 (g) 3 ΔH f o H 2 (g) = 2 ( 679 . 9kJ ) 226 . 77 kJ 3 ( 0 ) = 1586 . 6 kJ Δ H rxn o ( C 2 H 2 ) = 793.3 kJ/mol C 2 H 2

    8.106

    1. Assume 100 g of compound

      69.6 g S × 1 mol S 32.07 g = 2.17 mol S 30 .4 g N × 1 mol N 14 .01 g = 2.17 mol N

      S and N are present in a 1:1 mol ratio, so the empirical formula is SN. The empirical formula mass is 46. MM/FW = 184.3/46 = 4 The molecular formula is S 4 N 4 .

    2. 44 e , 22 e pairs. Because of its small radius, N is unlikely to have an expanded octet. Begin with alternating S and N atoms in the ring. Try to satisfy the octet rule with single bonds and lone pairs. At least two double bonds somewhere in the ring are required.
      image

      These structures carry formal charges on S and N atoms as shown. Other possibilities include:

      image

      These structures have zero formal charges on all atoms and are likely to contribute to the true structure. Note that the S atoms that are shown with two double bonds are not necessarily linear because S has an expanded octet. Other resonance structures with four double bonds are:

      image

      In either resonance structure, the two “extra” electron pairs can be placed on any pair of S atoms in ring, leading to a total of 10 resonance structures. The sulfur atoms alternately carry formal charges of +1 and –1. Without further structural information, it is not possible to eliminate any of the above structures. Clearly, the S 4 N 4 molecule stretches the limits of the Lewis model of chemical bonding.


    1. Each resonance structure has 8 total bonds and more than 8 but fewer than 16 bonding e pairs, so an “average” bond will be intermediate between a S–N single and double bond. We estimate an average S–N single bond length to be 1.77 Å (sum of bonding atomic radii from Figure 7.7). We do not have a direct value for a S–N double bond length. Comparing double and single bond lengths for C–C (1.34 Å, 1.54 Å), N–N (1.24 Å, 1.47 Å), and O–O (1.21 Å, 1.48 Å) bonds from Table 8.4, we see that, on average, a double bond is approximately 0.23 Å shorter than a single bond. Applying this difference to the S–N single bond length, we estimate the S–N double bond length as 1.54 Å. Finally, the intermediate S–N bond length in S 4 N 4 should be between these two values, approximately 1.60–165 Å. (The measured bond length is 1.62 Å.)
    1. S 4 N 4 → 4 S(g) + 4 N(g)

      Δ H = 4 Δ H f o S(g) + 4 Δ H f o N(g) Δ H f o S 4 N 4

      ΔH = 4(222.8 kJ) + 4(472.7 kJ) – 480 kJ = 2302 kJ

      This energy, 2302 kJ, represents the dissociation of 8 S–N bonds in the molecule; the average dissociation energy of one S–N bond in S 4 N 4 is then 2302 kJ/8 bonds = 287.8 kJ.

    8.107

    1. Yes. In the structure shown in the exercise, each P atom needs 1 unshared pair to complete its octet. This is confirmed by noting that only 6 of the 10 valence e pairs are bonding pairs.
    2. There are 6 P–P bonds in P 4 .
    3. 20 e , 10 e pr
      image

      There are no other resonance forms for this structure. The octet rule is satisfied for all atoms. However, it requires P=P, which is uncommon because P has a covalent radius that is too large to accommodate the side-to-side π overlap of parallel p orbitals required for double bond formation.

    4. From left to right, the formal charges are on the P atoms in the linear structure are −1, +1, +1, −1. In the tetrahedral structure, all formal charges are zero. Clearly the linear structure does not minimize formal charge and is probably less stable than the tetrahedral structure, owing to the difficulty of P=P bond formation (see above).

    8.108

    1. HCOOH, 18 e , 9 e pr
      image

      The resonance structure on the right is the dominant form. The structure on the left can be drawn; no atom violates the octet rule, but formal charges on O are not minimized and it is a very minor form.


    1. HCOOH(aq) + NaOH(aq) →HCOO–(aq) + H 2 O(l) + Na + (aq)
    1. image
    1. 0 .785 mL HCOOH × 1 .220 g mL × 1 mol HCOOH 46.025 g HCOOH = 0.02081 = 0.0208 mol HCOOH

      0 .02081 moL HCOOH × 1 mol NaOH 1 mol HCOOH × 1 L NaOH 0.100 M NaOH × 1000 mL L = 208.08 = 208 mL of 0.100 M NaOH

    8.109

    1. NH 3 BF 3 , 32 e , 16 e pr
      image
    2. Moving from left to right across a row of the periodic table, electronegativity increases. The electron density will be greater around the atom with greater electronegativity, in this case N.
      image
    3. The difference between NH 3 BCl 3 and NH 3 BF 3 is that Cl has replaced F as the element bound to B. Chlorine is less electronegative and less electron withdrawing than fluorine. This increases the electron density at B and renders the B–N bond less polar in NH 3 BCl 3 , than NH 3 BF 3 .

    8.110

    1. NH 4 + , 8 e , 4 e pr; Cl , 8 e , 4 e pr
      image
    2. No. The bond in NH 4 Cl(s) is ionic; it is electrostatic attraction among oppositely charged ions.
    3. 14 g NH 4 Cl × 1 mol NH 4 Cl 53.495 g NH 4 Cl × 1 0.5000 L = 0.5234 = 0.52 M
    4. NH 4 Cl(aq) + AgNO 3 (aq) → AgCl(s) + NH 4 NO 3 (aq)

      14 g NH 4 Cl × 1 mol NH 4 Cl 53.495 g NH 4 Cl × 1 mol AgNO 3 1 mol NH 4 Cl × 169.88 g AgNO 3 1 mol AgNO 3 = 44.459 = 44 g AgNO 3


     

    9 Molecular Geometry and Bonding Theories

    Visualizing Concepts

    9.1   Removing an atom from the equatorial plane of trigonal bipyramid in Figure 9.3 creates a seesaw shape. It might appear that you could also obtain a seesaw by removing two atoms from the square plane of the octahedron. However, one of the B–A–B angles in the seesaw is 120°, so it must be derived from a trigonal bipyramid.

    9.2

    1. 120°
    2. If the blue balloon expands, the angle between red and green balloons decreases.
    3. (ii)

    9.3 Analyze/Plan . Visualize the molecular geometry and the electron-domain geometries that could produce it. Confirm your choices with Tables 9.2 and 9.3. In Table 9.3, note that octahedral electron-domain geometry results in only 3 possible molecular geometries: octahedral, square pyramidal, and square planar (not T-shaped, bent, or linear). Solve.

    1. 2. Molecular geometry: linear. Possible electron-domain geometries: linear, trigonal bipyramidal
    2. 1. Molecular geometry, T-shaped. Possible electron-domain geometries: trigonal bipyramidal
    3. 1. Molecular geometry, octahedral. Possible electron-domain geometries: octahedral
    4. 1. Molecular geometry, square-pyramidal. Possible electron-domain geometries: octahedral
    5. 1. Molecular geometry, square planar. Possible electron-domain geometries: octahedral
    6. 1. Molecular geometry, triangular pyramid. Possible electron-domain geometries: trigonal bipyramidal. This is an unusual molecular geometry that is not listed in Table 9.3. It could occur if the equatorial substituents on the trigonal bipyramid were extremely bulky, causing the nonbonding electron pair to occupy an axial position.

    9.4

    1. 4 e domains
    2. The molecule has a nonzero dipole moment, because the C–H and C–F bond dipoles do not cancel each other.
    3. (ii)

    9.5

    1. Zero. Moving from left to right along the x-axis of the plot, the distance between the Cl atoms increases. At very large separation, the potential energy of interaction approaches zero.
    2. The Cl–Cl bond distance is approximately 2.0 Å. The Cl–Cl bond energy is approximately 240 kJ/mol. The minimum energy for the two atoms represents the stabilization obtained by bringing two Cl atoms together at the optimum (bond) distance. The x-coordinate of the minimum point on the plot is the Cl–Cl bond length; the y-coordinate is the bond strength or enthalpy.
    3. Weaker. Under extreme pressure, assume the Cl–Cl separation gets shorter. According to the plot, the potential energy of the atom pair increases and the bond gets weaker as the separation becomes shorter than the optimum bond distance.

    9.6

    1. (iii)
    2. sp 3

    9.7

    1. Recall that π bonds require p atomic orbitals, so the maximum hybridization of a C atom involved in a double bond is sp 2 and in a triple bond is sp. There are 6 C atoms in the molecule. Starting on the left, the hybridizations are: sp 2 , sp 2 , sp 3 , sp, sp, sp 3 .
    2. All single bonds are σ bonds. Double and triple bonds each contain 1 σ bond. This molecule has 8 C–H σ bonds and 5 C–C σ bonds, for a total of 13 σ bonds.
    3. Double bonds have 1 π bond and triple bonds have 2 π bonds. This molecule has a total of 3 π bonds.
    4. Any central atom with sp 2 hybridization will have bond angles of 120° around it. The two left-most C atoms are sp 2 hybridized, so any angle with one of these C atoms central will be 120°. This amounts to 1 H–C–H, 4 H–C–C and 1 C–C–C angle.

    9.8

    1. (i)
    2. (iii)

    9.9

    1. C 4 H 4 O
    2. 26 valence e
    3. sp 2
    4. 4 e
    5. (iii)

    9.10

    1. The lower-energy MO is σ 1s , the higher-energy MO is σ* 1s .
    2. H 2 +
    3. BO = ½
    4. σ 1s (the lowest energy available orbital)

    9.11 Analyze/Plan .  σ molecular orbitals (MOs) are symmetric about the internuclear axis, π MOs are not. Bonding MOs have most of their electron density in the area between the nuclei, antibonding MOs have a node between the nuclei.

      1. Two s atomic orbitals (electron density at each nucleus).
      2. Two p atomic orbitals overlapping end to end (node near each nucleus).
      3. Two p atomic orbitals overlapping side to side (node near each nucleus).
      1. σ-type (symmetric about the internuclear axis, s orbitals can produce only σ overlap).
      2. σ-type (symmetric about internuclear axis)
      3. π-type (not symmetric about internuclear axis, side-to-side overlap)

      1. antibonding (node between nuclei)
      2. bonding (concentration of electron density between nuclei)
      3. antibonding (node between nuclei)
      1. The nodal plane is between the atom centers, perpendicular to the interatomic axis and equidistant from each atom.
      2. There are two nodal planes; both are perpendicular to the interatomic axis. One is left of the left atom and the second is right of the right atom.
      3. There are two nodal planes; one is between the atom centers, perpendicular to the interatomic axis and equidistant from each atom. The second contains the interatomic axis and is perpendicular to the first.

    9.12

    1. The diagram has five electrons in MOs formed by 2p atomic orbitals. C has two 2p electrons, so X must have three 2p electrons. X is N.
    2. The molecule has an unpaired electron, so it is paramagnetic.
    3. Atom X is N, which is more electronegative than C. The atomic orbitals of the more electronegative N are slightly lower in energy than those of C. The lower- energy π 2p bonding molecular orbitals will have a greater contribution from the lower-energy N atomic orbitals. (Higher energy π 2p * MOs will have a greater contribution from higher-energy C atomic orbitals.)

    Molecular Shapes; the VSEPR Model (Sections 9.1 and 9.2)

    9.13

    1. It is not possible to tell the number of nonbonding electron pairs about the A atom from this information. If AB 2 obeys the octet rule, A would have 0 nonbonding pairs around it, as in CO 2 . If AB 2 does not obey the octet rule, there could be 0 or 3 nonbonding pairs around A. Examples are BeH 2 and XeF 2 .
    2. Three. XeF 2 has 11 electron pairs. Of these, 2 are bonding pairs between Xe and F and 6 are nonbonding pairs around the two F atoms. This leaves three nonbonding electron pairs around Xe.
    3. Yes. The electron domain geometry of XeF 2 is trigonal bipyramidal; the 3 nonbonding pairs are equatorial, the 2 bonding pairs are axial, and the molecular geometry is linear.

    9.14

    1. In a symmetrical tetrahedron, the four bond angles are equal to each other, with values of 109.5°. The H–C–H angles in CH 4 and the O–Cl–O angles in ClO 4 will have values close to 109.5°.
    2. ‘Planar’ molecules are flat, so trigonal planar BF 3 is flat. In the trigonal pyramidal NH 3 molecule, the central N atom sits out of the plane of the three H atoms; this molecule is not flat.

    9.15   A molecule with tetrahedral molecular geometry has an atom at each vertex of the tetrahedron. A trigonal pyramidal molecule has one vertex of the tetrahedron occupied by a nonbonding electron pair rather than an atom. That is, a trigonal pyramid is a tetrahedron with one vacant vertex.


    9.16

    1. three coplanar 120° angles
    2. four 109.5° angles
    3. 90° angles in the equatorial square plane and between axial atoms and those in the square plane, 12 in all; 180° angles between atoms opposite each other, 3 in all
    4. one 180° angle

    9.17

    1. Octahedral. There are 6 electron domains around A in an AB 6 molecule. Because none of the 6 electron domains are nonbonding, the electron domain geometry and molecular geometry are octahedral.
    2. Octahedral. An AB 4 molecule has 4 bonding electron domains around A. Additionally, this molecules has two nonbonding domains, for a total of 6 electron domains around A. A total of six electron pairs dictates octahedral electron domain geometry.
    3. Square planar. For an octahedral electron domain geometry that includes two nonbonding domains, the nonbonding domains are opposite each other to minimize repulsions. The four bonding domains occupy the remaining positions in the octahedron, forming a square plane.

    9.18   We expect the nonbonding electron domain in NH 3 to occupy a smaller volume than the one in PH 3 . The electronegativity of N, 3.0, is larger than that of P, 2.1. The nonbonding electrons will be more strongly attracted to N than to P, and the volume of the domain will be smaller. This means that the charge density of the nonbonding domain in NH 3 will be greater and it will experience stronger repulsions than the nonbonding domain in PH 3 .

    9.19 Analyze/Plan .  Draw the Lewis structure of each molecule and take note of nonbonding (lone) electron pairs about the central atom. Solve .

    1. SiH 4 , 8 valence e , 4 e pr, 0 nonbonding pairs, no effect on molecular shape
      image
    2. PF 3 , 26 valence e , 13 e pr, 1 nonbonding pair on P, influences molecular shape
      image
    3. HBr, 8 valence e , 4 e pr, 3 nonbonding pairs on Br, no effect on molecular shape because Br is not “central”
      image
    4. HCN, 10 valence e , 5 e pr, 0 nonbonding pairs on C, no effect on molecular shape
      image
    5. SO 2 , 18 valence e , 9 e pr, 1 nonbonding pair on S, influences molecular shape
      image

    9.20   Draw the Lewis structure of each molecule. If it has nonbonding electron pairs on the central atom, decide whether they will cause bond angles to deviate from ideal values for the particular electron-domain geometry.

    1. H 2 S, 8 valence e , 4 e pr, tetrahedral electron-domain geometry with 2 nonbonding electron pairs on S will cause the bond angle to deviate from ideal 109.5° angles
      image
    2. BCl 3 , 24 valence e , 12 e pr, trigonal planar electron-domain geometry with zero nonbonding pairs on B. We confidently predict 120° angles.
      image
    3. CH 3 I, 14 valence e , 7 e pr, tetrahedral electron-domain geometry with zero nonbonding pairs on C. Because the bonding electron domains are not exactly the same, we predict some deviation from ideal 109.5° angles.
      image
    4. CBr 4 , 32 valence e , 16 e pr, tetrahedral electron-domain geometry with zero nonbonding pairs on C. We confidently predict 109.5° angles.
      image
    5. TeBr 4 , 34 valence e , 17 e pr, trigonal bipyramidal electron-domain geometry with one nonbonding pair on Te. The structure is similar to SF 4 shown in Sample Exercise 9.2. The bond angles will deviate from ideal values, but perhaps not as much as in SF 4 .
      image

    9.21 Analyze/Plan .  Draw the Lewis structure of each molecule and count the number of nonbonding (lone) electron pairs. Note that the question asks ‘in the molecule’ rather than just around the central atom. Solve .

    1. (CH 3 ) 2 S, 20 valence e , 10 e pr, 2 nonbonding pairs
      image
    2. HCN, 10 valence e , 5 e pr, 1 nonbonding pair
      image

    1. H 2 C 2 , 10 valence e , 5 e pr, 0 nonbonding pairs
      image
    1. CH 3 F, 14 valence e , 7 e pr, 3 nonbonding pairs
      image

    9.22 Analyze/Plan .   See Table 9.1. Solve .

    1. trigonal planar
    2. tetrahedral
    3. trigonal bipyramidal
    4. octahedral

    9.23 Analyze/Plan . See Tables 9.2 and 9.3. Solve .

    1. image
    2. image
    3. image
    4. image

    9.24

    1. image
    2. image

    1. image

    9.25 Analyze/Plan .  Follow the logic in Sample Exercises 9.1 and 9.2 . Solve .

    bent (b), linear (l), octahedral (oh), seesaw (ss), square pyramidal (sp),

    square planar (spl), tetrahedral (td), trigonal bipyramidal (tbp), trigonal planar (tr), trigonal pyramidal (tp), T-shaped (T)

    image

    *More than one resonance structure is possible. All equivalent resonance structures predict the same molecular geometry.


    9.26   bent (b), linear (l), octahedral (oh), seesaw (ss), square pyramidal (sp), square planar (spl), tetrahedral (td), trigonal bipyramidal (tbp), trigonal planar (tr), trigonal pyramidal (tp), T-shaped (T). bent (b), linear (l), octahedral (oh), seesaw (ss) square pyramidal (sp), square planar (spl), tetrahedral (td), trigonal bipyramidal (tbp), trigonal planar (tr), trigonal pyramidal (tp), T-shaped (T)

    image
    image

    *More than one resonance structure is possible. All equivalent resonance structures predict the same molecular geometry.

    9.27 Analyze/Plan. Work backward from molecular geometry, using Tables 9.2 and 9.3. Solve .

    1. Electron-domain geometries: (i), trigonal planar; (ii), tetrahedral; (iii), trigonal bipyramidal
    2. nonbonding electron domains: (i), 0; (ii), 1; (iii), 2
    3. N and P. Shape (ii) has three bonding and one nonbonding electron domains. Li and Al would form ionic compounds with F, so there would be no nonbonding electron domains. Assuming that F always has three nonbonding domains, BF 3 and ClF 3 would have the wrong number of nonbonding domains to produce shape ii.
    4. Cl (also Br and I, because they have seven valence electrons). This T-shaped molecular geometry arises from a trigonal bipyramidal electron-domain geometry with two nonbonding domains (Table 9.3). Assuming each F atom has three nonbonding domains and forms only single bonds with A, A must have seven valence electrons to produce these electron-domain and molecular geometries. It must be in or below the third row of the periodic table, so that it can accommodate more than four electron domains.

    9.28

    1. Electron-domain geometries: (i), octahedral; (ii), tedrahedral; (iii), trigonal bipyramidal
    2. nonbonding electron domains: (i), 2; (ii), 0; (iii), 1
    3. S or Se. Shape (iii) has five electron domains, so A must be in or below the third row of the periodic table. This eliminates Be and C. Assuming each F atom has three nonbonding electron domains and forms only single bonds with A, A must have six valence electrons to produce these electron-domain and molecular geometries.
    4. Xe. (See Table 9.3.) Assuming F behaves typically, A must be in or below the third row and have eight valence electrons. Only Xe fits this description. (Noble- gas elements above Xe have not been shown to form molecules of the type AF 4 . See Section 7.8.)

    9.29 Analyze/Plan .  Follow the logic in Sample Exercise 9.3. Solve .

    1. 1, less than 109.5°; 2, less than 109.5°
    2. 3, different than 109.5°; 4, less than 109.5°
    3. 5 – 180°
    4. 6, slightly more than 120°; 7, less than 109.5°; 8, slightly different than 109.5°

    9.30

    1. 1, less than 109.5°; 2, less than 120°
    2. 3, close to 109.5°; 4, slightly greater than 120°
    3. 5, less than 109.5°; 6, less than 109.5°
    4. 7, 180°; 8, close to 109.5°

    9.31 Analyze/Plan . Draw correct Lewis structures for NH 2 , NH 3 , and NH 4 + . The more nonbonding electron domains (lone pairs) around N, the smaller the H–N–H bond angles. Solve .

    1. image
    2. NH 4 + . There are no lone pairs on N, so this ion has the largest bond angles.
    3. NH 2 . Amide ion has two bonding and two nonbonding domains around N. The two lone pairs compress the H–N–H bond angle to its smallest value.

    9.32 Analyze/Plan .  Given the formula of each molecule or ion, draw the correct Lewis structure and use principles of VSEPR to answer the question. Solve .

    1. image

    The three nonbonded electron pairs on each F atom have been omitted for clarity.

    The two molecules with trigonal bipyramidal electron-domain geometry, PF 5 and SF 4 , have more than one F–A–F bond angle.


    9.33 Analyze .  Given: molecular formulas. Find: explain features of molecular geometries.

    Plan .  Draw the correct Lewis structures for the molecules and use VSEPR to predict and explain observed molecular geometry. Solve .

    1. image

      The fundamental feature that determines molecular geometry is the number of electron domains around the central atom, and the number of these that are bonding domains. Although BrF 4 and BF 4 are both of the form AX 4 , the central atoms, and thus the number of valence electrons in the two ions, are different. This leads to different numbers of e domains about the two central atoms. Even though both ions have four bonding electron domains, the six total domains around Br require octahedral domain geometry and square planar molecular geometry, whereas the four total domains about B lead to tetrahedral domain and molecular geometry.

    2. H 2 X, 8 e , 4 e pr
      image

      All molecules in the series have tetrahedral electron-domain geometry and bent molecular structure. To a first approximation, the H–X–H angles will be less than 109.5°. Any variation will be because of differences in repulsion among the nonbonding and bonding electron domains. The less electronegative the central atom, the larger the nonbonding electron domain, the greater the effect of repulsive forces on adjacent bonding domains. The less electronegative the central atom, the larger the deviation from ideal tetrahedral angles. The angles will vary as H 2 O > H 2 S > H 2 Se.

    9.34

    1. ClO 2– , 20 e , 10 e pr
      image

      (More than one resonance structure is possible. All have 2 bonding and two nonbonding domains around Cl and predict bent shape.)


    1. SO 4 2– , 32 e , 16 e pr
      image

      (Many equivalent resonance structures for SO 4 2– are possible, but all have 4 bonding e domains around S and tetrahedral shape.)

    1. NF 3 , 26 e , 13 e pr
      image
    1. CCl 2 Br 2 , 32 e , 16 e pr
      image
    1. SF 4 2+ , 32 e , 16 e pr
      image

    Shapes and Polarity of Polyatomic Molecules (Section 9.3)

    9.35   A bond dipole is the asymmetric charge distribution between two bonded atoms with unequal electronegativities. A molecular dipole moment is the three-dimensional sum of all the bond dipoles in a molecule. (A molecular dipole moment is a measurable physical property; a bond dipole is not measurable, unless the molecule is diatomic.)

    9.36   For a polar A–X bond in an AX 3 molecule, as the X–A–X bond angle increases from 100° to 120°, the molecular dipole moment decreases. In a symmetrical AX 3 molecule with 120° bond angles, bond dipoles cancel and the molecule is nonpolar. As the bond angle decreases, the resultant of the three bond dipoles becomes larger, and the dipole moment increases.


    9.37 Analyze/plan . Follow the logic in Sample Exercise 9.4. Solve .

    1. SCl 2 , 20 e , 10 e pr
      image

      tetrahedral e domain geometry

      bent molecular geometry

      S and Cl have different electronegativities; the S−Cl bonds are polar. The bond dipoles are not opposite each other, so the molecule is polar. The dipole moment vector bisects the Cl−S−Cl bond angle. (A more difficult question is which end of the dipole moment vector is negative. The resultant of the two bond dipoles has its negative end toward the Cl atoms. However, the partial negative charge because of the lone pairs on S points opposite to the negative end of the resultant. A reasonable guess is that the negative end of the dipole moment vector is in the direction of the lone pairs.)

  • BeCl 2 , 16 e , 8 e pr
    image

    linear electron-domain and molecular geometry

    (Resonance structures with Be=Cl can be drawn, but electronegativity arguments predict that most electron density will reside on Cl and that the structure above is the main resonance contributor.) Be and Cl have very different electro-negativities, so the Be−Cl bonds are polar. The individual bond dipoles are equal and opposite, so the net molecular dipole moment is zero.

  • 9.38

    1. If PH 3 is polar, it must have a measurable dipole moment. This means that the three P−H bond dipoles do not cancel. If PH 3 were planar, the P−H bond dipoles would cancel, and the molecule would be nonpolar. The measurable dipole moment of PH 3 is experimental evidence that the molecule cannot be planar.
    2. image

      trigonal planar e domain geometry

      bent molecular geometry

      Because all atoms are the same, the individual bond dipoles are zero. However, the central O atom has a lone pair of electrons that cause an unequal electron (and charge) distribution in the molecule. This lone pair is the source of the dipole moment in O 3 .

    9.39

    1. Nonpolar. The BF 3 molecule has polar B–F bonds, but they are arranged in a symmetrical trigonal plane. The individual bond dipoles cancel, leaving the molecule with zero net dipole moment.
    2. No. The BF 3 2– ion has 3 bonding and 1 nonbonding electron pairs around B. The added nonbonding electron pair requires that the electron domain geometry is tetrahedral and the shape is a trigonal pyramid.
    3. Yes. In BF 2 Cl the B–F bond dipoles do not exactly cancel with the B–Cl bond dipole, resulting in a net dipole moment.

    9.40

    1. In Exercise 9.27, molecules (ii) and (iii) will have nonzero dipole moments. Molecule (i) has zero nonbonding electron pairs on A, and the 3 A–F dipoles are oriented so that the sum of their vectors is zero (the bond dipoles cancel). Molecules (ii) and (iii) have nonbonding electron pairs on A and their bond dipoles do not cancel. A nonbonding electron pair (or pairs) on a central atom almost guarantees at least a small molecular dipole moment, because no bond dipole exactly cancels a nonbonding pair. (Exceptions are molecular geometries with nonbonding electron domains 180 o apart.)
    2. In Exercise 9.28, molecules (i) and (ii) have zero dipole moments and are nonpolar. AF 4 molecules will have a zero dipole moment if the 4 A–F bond dipoles are arranged (symmetrically) so that they cancel, and any nonbonding pairs are arranged so that they cancel.

    9.41 Analyze/Plan .  Given molecular formulas, draw correct Lewis structures, determine molecular structure and polarity. Solve .

    1. Polar, ΔEN > 0

      I–F

    2. Nonpolar, the molecule is linear and the bond dipoles cancel.
      image
    3. Nonpolar, in a symmetrical trigonal planar structure, the bond dipoles cancel.
      image
    4. Polar, although the bond dipoles are essentially zero, there is an unequal charge distribution because of the nonbonded electron pair on P.
      image
    5. Nonpolar, symmetrical octahedron
      image
    6. Polar, square pyramidal molecular geometry, bond dipoles do not cancel.
      image

    9.42

    1. Nonpolar, in a symmetrical tetrahedral structure (Figure 9.1) the bond dipoles cancel.
      image

    1. Polar, there is an unequal charge distribution because of the nonbonded electron pair on N.
      image
    1. Polar, there is an unequal charge distribution because of the nonbonded electron pair on S.
      image
    1. Nonpolar, the bond dipoles and the nonbonded electron pairs cancel.
      image
    1. Polar, the C–H and C–Br bond dipoles are not equal and do not cancel.
      image
    1. Nonpolar, in a symmetrical trigonal planar structure, the bond dipoles cancel.
      image

    9.43 Analyze/Plan .  Given molecular formulas, draw correct Lewis structures, analyze molecular structure and determine polarity. Solve .

    1. C 2 H 2 Cl 2 , each isomer has 24 e , 12 e pr. Lewis structures:
      image

      Molecular geometries:

      image
    2. All three isomers are planar. The molecules on the left and right are polar because the C–Cl bond dipoles do not point in opposite directions. In the middle isomer, the C–Cl bonds and dipoles are pointing in opposite directions (as are the C–H bonds), the molecule is nonpolar and has a measured dipole moment of zero.

    1. C 2 H 3 Cl (lone pairs on Cl omitted for clarity)

      There are four possible placements for Cl:

      image

      By rotating each of these structures in various directions, it becomes clear that the four structures are equivalent; C 2 H 3 Cl has only one isomer. Because C 2 H 3 Cl has only one C−Cl bond, the bond dipoles do not cancel, and the molecule has a dipole moment.

    9.44   Each C–Cl bond is polar. The question is whether the vector sum of the C–Cl bond dipoles in each molecule will be nonzero. In the ortho and meta isomers, the C–Cl vectors are at 60° and 120° angles, respectively, and their resultant dipole moments are nonzero. In the para isomer, the C–Cl vectors are opposite, at an angle of 180°, with a resultant dipole moment of zero. The ortho and meta isomers are polar, the para isomer is nonpolar.

    Orbital Overlap; Hybrid Orbitals (Sections 9.4 and 9.5)

    9.45

    1. True.
    2. False. Examples of bonds that could involve an s orbital on one atom and a p orbital on another are H–F, H–Cl, etc.
    3. True. See Tables 9.2 and 9.3.
    4. False. A 1s orbital is shaped like a sphere; there are no nodal planes.
    5. True. All p orbitals have a nodal plane.

    9.46

    1. image
    2. image
    3. image

    9.47

    1. False. The more orbital overlap in a bond, the stronger the bond.
    2. True.
    3. False. Hybrid orbitals are combinations of atomic orbitals on the same atom.
    4. False. Nonbonding electron pairs (lone pairs) can occupy hybrid orbitals.

    9.48   By analogy to the H 2 molecule shown in Figure 9.13, as the distance between the atoms decreases, the overlap between their bonding orbitals increases. According to Figure 7.7, the bonding atomic radius for the halogens is in the order F < Cl < Br < I. The order of bond lengths in the molecules is I–F < I–Cl < I–Br < I–I. If the extent of orbital overlap increases as the distance between atoms decreases, I–F has the greatest overlap and I 2 the least. The order for extent of orbital overlap is I–I < I–Br < I–Cl < I–F.

    9.49

    1. B: [He]2s 2 2p 1
    2. F, [He]2s 2 2p 5
    3. BF 3 , 24 e , 12 e pairs
      image

      sp 2 . The three electron domains around B require sp 2 hybrid orbitals.

    4. A single 2p orbital is unhybridized. It lies perpendicular to the trigonal plane of the sp 2 hybrid orbitals.

    9.50

    1. S: [Ne]3s 2 3p 4
    2. Cl, [Ne]3s 2 2p 5
    3. SCl 2 , 20 e , 10 e pr
      image

      sp 3 . The four electron domains (two bonding and two nonbonding) around S require sp 3 hybrid orbitals.

    4. No valence atomic orbitals on S remain unhybridized. The sp 3 hybrids use all the 3s and 3p valence orbitals. There are 3d orbitals on S, but these are not considered valence orbitals.

    9.51 Analyze/Plan .  Given the molecular (or ionic) formula, draw the correct Lewis structure and determine the electron-domain geometry, which determines hybridization. Solve .

    1. 24 e , 12 e pairs
      image

      3 e pairs around B, trigonal planar e domain geometry, sp 2 hybridization

    2. 32 e , 16 e pairs
      image

      4 e domains around Al, tetrahedral e domain geometry, sp 3 hybridization


    1. 16 e , 8 e pairs
      image

      2 e domains around C, linear e domain geometry, sp hybridization

    1. 8 e , 4 e pairs
      image

      4 e pairs around Ge, tetrahedral e domain geometry, sp 3 hybridization

    9.52

    1. 32 e , 16 e pairs
      image

      4 e pairs around Si, tetrahedral e domain geometry, sp 3 hybridization

    2. 10 e , 5 e pairs
      image

      2 e domains around C, linear e domain geometry, sp hybridization

    3. 24 e , 12 e pairs
      image

      (other resonance structures are possible)

      3 e domains around S, trigonal planar e domain geometry, sp 2 hybridization

    4. 20 e , 10 e pairs
      image

      4 e domains around Te, tetrahedral e domain geometry, sp 3 hybridization

    9.53

    Left: No hybrid orbitals discussed in this chapter have angles of 90°; p atomic orbitals are perpendicular to each other.
    Center: Angles of 109.5° are characteristic of sp 3 hybrid orbitals.
    Right: Angles of 120° can be formed by sp 2 hybrids.

    9.54

    1. The three moieties, BH 4 , CH 4 , and NH 4 + , each have 8 valence e , 4 e pairs, 4 bonding e domains, tetrahedral e domain and molecular geometry, and sp 3 hybridization at the central atom.
    2. The electronegativity of the central atoms decreases in the series N > C > B. The question is: where does the electronegativity of H lie in this series? By examination of electronegativity values in Figure 8.8, H is slightly less electronegative than C, and almost the same as B. The magnitude of the bond dipole decreases in the series N−H > C−H > B−H. The negative end of the dipole is toward N, C, and H, respectively248.

    1. AlH 4 , SiH 4 , and PH 4 + . By the same arguments used in part (a), we expect these three moieties to have the same tetrahedral e domain and molecular geometry and sp 3 hybridization at the central atom as the species in part (a).

    Multiple Bonds (Section 9.6)

    9.55

    1. image
    2. image
    3. A σ bond is generally stronger than a π bond, because there is more extensive orbital overlap.
    4. Two s orbitals cannot form a π bond. A π bond has no electron density along the internuclear axis. Overlap of s orbitals results in electron density along the internuclear axis. (Another way to say this is that s orbitals have the wrong symmetry to form a π bond.)

    9.56

    1. Two unhybridized p orbitals remain, and the atom can form two pi bonds.
    2. It would be much easier to twist or rotate around a single sigma bond. Sigma bonds are formed by end-to-end overlap of orbitals and the bonding electron density is symmetric about the internuclear axis. Rotating (twisting) around a sigma bond can be done without disrupting either the orbital overlap or bonding electron density, without breaking the bond.

      The π part of a double bond is formed by side-to-side overlap of p atomic orbitals perpendicular to the internuclear axis. This π overlap locks the atoms into position and makes twisting difficult. Also, only a small twist (rotation) destroys overlap of the p orbitals and breaks the π bond.

    9.57 Analyze/Plan . Draw the correct Lewis structures, count electron domains and decide hybridization. Molecules with π bonds that require all bonded atoms to be in the same plane are planar. For bond-type counting, single bonds are σ bonds, double bonds consist of one σ and one π bond, triple bonds consist of one σ and two π bonds. Solve .

    1. image
    2. sp 3 sp 2 sp
    3. nonplanar planar planar
    4. 7 σ, 0 π 5 σ, 1 π 3 σ, 2 π

    9.58

    1. image
    2. The N atoms in N 2 H 4 are sp 3 hybridized; there are no unhybridized p orbitals available for π bonding. In N 2 , the N atoms are sp hybridized, with two unhybridized p orbitals on each N atom available to form the two π bonds in the N≡N triple bond.
    3. The N−N triple bond in N 2 is significantly stronger than the N−N single bond in N 2 H 4 , because it consists of one σ and two π bonds, rather than a ‘plain’ sigma bond. Generally, bond strength increases as the extent of orbital overlap increases. The additional overlap from the two π bonds adds to the strength of the N−N bond in N 2 .

    9.59 Analyze/Plan .  Single bonds are σ bonds, double bonds consist of 1 σ and 1 π bond. Each bond is formed by a pair of valence electrons. Solve .

    1. C 3 H 6 has 3(4) + 6(1) = 18 valence electrons
    2. 8 pairs or 16 total valence electrons form σ bonds
    3. 1 pair or 2 total valence electrons form π bonds
    4. no valence electrons are nonbonding
    5. The left and central C atoms are sp 2 hybridized; the right C atom is sp 3 hybridized.

    9.60

    1. The C with a double bond to O has three electron domains and is sp 2 hybridized; the other three C atoms are sp 3 hybridized.
    2. C 4 H 8 O 2 has 4(4) + 8(1) + 2(6) = 36 valence electrons.
    3. 13 pairs or 26 total valence electrons form σ bonds
    4. 1 pair or 2 total valence electrons form π bonds
    5. 4 pairs or 8 total valence electrons are nonbonding

    9.61 Analyze/Plan .  Given the correct Lewis structure, analyze the electron domain geometry at each central atom. This determines the hybridization and bond angles at that atom.

    Solve .

    1. ~109° bond angles about the left most C, sp 3 ; ~120° bond angles about the right-hand C, sp 2
    2. The doubly bonded O can be viewed as sp 2 , the other as sp 3 ; the nitrogen is sp 3 with approximately less than 109.5° bond angles.
    3. 9 σ bonds, 1 π bond

    9.62

    1. 1, ~120°; 2, ~120°; 3, less than 109.5°
    2. 1, sp 2 ; 2, sp 2 ; 3, sp 3
    3. 21 σ bonds

    9.63

    1. In a localized π bond, the electron density is concentrated strictly between the two atoms forming the bond. In a delocalized π bond, parallel p orbitals on more than two adjacent atoms overlap and the electron density is spread over all the atoms that contribute p orbitals to the network. There are still two regions of overlap, above and below the σ framework of the molecule.
    2. The existence of more than one resonance form is a good indication that a molecule will have delocalized π bonding.
    3. image

      The existence of more than one resonance form for NO 2 indicates that the π bond is delocalized. From an orbital perspective, the electron-domain geometry around N is trigonal planar, so the hybridization at N is sp 2 . This leaves a p orbital on N and one on each O atom perpendicular to the trigonal plane of the molecule, in the correct orientation for delocalized π overlap. Physically, the two N–O bond lengths are equal, indicating that the two N–O bonds are equivalent, rather than one longer single bond and one shorter double bond.

    9.64   (a,b) 24 e , 12 e pairs

    1. image

      3 electron domains around S, trigonal planar electron-domain geometry,

      sp 2 hybrid orbitals

    1. The multiple resonance structures indicate delocalized π bonding. All four atoms lie in the trigonal plane of the sp 2 hybrid orbitals. On each atom there is a

      p atomic orbital perpendicular to this plane in the correct orientation for π overlap. The resulting delocalized π electron cloud is Y-shaped (the shape of the molecule) and has electron density above and below the plane of the molecule.

    9.65 Analzye/Plan. Follow the logic in Sample Exercise 9.7.

    1. 18 e , 9 e pairs
      image
    2. sp 2
    3. Yes, there is one other resonance structure.
      image
    4. There are four electrons in the π system of the molecule. If the C and both O atoms are sp 2 hybridized, there are three bonding electron pairs and four nonbonding electron pairs in the σ system. This leaves two electron pairs or four electrons in the π system.

    9.66

    1. The Lewis structure depicts an anion with a 1– charge. The chemical formula of the given structure is C 3 H 3 O 2 . This grouping of atoms has 27 valence electrons, whereas the structure shown has 14 electron pairs or 28 electrons. This means that the structure is an anion with a 1– charge.
    2. sp 2
    3. Yes, there is one other resonance structure.
      image
    4. There are six electrons in the π system of the molecule. If all the C and O atoms are sp 2 hybridized, there are seven bonding electron pairs and four nonbonding electron pairs in the σ system. This leaves three electron pairs or six electrons in the π system.

    9.67 Analyze/Plan .  Count valence e and e pairs in each molecule. Complete the Lewis structure by placing nonbonding electron pairs. Analyze the electron-domain geometry at each central atom; visualize and describe the molecular structure. Solve .

    1. 26 e , 13 e pairs
      image

      The molecule is linear. Each C atom has 2 bonding e domains, linear geometry, and sp hybridization. This requires that all atoms not only lie in the same plane, but in a line.

    2. 34 e , 17 e pairs
      image

      The two central C atoms each have 3 bonding e domains, trigonal planar geometry, and sp 2 hybridization. Each O−C−O group is planar, whereas the terminal H atoms can rotate out of these planes. In principle, there is free rotation about the C−C σ bond, but delocalization of the π electrons is possible if the two planes are coincident. It is possible to put all 8 atoms in the same plane.

    3. 12 e , 6 e pairs
      image

      The molecule is planar. Each N atom has 3 bonding e domains, trigonal planar geometry, and sp 2 hybridization. Because the N atoms share a π bond, the planes must be coincident and all 4 atoms are required to lie in this plane. [The structure shown here has H atoms on the same side of the double bond. There is another isomer that has H atoms on opposite sides of the double bond. Both compounds are planar with 120 degree bond angles.]


    9.68

    1. 24 e , 12 e pairs
      image

      The designated C atom has 3 bonding e domains and sp 2 hybridization.

    2. 8 e , 4 e pairs
      image

      The P atom has 4 bonding e domains and sp 3 hybridization

    3. 24 e , 12 e pairs
      image

      The Al atom has 3 bonding e domains and sp 2 hybridization

    4. 16 e , 8 e pairs
      image

      The designated C atom has 3 bonding e domains and sp 2 hybridization.

    Molecular Orbitals and Period 2 Diatomic Molecules (Sections 9.7 and 9.8)

    9.69

    1. Hybrid orbitals are mixtures (linear combinations) of atomic orbitals from a single atom; the hybrid orbitals remain localized on that atom. Molecular orbitals are combinations of atomic orbitals from two or more atoms. They are associated with the entire molecule, not a single atom.
    2. Each MO, like each AO or hybrid, can hold a maximum of two electrons.
    3. Yes, antibonding MOs can have electrons in them.

    9.70

    1. An MO, because the AOs come from two different atoms.
    2. A hybrid orbital, because the AOs are on the same atom.
    3. Yes. The Pauli exclusion principle, that no two electrons can have the same four quantum numbers, means that an orbital can hold at most two electrons. (Because n , l , and m l are the same for a particular orbital and m s has only two possible values, an orbital can hold at most two electrons). This is true for atomic and molecular orbitals.

    9.71

    1. image
    2. There is one electron in H 2 + .
    3. image
    4. Bond order = 1/2 (1 − 0) = ½
    5. Fall apart. The stability of H 2 + is because of the lower-energy state of the σ bonding molecular orbital relative to the energy of a H 1s atomic orbital. If the single electron in H 2 + is excited to the σ* 1s orbital, its energy is higher than the energy of an H 1s atomic orbital and H 2 + will decompose into a hydrogen atom and a hydrogen ion.

      H 2 h ν H + H + .

    6. Statement (i) is true.

    9.72

    1. image
    2. image
    3. Bond order = 1/2 (2 − 1) = ½
    4. If one electron moves from σ 1s to σ* 1s , the bond order becomes –½. There is a net increase in energy relative to isolated H atoms, so the ion will decompose.

      H 2 h ν H + H .

    5. Statement (i) is true.

    9.73

    image
    1. One. With three mutually perpendicular p orbitals on each atom, only one set can be oriented for end-to-end sigma overlap.
    2. Two. The 2 p orbitals on each atom not involved in σ bonding can be aligned for side-to-side π overlap.
    3. Three, 1 σ* and 2 π*. There are a total of 6 p orbitals on the two atoms. When combining AOs to form MOs, total number of orbitals is conserved. If 3 of the 6 MOs are bonding MOs, as described in (a) and (b), then the remaining 3 MOs must be antibonding. They will have the same symmetry as the bonding MOs, σ* and 2 π*.

    9.74

    1. True.
    2. False. Pi star molecular orbitals have a nodal plane through the nuclei. This means that there is zero probability of finding an electron in a pi star orbital at the nucleus.
    3. True.
    4. False. Electrons can and do occupy antibonding molecular orbitals.

    9.75

    1. When comparing the same two bonded atoms, the greater the bond order, the shorter the bond length and the greater the bond energy. That is, bond order and bond energy are directly related, whereas bond order and bond length are inversely related. When comparing different bonded nuclei, there are no simple relationships (see Solution 8.100).
    2. image

      Be 2 has a bond order of zero and is not energetically favored over isolated Be atoms; it is not expected to exist. Be 2 + has a bond order of 0.5 and is slightly lower in energy than isolated Be atoms. It will probably exist under special experimental conditions, but be unstable.

    9.76

    1. O 2 2– has a bond order of 1.0, whereas O 2 has a bond order of 1.5. For the same bonded atoms, the greater the bond order the shorter the bond, so O 2 has the shorter bond.

    1. The two possible orbital energy level diagrams are:
      image

      The magnetic properties of a molecule reveal whether it has unpaired electrons. If the σ 2p MOs are lower in energy, B 2 has no unpaired electrons. If the π 2p MOs are lower in energy than the σ 2p MO, there are two unpaired electrons. The magnetic properties of B 2 must indicate that it has unpaired electrons.

    1. According to Figure 9.43, the two highest-energy electrons of O 2 are in antibonding π 2 p * MOs and O 2 has a bond order of 2.0. Removing these two electrons to form O 2 2+ produces an ion with bond order 3.0. O 2 2+ has a stronger O−O bond than O 2 , because O 2 2+ has a greater bond order.

    9.77   (a,b)  Substances with no unpaired electrons are weakly repelled by a magnetic field. This property is called diamagnetism .

    1. O 2 2– , Be 2 2+ [see Figure 9.43 and Solution 9.75(b)]

    9.78

    1. Substances with unpaired electrons are attracted into a magnetic field. This property is called paramagnetism .
    2. Weigh the substance normally and in a magnetic field, as shown in Figure 9.44. Paramagnetic substances appear to have a larger mass when weighed in a magnetic field.
    3. See Figures 9.35 and 9.43. O 2 + , one unpaired electron; N 2 2– , two unpaired electrons; Li 2 + , one unpaired electron

    9.79

    1. image
      B 2 + increase
    2. image
      Li 2 + increase
    3. image
      N 2 + increase
    4. image
      Ne 2 2+ decrease

    Addition of an electron increases bond order if it occupies a bonding MO and decreases stability if it occupies an antibonding MO.

    9.80   Determine the number of “valence” (noncore) electrons in each molecule or ion. Use the homonuclear diatomic MO diagram from Figure 9.43 (shown below) to calculate bond order and magnetic properties of each species. The electronegativity difference between heteroatomics increases the energy difference between the 2s AO on one atom and the 2p AO on the other, rendering the “no interaction” MO diagram in Figure 9.43 appropriate.

    1. image
    2. CO + : 9 e , B.O. = (7 – 2) / 2 = 2.5, paramagnetic
    3. NO : 12 e , B.O. = (8 – 4) / 2 = 2.0, paramagnetic
    4. OF + : 12 e , B.O. = (8 – 4) / 2 = 2.0, paramagnetic
    5. NeF + : 14 e , B.O. = (8 – 6) / 2 = 1.0, diamagnetic

    9.81 Analyze/Plan .  Determine the number of “valence” (non-core) electrons in each molecule or ion. Use the homonuclear diatomic MO diagram from Figure 9.43 (shown below) to calculate bond order and magnetic properties of each species. The electronegativity difference between heteroatomics increases the energy difference between the 2s AO on one atom and the 2p AO on the other, rendering the “no interaction” MO diagram in Figure 9.43 appropriate. Solve .

    image

    CN: 9 e , B.O. = (7 – 2) / 2 = 2.5

    CN + : 8 e , B.O. = (6 – 2) / 2 = 2.0

    CN : 10 e , B.O. = (8 – 2) / 2 = 3.0

    1. CN– has the highest bond order and therefore the strongest C−N bond.
    2. CN and CN + . CN has an odd number of valence electrons, so it must have an unpaired electron. The electron configuration for CN is shown in the diagram. Removing one electron from the π 2p MOs to form CN + produces an ion with two unpaired electrons. Adding one electron to the π 2p MOs of CN to form CN produces an ion with all electrons paired.

    9.82

    1. Statement (ii) is the best explanation. The bond order of NO is [1/2 (8 – 3)] = 2.5. The electron that is lost is in an antibonding molecular orbital, so the bond order in NO + is 3.0. The increase in bond order is the driving force for the formation of NO + .
    2. To form NO , an electron is added to an antibonding orbital, and the new bond order is [1/2 (8 – 4)] = 2. The order of increasing bond order and bond strength is: NO < NO < NO + . NO and NO are paramagnetic with two and one unpaired electrons, respectively. NO + is diamagnetic.
    3. NO + is isoelectronic with N 2 , and NO is isoelectronic with O 2 .

    9.83

    1. 3s, 3p x , 3p y , 3p z
    2. π 3p
    3. Two. Note that there are two degenerate π 3p bonding molecular orbitals; each holds two electrons. A total of 4 electrons can be designated as π 3p , but no single molecular orbital can hold more than two electrons.
    4. If the MO diagram for P 2 is similar to that of N 2 , P 2 will have no unpaired electrons and be diamagnetic.

    9.84

    1. I: 5s, 5p x , 5p y , 5p z ; Br: 4s, 4p x , 4p y , 4p z
    2. By analogy to F 2 , the BO of IBr will be 1.
    3. I and Br have valence atomic orbitals with different principal quantum numbers. This means that the radial extensions (sizes) of the valence atomic orbital that contribute to the MO are different. The n = 5 valence AOs on I are larger than the n = 4 valence AOs on Br.
    4. σ np *
    5. None

    Additional Exercises

    9.85

    1. The physical basis of VSEPR is the electrostatic repulsion of like-charged particles, in this case groups or domains of electrons. That is, owing to electrostatic repulsion, electron domains will arrange themselves to be as far apart as possible.
    2. The σ-bond electrons are localized in the region along the internuclear axes. The positions of the atoms and geometry of the molecule are thus closely tied to the locations of these electron pairs. Because the π-bond electrons are distributed above and below the plane that contains the σ bonds, these electron pairs do not, in effect, influence the geometry of the molecule. Thus, all σ- and π-bond electrons localized between two atoms are located in the same electron domain.

    9.86

    1. Two. If the electron-domain geometry is trigonal bipyramidal, there are five total electron domains around the central atom. An AB 3 molecule has three bonding domains, so there must be two nonbonding domains on A.
    2. (iii) (See Table 9.3.)

    9.87

    image
    1. Three. The P, Br, and Cl central atoms have more than an octet of electrons about them.
    2. One, AlF 4
    3. BrF 4
    4. PF 4 and ClF 4 +

    9.88

    1. 40 e , 20 e pairs
      image

      5 e domains

      trigonal-bipyramidal electron-domain geometry

    2. The greater the electronegativity of the terminal atom, the larger the negative charge centered on the atom, the smaller the effective size of the P–X bonding electron domain. A P–F bond will produce a smaller (and shorter) electron domain than a P–Cl bond.
    3. The molecular geometry (shape) is also trigonal bipyramidal, because all five electron domains are bonding domains. Because we predicted the P–F electron domain to be smaller, the larger P–Cl bonding domain will occupy the equatorial plane of the molecule, minimizing the number of 90° P–Cl to P–F repulsions. This is the same argument that places a “larger” nonbonding domain in the equatorial position of a molecule like SF 4 .
    4. The molecular geometry is distorted from a perfect trigonal bipyramid because not all electron domains are alike. The 90° P–Cl to P–F repulsions will be greater than the 90° P–F to P–F repulsions, so the F(axial)–P–Cl angles will be greater than 90°. The equatorial F–P–F angles may distort slightly to “make room” for the axial F atoms that are “pushed away” from the equatorial Cl atom.

    9.89   For any triangle, the law of cosines gives the length of side c as c 2 = a 2 + b 2 – 2ab cosθ.

    Let the edge length of the cube (uy = vy = vz) = X

    The length of the face diagonal (uv) is

    (uv) 2 = (uy) 2 + (vy) 2 – 2(uy)(vy) cos 90

    (uv) 2 = X 2 + X 2 – 2(X)(X) cos 90

    ( uv) 2 = 2 X 2 ; uv = 2 X

    The length of the body diagonal (uz) is

    (uz) 2 = (vz 2 ) + (uv) 2 – 2(vz)(uv) cos 90

    ( uz) 2 = X 2 + ( 2 X) 2 2 ( X) ( 2 X) cos 90

    ( uz) 2 = 3 X 2 ; uz = 3 X

    image

    For calculating the characteristic tetrahedral angle, the appropriate triangle has vertices u, v, and w. Theta, θ, is the angle formed by sides wu and wv and the hypotenuse is side uv.

    wu = wv = uz/2 = 3 / 2 X; uv = 2 X

    ( 2 X) 2 = ( 3 / 2 X) 2 + ( 3 / 2 X) 2 2 ( 3 / 2 X) ( 3 / 2 ) cos θ

    2X 2 = 3/4 X 2 + 3/4 X 2 – 3/2 X 2 cos θ

    2X 2 = 3/2 X 2 – 3/2 X 2 cos θ

    1/2 X 2 = –3/2 X 2 cos θ

    cos θ = –(1/2 X 2 ) / (3/2 X 2 ) = –1/3 = –0.3333

    θ = 109.47°

    9.90 Analyze/Plan. For entries where the molecule is listed, follow the logic in Sample Exercises 9.4 and 9.5. For entries where no molecule is listed, decide electron-domain geometry from hybridization (or vice versa). If the molecule is nonpolar, the terminal atoms will be identical. If the molecule is polar, the terminal atoms will be different, or the central atom will have one or more lone pairs, or both. Solve .

    Molecule Molecular Structures Electron Domain Geometry Hybrdization of CentralAtom Dipole Moment Yes or No
    CO 2
    image
    linear sp no
    NH 3
    image
    tetrahedral sp 3 yes
    CH 4
    image
    tetrahedral sp 3 no
    BH 3
    image
    trigonal planar sp 2 no
    SF 4
    image
    trigonal bipyramidal not applicable yes
    SF 6
    image
    octahedral not applicable no
    H 2 CO
    image
    trigonal planar sp 2 yes
    PF 5
    image
    trigonal bipyramidal not applicable no
    XeF 2
    image
    trigonal bipyramidal not applicable no

    9.91

    1. CO 2 , 16 valence e
      image
    2. (CN) 2 , 18 valence e
      image
    3. H 2 CO, 12 valence e
      image
    4. HCOOH, 18 valence e
      image

    9.92

    1. image

      3(4) + 3(6) + 6(1) = 36 e , 18 e pr

    2. There are 11 σ and 1 π bonds.
    3. The C=O on the right-hand C atom is shortest. For the same bonded atoms, in this case C and O, the greater the bond order, the shorter the bond.
    4. The right-most C has three e domains, so the hybridization is sp 2 .
    5. Bond angles about the right-most C atom are approximately 120°. The middle and left-hand C atoms both have four bonding e domains and are sp 3 hybridized. Because the bonding domains about each C atom are not exactly the same, the bond angles will deviate somewhat from 109.5°; we expect larger deviations for the angles around the middle C atom.

    9.93

    1. Square pyramidal
    2. Yes, there is one nonbonding electron domain on A. If there were only five bonding domains, the shape would be trigonal bipyramidal. With five bonding and one nonbonding electron domains, the molecule has octahedral domain geometry.
    3. (iii). If the B atoms are halogens, each will have three nonbonding electron pairs; there are five bonding pairs, and A has one nonbonded pair, for a total of [5(3) + 5 + 1] = 21 e pairs and 42 electrons in the Lewis structure. If the five halogens contribute 35 e , A must contribute seven valence electrons. A is also a halogen.

    9.94

    1. Cisplatin has a dipole moment. In the square planar trans structure, all equivalent bond dipoles can be oriented opposite each other, for a net dipole moment of zero.
    2. Cisplatin is the cancer drug. The Cl atoms in cisplatin occupy bonding sites that are in the correct orientation to bind adjacent N atoms in DNA. That is, the Cl–Pt–Cl angle in cisplatin is about 90°, and the N–Pt–N angle in the new DNA complex is also about 90°.
      image

    9.95

    image
    1. The bond dipoles in H 2 O lie along the O–H bonds with the positive end at H and the negative end at O. The dipole moment vector of the H 2 O molecule is the resultant (vector sum) of the two bond dipoles. This vector bisects the H–O–H angle and has a magnitude of 1.85 D with the negative end pointing toward O.
    2. Because the dipole moment vector bisects the H–O–H bond angle, the angle between one H–O bond and the dipole moment vector is 1/2 the H–O–H bond angle, 52.25°. Dropping a perpendicular line from H to the dipole moment vector creates the right triangle pictured. If x = the magnitude of the O–H bond dipole, x cos (52.25) = 0.925 D. x = 1.51 D.

    1. The X–H bond dipoles (Table 8.2) and the electronegativity values of X (Figure 8.8) are
      Electronegativity Bond dipole
      F 4.0 1.82
      O 3.5 1.51
      Cl 3.0 1.08

      Because the electronegativity of O is midway between the values for F and Cl, the O–H bond dipole should be approximately midway between the bond dipoles of HF and HCl. The value of the O–H bond dipole calculated in part (b) is consistent with this prediction.

    9.96

    1. XeF 2 , 22 e , 11 e pr
      image

      5 electron domains around Xe, trigonal bipyramidal electron domain geometry

      XeF 4 , 36 e , 18 e pr

      image

      6 electron domains around Xe, octahedral electron domain geometry

      XeF 6 50 e , 25 e pairs

      image

      7 electron domains around Xe, pentagonal bipyramidal electron domain geometry

    2. IF 6 , 50 e , 25 e pairs

      IF 6 is isoelectronic with XeF 6 . The Lewis structure is similar to the one for XeF 6 , but with I as the central atom. The electron domain geometry is pentagonal bipyramidal. With respect to molecular geometry, the question is whether the nonbonding domain occupies an axial or equatorial position. The equatorial plane of a pentagonal bipyramid has F–I–F angles of 72°. Placing the nonbonded domain in the equatorial plane would create severe repulsions between it and the adjacent bonded domains. Thus, the nonbonded domain will reside in the axial position. The molecular geometry is a pentagonal pyramid.


    9.97   Statements (ii) and (iii) are true.

    9.98

    image
    1. The molecule is not planar. The CH 2 planes at each end are twisted 90° from one another.
    2. Allene has no dipole moment.
    3. The bonding in allene would not be described as delocalized. The π electron clouds of the two adjacent C=C are mutually perpendicular. The mechanism for delocalization of π electrons is mutual overlap of parallel p atomic orbitals on adjacent atoms. If adjacent π electron clouds are mutually perpendicular, there is no overlap and no delocalization of π electrons.

    9.99

    1. 9σ, 3π
      1. 2, the second and third C atoms from the left
      2. 2, the rightmost C atom and the N atom
      3. 1, the leftmost C atom

    9.100

    1. 16 e , 8 e pairs
      image

      There are 2 electron domains around the central nitrogen atom. The N–N–N angle is 180° and the ion is linear.

    2. The central N atom has 2 electron domains and sp hydridization.
    3. The central N atom forms 2 sigma and 2 pi bonds.

    9.101

    1. image

      To accommodate the π bonding by all 3 O atoms indicated in the resonance structures above, all O atoms are sp 2 hybridized.

    2. For the first resonance structure, both sigma bonds are formed by overlap of sp 2 hybrid orbitals, the π bond is formed by overlap of atomic p orbitals, one of the nonbonded pairs on the right terminal O atom is in a p atomic orbital, and the remaining five nonbonded pairs are in sp 2 hybrid orbitals.
    3. Only unhybridized p atomic orbitals can be used to form a delocalized π system.
    4. The unhybridized p orbital on each O atom is used to form the delocalized π system, and in both resonance structures one nonbonded electron pair resides in a p atomic orbital. The delocalized π system then contains four electrons, two from the π bond and two from the nonbonded pair in the p orbital.

    9.102

    1. Each C atom is surrounded by three electron domains (two single bonds and one double bond), so bond angles at each C atom will be approximately 120°.
      image

      Because rotation around the central C–C single bond is possible, other conformations can be drawn.

    2. Each C atom has 3 bonding electron domains and 120 o bond angles, so all C atoms have sp 2 hybridization.
    3. Stronger. When comparing C–C bonds, the shorter the bond, the stronger the bond.
    4. Each C atom in butadiene has sp 2 hybridization and one unhybridized 2p orbital. If the C atoms are coplanar, the 4 unhybridized 2p orbitals are parallel and in the correct orientation for overlap. This provides a mechanism for delocalization of the pi electrons across the entire molecule, resulting in the shorter central C–C bond.

    9.103

    1. 30 e , 15 e pairs
      image
    2. image
    3. In the Lewis structure in part (b), the N atoms have a +1 formal charge and the B atoms have a –1 formal charge. Because N is more electronegative than B, these formal charges do not seem favorable.
    4. The Lewis structure in part (b) has two resonance structures.
    5. In part (a), the B atoms are sp 2 hybridized and the N atoms are sp 3 hybridized. In part (b), both B and N are sp 2 hybridized. We would not expect the structure in part (a) to lead to a planar molecule, whereas the structure in (b) would be planar.
    6. The magnitude of the bond distance is between the values for single and double bonds, which favors multiple resonance structures with alternating single and double bonds, the structure in part (b). That the B–N bond lengths are identical also favors this structure.
    7. Six. There are 12 electron pairs in the σ system, which leaves 3 electron pairs in the π system.

    9.104  Refer to Figure 9.43 and the Chemistry Put to Work box on Orbitals and Energy.

    1. Ground state, σ 2s 2 σ * 2s 2 π 2p 4 σ 2p 2 ; excited state, σ 2s 2 σ* 2s 2 π 2p 4 σ 2p 1 π * 2p 1
    2. Paramagnetic. The first excited state has two unpaired electrons, one in the σ 2p and one in the π* 2p .
    3. σ 2p to π* 2p
    4. ΔE = hc/λ; λ = 170 nm = 170 × 10 −9 m

      Δ E 170 = hc λ = 6.626 × 10 34 J-s 170 × 10 9 m × 2.998 × 10 8 m s = 1.1685 × 10 18 = 1.2 × 10 18 J/photon

      1.1685 × 10 18 J 1 photon × 6.022 × 10 23 photons 1 mol = 7.04 × 10 5 J/mol = 7.0 × 10 2 kJ/mol

    5. Weaker. In the first excited state there is one fewer electron in a bonding MO and one more electron in an antibonding MO. The bond order of the N–N bond in the first excited state is smaller and the bond is weaker than in the ground state.

    9.105

    1. The orbital in the sketch is a σ antibonding MO.
    2. In H 2 , there is one electron in the σ* antibonding MO.
    3. In H 2 , BO = ½.
    4. (iv). For the same two bonded atoms, the smaller the bond order, the weaker and longer the bond.

    9.106  Use the MO diagrams in Figure 9.43 to calculate bond order, taking into account the correct number of electrons in each ion.

    Ne 2 (BO = 0) < H 2 + (BO = ½) < B 2 (BO = 1) < F 2 + (BO = 1.5) < N 2 + (BO = 2.5)

    9.107

    1. The diagram shows two s atomic orbitals with opposite phases. (See Figures 9.31 and 9.40.) Because they are spherically symmetric, the interaction of s orbitals can only produce a σ molecular orbital. Because the two orbitals in the diagram have opposite phases, the interaction excludes electron density from the region between the nuclei. The resulting MO has a node between the two nuclei and is labeled σ 2 s * . The principal quantum number designation is arbitrary, because it defines only the size of the pertinent AOs and MOs. Shapes and phases of MOs depend only on these same characteristics of the interacting AOs.
    2. The diagram shows two p atomic orbitals with oppositely phased lobes pointing at each other. (See Figure 9.36.) End-to-end overlap produces a σ-type MO; opposite phases mean a node between the nuclei and an antibonding MO. The interaction results in a σ 2 p * MO.
    3. The diagram shows parallel p atomic orbitals with like-phased lobes aligned. (See Figure 9.36.) Side-to-side overlap produces a π-type MO; overlap of like-phased lobes concentrates electron density between the nuclei and a bonding MO. The interaction results in a π 2p MO.

    9.108  The white solid has the larger HOMO-LUMO gap. The green solid absorbs visible red light and appears green. The white solid does not absorb visible light, but light of higher energy in the ultraviolet region of the spectrum. The white solid has the larger HOMO-LUMO gap.

    9.109  We will refer to azobenzene (on the left) as A and hydrazobenzene (on the right) as H.

    1. A: sp 2 ; H: sp 3
    2. A: Fourteen. Each N and C atom has one unhybridized p orbital. H: Twelve. Each C atom has one unhybridized p orbital, but the N atoms have no unhybridized p orbitals.
    3. A: ~120°; H: less than 109.5°
    4. Because all C and N atoms in A have unhybridized p orbitals, all can participate in delocalized π bonding. The delocalized π system extends over the entire molecule, including both benzene rings and the azo “bridge.” In H, the N atoms have no unhybridized p orbitals, so they cannot participate in delocalized π bonding. Each of the benzene rings in H is delocalized, but the network cannot span the N atoms in the bridge.
    5. This is consistent with the answer to (d). In order for the unhybridized p orbitals in A to overlap, they must be parallel. This requires a planar σ-bond framework where all atoms in the molecule are coplanar.
    6. For a molecule to be useful in a solar energy conversion device, it must absorb visible light. This requires a HOMO-LUMO energy gap in the visible region. For organic molecules, the size of the gap is related to the number of conjugated π bonds; the more conjugated π bonds, the smaller the gap and the more likely the molecule is to be colored. Azobenzene has seven conjugated π bonds (π network delocalized over the entire molecule) and appears red-orange. Hydrazobenzene has only three conjugated π bonds (π network on benzene rings only) and appears white. Thus, the smaller HOMO-LUMO energy gap in A causes it to be both intensely colored and a more useful molecule for solar energy conversion.

    9.110

    1. H: 1s 1 ; F: [He]2s 2 2p 5

      When molecular orbitals are formed from atomic orbitals, the total number of orbitals is conserved. Because H and F have a total of five valence AOs (H 1s + F 2s + 3F 2p ), the MO diagram for HF has five MOs.

    2. H and F have a total of eight valence electrons. Because each MO can hold a maximum of two electrons, four of the five MOs would be occupied.
    3. image

      If H and F lie on the z-axis, then the 2p z orbital of F will overlap with the 1s orbital of H.


    1. Because F is more electronegative than H, the valence orbitals on F are at lower energy than those on H.
      image

      The HF MO diagram has 6 nonbonding, 2 bonding, and 0 antibonding electrons. The BO = [2 – 0]/2 = 1. (Nonbonding electrons do not “count” toward bond order.)

    1. image

      In the Lewis structure for HF, the nonbonding electrons are on the (more electronegative) F atom, as they are in the MO diagram.

    9.111

    1. CO, 10 e , 5 e pair
      image
    2. The bond order for CO, as predicted by the MO diagram in Figure 9.46, is 1/2[8 – 2] = 3.0. A bond order of 3.0 agrees with the triple bond in the Lewis structure.
    3. Applying the MO diagram in Figure 9.46 to the CO molecule, the highest-energy electrons would occupy the π 2p MOs. That is, π 2p would be the HOMO, highest occupied molecular orbital. If the true HOMO of CO is a σ-type MO, the order of the π 2p and σ 2p orbitals must be reversed. Figure 9.42 shows how the interaction of the 2s orbitals on one atom and the 2p orbitals on the other atom can affect the relative energies of the resulting MOs. This 2s–2p interaction in CO is significant enough so that the σ 2p MO is higher in energy than the π 2p MOs, and the σ 2p is the HOMO. This is the same energy order of MOs shown for large 2s–2p interaction in homonuclear diatomic molecules in Figure 9.43.
    4. We expect the atomic orbitals of the more electronegative element to have lower energy than those of the less electronegative element. When atoms of the two elements combine, the lower-energy atomic orbitals make a greater contribution to the bonding MOs and the higher-energy atomic orbitals make a larger contribution to the antibonding orbitals. Thus, the π 2p bonding MOs will have a greater contribution from the more electronegative O atom.

    9.112

    1. π 2p . The 2p valence atomic orbitals on C form the π molecular orbitals in ethylene. The symbol for the bonding molecular orbital is π 2p .
    2. π* 2p . By analogy to diatomic molecular orbital diagrams, the MO closest in energy to π 2p is π* 2p . This is the LUMO.

    1. Weaker. In the excited state, there is one fewer electron in a bonding MO and one more electron in an antibonding MO. This reduces the bond order of the C–C bond and weakens the bond.
    1. The C–C bond in ethylene is easier to twist when the molecule is in the excited state. In the excited state the C–C bond order is decreased (see above) and the π bond is essentially “broken.” There is free rotation around a C–C single bond.

    Integrative Exercises

    9.113

    1. Assume 100 g of compound

      2.1 g H × 1 mol H 1 .008 g H = 2.1 mol H; 2 .1/2 .1 = 1

      29.8 g N × 1 mol N 14 .01 g N = 2.13 mol N; 2 .13/2 .1 1

      68.1 g O × 1 mol O 16 .00 g O = 4.26 mol O; 4 .26/2 .1 2

      The empirical formula is HNO 2 ; formula weight = 47. Because the approximate molar mass is 50, the molecular formula is HNO 2 .

    2. Assume N is central, because it is unusual for O to be central, and part (d) indicates as much. HNO 2 : 18 valence e
      image

      The second resonance form is a minor contributor because of unfavorable formal charges.

    3. The electron-domain geometry around N is trigonal planar with an O–N–O angle of approximately 120°. If the resonance structure on the right makes a significant contribution to the molecular structure, all four atoms would lie in a plane. If only the left structure contributes, the H could rotate in and out of the molecular plane. The relative contributions of the two resonance structures could be determined by measuring the O–N–O and N–O–H bond angles.
    4. 3 VSEPR e domains around N, sp 2 hybridization
    5. 3 σ, 1 π for both structures (or for H bound to N).

    9.114

    1. 2SF 4 (g) + O 2 (g) → 2OSF 4 (g)
    2. 40 e , 20 e pairs
      image

      There must be a double bond drawn between O and S in order for their formal charges to be zero.


    1. ΔH = 8D(S–F) + D(O=O) – 8D(S–F) – 2D(S=O)

      ΔH = D(O=O) – 2D(S=O) = 495 – 2(523) = –551 kJ, exothermic

    1. trigonal-bipyramidal electron-domain geometry
      image
    1. In the structure on the left, there are 3 equatorial and 1 axial fluorine atoms. In the structure on the right, there are 2 equatorial and 2 axial fluorine atoms.

    9.115

    1. PX 3 , 26 valence e , 13 pairs
      image

      4 electron domains around P, tetrahedral e domain geometry,

      If all bonding and nonbonding electron domains are the same size, perfect tetrahedral angles are 109.5°. If all bonding electron domains are the same size but the nonbonding domain is larger, bond angles are somewhat less than 109.5°.

    2. As electronegativity increases (I < Br < Cl < F), the X–P–X angles decreases.
    3. The greater the electronegativity of X, the larger the magnitude of negative charge centered on X. The more negative charge centered on X, the smaller the P–X bonding domains, the greater the effect of the nonbonding domain and the smaller the bond angle. Also, as the electronegativity of X decreases and the bonding domain size increases, the effect of the large nonbonding domain decreases.
    4. PBrCl 4 , 40 valence electrons, 20 e pairs. The molecule will have trigonal- bipyramidal electron-domain geometry (similar to PCl 5 in Table 9.3). Based on the argument in part (c), the P–Br bond will have greater repulsions with P–Cl bonds than P–Cl bonds have with each other. Therefore, the Br will occupy an equatorial position in the trigonal bipyramid, so that the more unfavorable P–Br to P–Cl repulsions can be situated at larger angles in the equatorial plane.
      image

    9.116

    1. Three electron domains around each central C atom, sp 2 hybridization
    2. A 180° rotation around the C=C double bond is required to convert the trans isomer into the cis isomer. A 90° rotation around the bond eliminates all overlap of the p orbitals that form the π bond and it is broken.

    1. average bond enthalpy

      C=C  614 kJ/mol

      C–C  348 kJ/mol

      The difference in these values, 266 kJ/mol, is the average bond enthalpy of a C–C π bond. This is the amount of energy required to break 1 mol of C–C π bonds. The energy per molecule is

      266 kJ/mol × 1000 J 1 kJ × 1 mol 6 .022 × 10 23 molecules = 4.417 × 10 19

      = 4.42 × 10 19 J/molecule

    1. λ = hc/E = 6.626 × 10 34 J-s × 2.998 × 10 8 m/s 4.417 × 10 19 J = 4.50 × 10 7 m = 450 nm
    1. Yes, 450 nm light is in the visible portion of the spectrum. A cis-trans isomerization in the retinal portion of the large molecule rhodopsin is the first step in a sequence of molecular transformations in the eye that leads to vision. The sequence of events enables the eye to detect visible photons, in other words, to see.

    9.117

    1. C ≡ C 839 kJ/mol (1 σ, 2 π)

      C=C 614 kJ/mol (1 σ, 1 π)

      C–C 348 kJ/mol (1 σ)

      The contribution from 1 π-bond would be (614–348) 266 kJ/mol. From a second π bond, (839 – 614), 225 kJ/mol. An average π bond contribution would be (266 + 225)/2 = 246 kJ/mol.

      This is 246 kJ/ π bond 348 kJ/ σ bond × 100 = 71 % of the average enthalpy of a σ bond.

    2. N≡N 941 kJ/mol

      N=N 418 kJ/mol

      N–N 163 kJ/mol

      first π = (418 – 163) = 255 kJ/mol

      second π = (941 – 418) = 523 kJ/mol

      average π-bond enthalpy = (255 + 523)/2 = 389 kJ/mol

      This is 389 kJ/ π bond 163 kJ/ σ bond × 100 = 240 % of the average enthalpy of a σ bond.

      N–N σ bonds are weaker than C–C σ bonds, whereas N–N π bonds are stronger than C–C π bonds. The relative energies of C–C σ and π bonds are similar, whereas N–N π bonds are much stronger than N–N σ bonds.


    1. N 2 H 4 , 14 valence e , 7 e pairs
      image

      4 electron domains around N, sp 3 hybridization

      N 2 H 2 , 12 valence e , 6 e pairs

      image

      3 electron domains around N, sp 2 hybridization

      N 2 , 10 valence e , 5 e pairs

      image

      2 electron domains around N, sp hybridization

    1. In the three types of N–N bonds, each N atom has a nonbonding or lone pair of electrons. The lone pair to bond pair repulsions are minimized going from less than 109.5° to 120° to 180° bond angles, making the π bonds stronger relative to the σ bond. In the three types of C–C bonds, no lone-pair to bond-pair repulsions exist, and the σ and π bonds have more similar energies.

    9.118

    image

    ΔH = 6D(C–H) + 3D(C–C) + 3D(C=C) – 0

    = 6(413 kJ) + 3(348 kJ) + 3(614 kJ)

    = 5364 kJ

    (The products are isolated atoms; there is no bond making.)

    According to Hess’ law:

    Δ H ° = 6 Δ H f ° C(g) + 6 Δ H f ° H(g) Δ H f ° C 6 H 6 (g)

    = 6(718.4 kJ) + 6(217.94 kJ) – (82.9 kJ)

    = 5535 kJ

    The difference in the two results, 171 kJ/mol C 6 H 6 is because of the resonance stabilization in benzene. That is, because the π electrons are delocalized, the molecule has a lower overall energy than that predicted for the presence of 3 localized C–C and C=C bonds. Thus, the amount of energy actually required to decompose 1 mole of C 6 H 6 (g), represented by the Hess’ law calculation, is greater than the sum of the localized bond enthalpies (not taking resonance into account) from the first calculation above.


    9.119

    1. 3 d z 2
    2. Ignoring the donut of the d z 2 orbital
      image
    3. A node is generated in σ 3 d * because antibonding MOs are formed when AO lobes with opposite phases interact. Electron density is excluded from the internuclear region and a node is formed in the MO.
    4. Sc: [Ar]4s 2 3d 1 Omitting the core electrons, there are six e in the energy-level diagram.
      image
    5. The bond order in Sc 2 is 1/2 (4 – 2) = 1.0.

    9.120

    1. The molecular and empirical formulas of the four molecules are:

      benzene: molecular, C 6 H 6 ; empirical, CH

      napthalene: molecular, C 10 H 8 ; empirical, C 5 H 4

      anthracene: molecular, C 14 H 10 ; empirical, C 7 H 5

      tetracene: molecular, C 18 H 12 , empirical, C 3 H 2

    2. Yes. Because the compounds all have different empirical formulas, combustion analysis could in principle be used to distinguish them. In practice, the mass % of C in the four compounds is not very different, so the data would have to be precise to at least 3 decimal places and 4 would be better.
    3. C 10 H 8 (s) + 12O 2 (g) → 10CO 2 (g) + 4H 2 O(g)

    1. ΔH comb = 5D(C=C) + 6D(C–C) + 8D(C–H) + 12D(O=O) – 20D(C=O) – 8D(O–H)

      = 5(614) + 6(348) + 8(413) + 12(495) – 20(799) – 8(463)

      = –5282 kJ/mol C 10 H 8

    1. Yes. For example, the resonance structures of naphthalene are:
      image
    1. Colored compounds absorb visible light and appear the color of the visible light that they reflect. Colorless compounds typically absorb shorter wavelength, higher-energy light. The energy of light absorbed corresponds to the energy gap between the HOMO and LUMO of the molecule. That tetracene absorbs longer wavelength, lower-energy visible light indicates that it has the smallest HOMO-LUMO energy gap of the four molecules. Tetracene also has the most conjugated double bonds of the four molecules. We might conclude that the more con-jugated double bonds in an organic molecule, the smaller the HOMO-LUMO energy gap. More information about the absorption spectra of anthracene, naphthalene, and benzene is needed to confirm this conclusion.

    9.121  (a,b)

    image
    1. The two lobes of a p AO have opposite phases. These are shown on the diagram as + and −. An antibonding MO is formed when p AOs with opposite phases interact.
      image
    1. Note that the d xy AO has lobes that lie between, not on, the x and y axes.
    1. A π bond forms by overlap of orbitals on M and C. There is electron density above and below, but not along, the M−C axis.
    1. According to Exercise 9.111, the HOMO of CO is a σ-type MO. So the appropriate MO diagram is shown on the left side of Figure 9.43. A lone CO molecule has 10 valence electrons, the HOMO is σ 2 p and the bond order is 3.0. The LUMO is π 2p * .

      When M and CO interact as shown in the π 2 p * diagram, d-π back bonding causes the π 2 p * to become partially occupied. Electron density in the π 2 p * decreases electron density in the bonding molecular orbitals and decreases the BO of the bound CO. The strength of the C–O bond in a metal–CO complex decreases relative to the strength of the C–O bond in an isolated CO molecule.


    9.122

    1. 22 valence e , 11 e pairs
      image

      (The structure on the right does not minimize formal charges and will make a minor contribution to the true structure.)

    2. Both resonance structures predict the same bond angles. We expect H–C–N angles close to 109.5°.
      image
    3. The two extreme Lewis structures predict different bond lengths. As the true bonding model is some blend of the extreme Lewis structures, the true bond lengths are a blend of the extreme values. Our bond length estimates take into account that the structure minimizing formal charge makes a larger contribution to the true stru275cture.

      From Figure 7.7, the sum of the covalent radii for C and H is 1.07Å. We list this value in the table. As corroboration, the value of the O–H bond length given in Exercise 9.95 is 0.96 Å. According to Figure 7.7, the covalent radius of C is 0.10 Å greater than that of O, so we expect the C–H bond length to be approximately 0.10 Å greater than the measured O–H distance, about 1.06 Å. The lengths for isolated C–N, N–C, and C–O bonds are taken from Table 8.4.

      Bond Length (Å) N=C=O Length (Å) N≡C–O Length (Å) estimated
      C–H 1.07 1.07 1.07
      C–N 1.43 1.43 1.43
      N–C 1.38 1.16 1.33
      C–O 1.23 1.43 1.28
    4. The molecule will have a dipole moment. The N–C and C–O bond dipoles are opposite each other, but they are not equal. And, there are nonbonding electron pairs that are not directly opposite each other (in either structure) and will not cancel. There will be a resulting dipole.

     

    10 Gases

    Visualizing Concepts

    10.1   It would be much easier to drink from a straw on Mars. When a straw is placed in a glass of liquid, the liquid level in the straw equals the liquid level in the glass. The atmosphere exerts equal pressure inside and outside the straw. When we drink through a straw, we withdraw air, thereby reducing the pressure on the liquid inside. If only 0.007 atm is exerted on the liquid in the glass, a very small reduction in pressure inside the straw will cause the liquid to rise.

    Another approach is to consider the gravitational force on Mars. Because the pull of gravity causes atmospheric pressure, the gravity on Mars must be much smaller than that on Earth. With a very small Martian gravity holding liquid in a glass, it would be very easy to raise the liquid through a straw.

    10.2

    1. V 1 / T 1 = V 2 / T 2 ( Charles’ law ) V 1 / 300 K = V 2 / 500 K V 2 = 5 / 3 V 1
    2. P 1 V 1 = P 2 V 2 ( Boyle’s law ) 1 atm × V 1 = 2 atm × V 2 V 2 = 1 / 2 V 1
      image
    3. V 1 / T 1 = V 2 / T 2 ( Charles' law ) V 1 / 300 K = V 2 / 200 K V 2 = 2 / 3 V 1
      image

    10.3   Statement (b) is correct. At constant volume and temperature, pressure depends on total number of particles. To reduce the pressure by a factor of 2, the number of particles must be reduced by a factor of 2. At the lower pressure, the container would have half as many particles as at the higher pressure.

    Compare the two situations using the ideal-gas law, PV = nRT. P = nRT/V.

    P 2 = P 1 /2. V, R, and T are the same for the two states.

    n 2 RT/V = (n 1 RT/V)/2; n 2 = n 1 /2

    10.4   Statement (d) describes the volume change. At constant pressure and temperature, the container volume is directly proportional to the number of particles present (Avogadro’s law). As the reaction proceeds, 3 gas molecules are converted to 2 gas molecules, so the number of particles and the container volume decrease by 33%.

    10.5   PV = nRT (ideal-gas equation). In the ideal-gas equation, R is a constant. Given constant V and n (fixed amount of ideal gas), P and T are directly proportional. If P is doubled, T is also doubled. That is, if P is doubled, T increases by a factor of 2.

    10.6   Over time, the gases will mix perfectly. Each bulb will contain 4 blue and 3 red atoms. The “blue” gas has the greater partial pressure after mixing, because it has the greater number of particles (at the same T and V as the “red” gas.)

    10.7

    1. Partial pressure depends on the number of particles of each gas present. Red has the fewest particles, then yellow, then blue. P red < P yellow < P blue
    2. P gas = X gas P t . Calculate the mole fraction, X gas = [mol gas / total moles] or [particles gas / total particles]. This is true because Avogadro’s number is a counting number, and mole ratios are also particle ratios.

      X red = 2 red atoms / 10 total atoms = 0.2; P red = 0.2(1.40 atm) = 0.28 atm

      X yellow = 3 yellow atoms / 10 total atoms = 0.3; P yellow = 0.3(1.40 atm) = 0.42 atm

      X blue = 5 blue atoms / 10 total atoms = 0.5; P blue = 0.5(1.40 atm) = 0.70 atm

      Check . (0.28 atm + 0.42 atm + 0.70 atm) = 1.40 atm. The sum of the calculated partial pressures equals the given total pressure.

    10.8

    image

    10.9

    1. Curve B. At constant temperature, the root-mean-square (rms) speed (as well as the average speed) of a collection of gas particles is inversely related to molar mass; the lighter the particle, the faster it moves. Therefore, curve B represents He and curve A represents O 2 . Curve B has the higher rms speed and He is the lighter gas. Curve A has the lower rms speed and O 2 is the heavier gas.
    2. B. For the same gas, average kinetic energy (1/2 m u rms 2 ) , and therefore root-mean-square speed ( u rms ) is directly related to Kelvin temperature. Curve A is the lower temperature and curve B is the higher temperature.
    3. The root-mean-square speed. According to Figure 10.12(b), the most probable speed is lowest, then the average speed, and the rms speed is highest. This is true for the distribution of speeds for any gas, including the two in this exercise.

    10.10

    1. Total pressure is directly related to total number of particles (or total mol particles). P(ii) < P(i) = P(iii)
    2. Partial pressure of He is directly related to number of He atoms (yellow) or mol He atoms. P He (iii) < P He (ii) < P He (i)
    3. Density is total mass of gas per unit volume. We can use the atomic or molar masses of He (4) and N 2 (28), as relative masses of the particles.

      mass(i) = 5(4) + 2(28) = 76

      mass(ii) = 3(4) + 1(28) = 40

      mass(iii) = 2(4) + 5(28) = 148

      Because the container volumes are equal, d(ii) < d(i) < d(iii).

    4. At the same temperature, all gases have the same average kinetic energy. The average kinetic energies of the particles in the three containers are equal.

    10.11  The NH 4 Cl(s) ring will form at location a. The process described in this exercise is diffusion, rather than simple effusion. According to Section 10.8, Graham’s law approximates (but does not exactly describe) the diffusion rates of two gases under identical conditions. According to Graham’s law, the ratio of rates is inversely related to the ratio of molar masses of the two gases. That is, the lighter gas moves faster than the heavier gas. When introduced into the tube, NH 3 , MM = 17, moves faster and therefore farther than HCl, MM = 36. If NH 3 moves farther than HCl, the two gases meet and form NH 4 Cl(s) nearer the end where HCl was introduced; this is in the vicinity of location a.

    10.12

    1. The van der Waals constant a accounts for the influence of intermolecular forces in lowering the pressure of a real gas.
    2. According to the plot, Gas B is closest to ideal behavior, then Gas A and Gas C. Gas B will have the smallest a value, then A and C. From the values in Table 10.3, Gas B is N 2 , a = 1.39 ; Gas A is CO 2 , a = 3.59; Gas C = Cl 2 , a = 6.49.

    Gas Characteristics; Pressure (Sections 10.1 and 10.2)

    10.13  Statement (c) is false. Gaseous molecules are so far apart that there is no barrier to mixing, regardless of the identity of the molecules.

    10.14

    1. g/L. Gas samples are mostly empty space; their volume is large relative to the mass of the molecules present. The unit g/L has a relatively small mass and large volume.
    2. Pa and atm. Pressure is defined as force per unit area. The unit N is only a unit of force. The unit kg/m 2 is a unit of mass per unit area, not force per unit area.
    3. F 2 is most likely to be a gas at room temperature. K 2 O is an ionic compound; all simple ionic compounds are solids. Br 2 is a covalent compound with high molar mass. The greater the molar mass, the stronger the dispersion forces and the less likely the compound is to be a gas. (In the series of diatomic halogens, F 2 is a gas, Br 2 is a liquid, and I 2 is a solid at room temperature and ordinary atmospheric pressure.)

    10.15 Analyze .  Given: mass, area. Find: pressure. Plan . P = F/A. Data is given in terms of lb and in. 2 ; calculate pressure in lb/in. 2 (or psi) for part (c), then use this value to find pressure in and atm for part (b) and kPa for part (a). Solve .

    1. P = F A = 130 lb 0 .50 in . 2 = 260 = 2 .6 × 10 2 lb/in . 2
    2. 1 atm = 101.325 kPa

      17 .687 atm × 1 01 .325 kPa 1 atm = 1792 = 1 .8 × 10 3 kPa

    3. 14.70 lb/in. 2 = 1 atm

      260 lb/in . 2 × 1 atm 14.70 lb/in . 2 = 17 .687 = 18 atm

    10.16  P = m × a/A; 1 Pa = 1 kg/m-s 2 ; A = 3.0 cm × 4.1 cm × 4 = 49.2 = 49 cm 2

    262 kg 49 .2 cm 2 × 9.81 m s 2 × ( 100 ) 2 cm 2 1 m 2 = 5.224 × 10 5 kg m-s 2 = 5.2 × 10 5 Pa

    10.17 Analyze .  Given: 760 mm column of Hg, densities of Hg and glycerol. Find: height of a column of glycerol at the same pressure.

    Plan .  Develop a relationship between pressure, height of a column of liquid, and density of the liquid. Relationships that might prove useful: P = F/A; F = m × a; m = d × V(density)(volume); V = A × height Solve .

    P = F A = m × a A = d × V × a A = d × A × h × a A = d × h × a


    1. P Hg = P glycerol ; Using the relationship derived above: ( d × h × a) glycerol = ( d × h × a) Hg

      Because a, the acceleration due to gravity, is equal in both liquids, ( d × h) glycerol = ( d × h) Hg

      1.26 g/mL × h glycerol = 13.6 g/mL × 760 mm

      h glycerol = 13 .6 g / mL × 760 mm 1.26 g / mL = 8.203 × 10 3 = 8.20 × 10 3 mm = 8 .20 m

    2. Plan. Calculate the height of a column of Hg that exerts the same pressure as 15 ft of water. The density of water is 1.00 g/mL. Then calculate the total pressure, atmospheric plus water, in atmospheres.

      15 ft H 2 O × 12 in . 1 ft × 2.54 cm 1 in . × 10 mm 1 cm = 4572 = 4.6 × 10 3 mm H 2 O

      ( d × h) H 2 O = ( d × h) Hg

      1.00 g/mL × 4572 mm H 2 O = 13.6 g/mL × ? mm Hg

      h Hg = 1 .00 g / mL × 4572 mm 13.6 g / mL = 336.2 = 3.4 × 10 2 mm Hg

      P total = P atm + P H 2 O = 750 mm Hg + 336 mm Hg = 1086 mm Hg = 1 .1 × 10 2 mm Hg

      P total = 1086 mm Hg × 1 atm 760 mm Hg = 1.429 = 1.4 atm

    10.18

    1. Abbreviate 1-iododecane as 1id. Using the relationship derived in Solution 10.17 for two liquids under the influence of gravity, (d × h) 1id = (d × h) Hg . At 749 torr, the height of an Hg barometer is 749 mm.

      1.20 g 1 mL × h 1id = 13.6 g 1 mL × 749 mm; h 1id = 13.6 g/mL × 749 mm 1.20 g/mL = 8.49 × 10 3 mm = 8 .49 m

    2. 21 ft H 2 O × 12 in . 1 ft × 2.54 cm 1 in . × 10 mm 1 cm = 6401 = 6.4 × 10 3 mm H 2 O

      ( d × h) H 2 O = ( d × h) Hg

      1.00 g/mL × 6401 mm H 2 O = 13.6 g/mL × ? mm Hg

      h Hg = 1 .00 g / mL × 6401 mm 13.6 g / mL = 470.7 = 4.7 × 10 2 mm Hg

      P total = P atm + P H 2 O = 742 mm Hg + 471 mm Hg = 1213 mm Hg = 1 .2 × 10 3 mm Hg

      P total = 1213 mm Hg × 1 atm 760 mm Hg = 1.596 = 1.6 atm

    10.19 Analyze/Plan .  We are given pressure in one unit and asked to change it to another unit. Select appropriate conversion factors and use dimensional analysis to find the desired quantities. Solve .

    1. 265 torr × 1 atm 760 torr = 0.349 atm

    1. 265 torr × 1 mm Hg 1 torr = 265 mm Hg
    1. 265 torr × 1 .01325 × 10 5 Pa 760 torr = 3.53 × 10 4 Pa
    1. 265 torr × 1 .01325 × 10 5 Pa 760 torr × 1 bar 1 × 10 5 Pa = 0.353 bar
    1. 265 torr × 1 atm 760 torr × 14.70 psi 1 atm = 5.13 psi

    10.20

    1. 0.912 atm × 760 torr 1 atm = 693 torr
    2. 0. 685 bar × 1 × 10 5 Pa 1 bar × 1 kPa 1 × 10 3 Pa = 68.5 kPa
    3. 655 mm Hg × 1 atm 760 mm Hg = 0.862 atm
    4. 1.323 × 10 5 Pa × 1 atm 1 .01325 × 10 5 Pa = 1.3057 = 1.306 atm
    5. 2.50 atm × 1 4 .70 psi 1 atm = 36.75 = 36.8 psi

    10.21 Analyze/Plan .  Follow the logic in Solution 10.19. Solve .

    1. 30.45 in . Hg × 25 .4 mm 1 in . × 1 torr 1 mm Hg = 773.4 torr

      [The result has 4 sig figs because 25.4 mm/in. is considered to be an exact number.]

    2. 30.45 in Hg = 773.4 torr; 773.4 torr × 1 atm 760 torr = 1.018 atm

    10.22  882 mbar = 0.882 bar

    1. 0. 882 bar × 1 × 10 5 Pa 1 bar × 1 atm 101 , 325 Pa = 0.8705 = 0.871 atm
    2. 0.882 bar = 0.8705 atm × 760 torr 1 atm = 661.55 = 662 torr
    3. 0.882 bar = 661 .55 torr × 1 mm Hg 1 torr × 1 cm Hg 10 mm  Hg × 1 in . Hg 2 .54 cm Hg = 26.045 = 26.0 in . Hg

    10.23 Analyze/Plan .  Follow the logic in Sample Exercise 10.2. Solve .

    1. The Hg level is lower in the open end than the closed end, so the gas pressure is less than atmospheric pressure.

      P gas = 0.995 atm ( 52 cm × 1 atm 76 .0 cm ) = 0.31 atm

    2. The Hg level is higher in the open end, so the gas pressure is greater than atmospheric pressure.

      P gas = 0.995 atm + ( 67 cm Hg × 1 atm 76 .0 cm Hg ) = 1.8766 = 1.88 atm


    1. This is a closed-end manometer, so P gas = h.

      P gas = 10.3 cm × 1 atm 76 .0 cm = 0.136 atm

    10.24

    1. The atmosphere is exerting 15.4 cm = 154 mm Hg (torr) more pressure than the gas.

      P gas = P atm 15.4 torr = ( 0 .985 atm × 760 torr 1 atm ) 15.4 torr = 7 33 torr

    2. The gas is exerting 12.3 mm Hg (torr) more pressure than the atmosphere.

      P gas = P atm + 12.3 torr = ( 0 .99 atm × 760 torr 1 atm ) + 12.3 torr = 764 .7 torr = 7.6 × 10 2 torr

      (Atmospheric pressure of 0.99 atm determines that the result has 2 sig figs.)

    The Gas Laws (Section 10.3)

    10.25 Analyze/Plan .  Given certain changes in volume and temperature of a gas contained in a cylinder with a moveable piston, predict which will double the gas pressure. Consider the gas law relationships in Section 10.3. Solve .

    1. No. P and V are inversely proportional at constant T. If the volume increases by a factor of 2, the pressure decreases by a factor of 2.
    2. No. P and Kelvin T are directly proportional at constant V. Doubling °C does increase V, but it does not double it.
    3. Yes. P and V are inversely proportional at constant T. If the volume decreases by a factor of 2, the pressure increases by a factor of 2.

    10.26 Analyze .  Given: initial P, V, T. Find: final values of P, V, T for certain changes of condition. Plan . Select the appropriate gas law relationships from Section 10.3; solve for final conditions, paying attention to units. Solve .

    1. P 1 V 1 = P 2 V 2 ; the proportionality holds true for any pressure or volume units.

      P 1 = 752 torr, V 1 = 5.12 L, P 2 = 1.88 atm

      V 2 = P 1 V 1 P 2 = 7 52 torr × 5 .12 L 1.88 atm × 1 atm 760 torr = 2.69 L

      Check .  As pressure increases, volume should decrease; our result agrees with this.

    2. V 1 /T 1 = V 2 /T 2 ; T must be in Kelvins for the relationship to be true.

      V 1 = 5.12 L, T 1 = 21 C = 294 K, T 2 = 175 C = 448 K

      V 2 = V 1 T 2 T 1 = 5. 12 L × 448 K 294 K = 7.80 L

      Check .  As temperature increases, volume should increase; our result is consistent with this.


    10.27

    1. Analyze/Plan . Use Boyle’s law, PV = constant or P 1 V 1 = P 2 V 2 , and Charles’ law, V/T = constant or V 1 /T 1 = V 2 /T 2 , to derive Amonton’s law for the relationship between P and T at constant V. Solve.

      Boyle’s law: P 1 V 1 = P 2 V 2 or P 1 P 2 = V 2 V 1 . At constant V, V 2 V 1 = 1 and P 1 P 2 = 1 .

      Charles’ law: V 1 T 1 = V 2 T 2 or V 1 V 2 = T 1 T 2 .  At constant V, V 1 V 2 = 1 and T 1 T 2 = 1 .

      P 1 P 2 = 1 and T 1 T 2 = 1, then P 1 P 2 = T 1 T 2 or P 1 T 1 = P 2 T 2 or  P/T = constant .

      Amonton’s law is that pressure and temperature are directly proportional at constant volume.

    2. Analyze/Plan . Recall that the proportional relationships for gases are true for any units of pressure (or volume) but only for Kelvin temperatures. Change °F to K, then use the relationship derived in (a) to calculate the new tire pressure.

      °C = 5/9(°F – 32); K = °C + 273. Solve.

      T 1 : °C = 5/9(75 – 32) = 23.89 = 24 °C. K = 24 °C + 273 = 297 K. T 1 = 32.0 psi

      T 2 : °C = 5/9(120 – 32) = 48.89 = 49 °C. K = 49 °C + 273 = 322 K. P 2 = ?

      P 1 T 1 = P 2 T 2 or  P 2 = P 1 T 2 T 1 = 32.0 psi × 322 K 297 K = 34.694 = 34.7 psi

    10.28  According to Avogadro’s hypothesis, the mole ratios in the chemical equation will be volume ratios for the gases if they are at the same temperature and pressure.

    N 2 (g) + 3H 2 (g) → 2NH 3 (g)

    The volumes of H 2 and N 2 are in a stoichiometric 3.6 L 1.2 L or 3 vol H 2 1 vol N 2 ratio, so either can be used to determine the volume of NH 3 (g) produced.

    1.2 L N 2 × 2 mol NH 3 1 mol N 2 = 2.4 L NH 3 ( g)  produced .

    The Ideal-Gas Equation (Section 10.4)

    (In Solutions to Exercises , the symbol for molar mass is MM.)

    10.29

    1. STP stands for standard temperature, 0 C (or 273 K), and standard pressure, 1 atm.
    2. V = nRT P ; V = 1 mol × 0 . 0 8 2 0 6 L - a t m m ol- K × 273 K 1 atm

      V = 22.4 L for 1 mole of gas at STP

    3. 25 C + 273 = 298 K

      V = nRT P ; V = 1 mol × 0 .08206 L-atm mol-K × 298 K 1 atm

      V = 24.5 L for 1 mol of gas at 1 atm and 25 C

    4. 0 .08206 L-atm mol-K × 1.01325 × 10 5 Pa 1 atm × 1 bar 10 5 Pa = 0 .08315 L-bar mol-K

    10.30  Assume 1 mole of Ne at STP. Sample volume = 22.4 L. Calculate the volume occupied by 1 mole of Ne atoms. 1 L = 1 dm 3 , r = 0.69 Å, V = 4πr 3 /3

    r = 0 .69 Å × 1 × 10 10 m 1 Å × 1 0 dm m = 6.9 × 10 10 dm

    V = 4 π r 3 3 ; 4 π (6 .9 × 10 10 dm) 3 3 × 6.022 × 10 23 Ne atoms mol = 8 .287 × 10 4 dm 3 = 8 .3 × 10 4 L

    8.3 × 10 4 L occupied by Ne atoms 22.4 L total gas volume = 1/27,000 of the total space occupied by Ne atoms

    10.31 Analyze/Plan. PV = nRT. At constant volume and temperature, P is directly proportional to n.

    Solve. For samples with equal masses of gas, the gas with MM = 30 will have twice as many moles of particles and twice the pressure. Thus, flask A contains the gas with MM = 30 and flask B contains the gas with MM = 60.

    10.32  n = g/MM; PV = nRT = gRT/MM; MM = gRT/PV.

    2-L flask: MM = 4.8 RT/2.0(x) = 2.4 RT/x

    3-L flask: MM = 0.36 RT/3.0 (0.1x) = 1.2 RT/x

    The molar masses of the two gases are not equal. The gas in the 2-L flask has a molar mass that is twice as large as the gas in the 3-L flask.

    10.33 Analyze/Plan .  Follow the strategy for calculations involving many variables given in Section 10.4. Solve .

    T = PV nR = 2.00 atm × 1 .00 L 0 .500 mol × mol-K 0 .08206 L-atm = 48.7 K

    K = 27 C + 273 = 300 K

    n = PV RT = 0.300 atm × 0 .250 L 300 K × mol-K 0 .08206-atm = 3.05 × 10 3 mol

    650 torr × 1 atm 760 torr = 0.85526 = 0.855 atm

    V = nRT P = 0.333 mol × 350 K 0 .85526 atm × 0.08206 L-atm mol-K = 11.2 L

    585 mL = 0.585 L

    P = nRT V = 0.250 mol × 295 K 0 .585 L × 0.08206 L-atm mol-K = 10.3 atm

    P V n T
    2.00 atm 1.00 L 0.500 mol 48.7 K
    0.300 atm 0.250 L 3.05 × 10 –3 mol 27 C
    650 torr 11.2 L 0.333 mol 350 K
    10.3 atm 585 mL 0.250 mol 295 K

    10.34 Analyze/Plan .  Follow the strategy for calculations involving many variables given in Section 10.4. Solve .

    1. n = 1.50 mol, P = 1.25 atm, T = –6 C = 267 K

      V = nRT P = 1.50 mol × 0 .08206 L-atm mol-K × 267 K 1 .25 atm = 26.3 L

    2. n = 3.33 × 10 –3 mol, V = 478 mL = 0.478 L

      P = 750 torr × 1 atm 760 torr = 0.9868 = 0.987 atm

      T = PV nR = 0.9868 atm × 0 .478 L 3 .33 × 10 3 mol × 1 mol-K 0.08206 L-atm = 1726 = 1.73 × 10 3 K

    3. n = 0.00245 mol, V = 413 mL = 0.413 L, T = 138 C = 411 K

      P = nRT V ; = 0.00245 mol × 0 .08206 L-atm mol-K × 411 K 0 .413 L = 0.200 atm

    4. V = 126.5 L, T = 54 C = 327 K,

      P = 11 .25 kPa × 1 atm 101 .325 kPa = 0.11103 = 0.1110 atm

      n = PV RT = 0.11103 atm × mol-K 0 .08206 L-atm × 126.5 L 327 K = 0.523 mol

    10.35 Analyze/Plan .  Follow the strategy for calculations involving many variables. Solve .

    n = g/MM ; PV = nRT; PV = gRT/MM ; g = MM × PV/RT

    P = 1.0 atm, T = 23 C = 296 K, V = 1.75 × 10 5 ft 3 . Change ft 3 to L, then calculate grams (or kg).

    1.75 × 10 5 ft 3 × ( 12 ) 3 in 3 ft 3 × ( 2.54 ) 3 cm 3 in 3 × 1 L 1 × 10 3 cm 3 = 4.9554 × 10 6 = 4 .96 × 10 6 L

    g = 4.003 g He 1 mol He × mol-K 0 .08206 L-atm × 1.0 atm × 4 .955 × 10 6 L 296 K = 8.2 × 10 5 g = 820 kg He

    10.36  Find the volume of the tube in cm 3 ; 1 cm 3 = 1 mL.

    r = d/2 = 2.5 cm/2 = 1.25 = 1.3 cm; h = 5.5 m = 5.5 × 10 2 cm

    V = πr 2 h = 3.14159 × (1.25 cm) 2 × (5.5 × 10 2 cm) = 2.700 × 10 3 cm 3 = 2.7 L

    PV = g MM RT;  g = MM × PV RT ; P = 1 .78 torr × 1 atm 760 torr = 2.342 × 10 3 = 2.34 × 10 3 atm

    g = 20.18 g Ne 1 mol Ne × mol-K 0 .08206 L-atm × 2.342 × 10 3 atm × 2 .700 L 308 K = 5.049 × 10 3 = 5.0 × 10 3 g Ne

    10.37 Analyze/Plan .  Follow the strategy for calculations involving many variables. Solve .

    1. V = 2 .25 L; T = 273 + 37 ° C = 310 K; P = 735 torr × 1 atm 760 torr = 0.96710 = 0.967 atm

      PV = nRT, n = PV/RT, number of molecules (#) = n × 6.022 × 10 23

      # = 0.9671 atm × 2 .25 L 310 K × mol-K 0 .08206 L-atm × 6.022 × 10 23 molecules mol

      = 5.15 × 10 22 molecules


    1. V = 5.0 × 10 3 L; T = 273 + 0 C = 273 K; P = 1.00 atm; MM = 28.98 g/mol

      PV = g MM RT; g = MM × PV RT

      g = 28.98 g air 1 mol air × mol-K 0 .08206 L-atm × 1. 00 atm × 5.0 × 10 3 L 2 73 K = 6468 g = 6.5 × 10 3 g air = 6.5 kg air

    10.38

    1. P O 3 = 3.0 × 10 3 atm; T = 250 K; V = 1 L (exact)

      # of O 3 molecules = PV RT × 6.022 × 10 23

      # = 3.0 × 10 3 atm × 1 L 250 K × mol-K 0 .08206 L-atm × 6.022 × 10 23 molecules mol = 8.8 × 10 19 O 3 molecules

    2. # of CO 2 molecules = PV RT × 6.022 × 10 23 × 0. 0004

      # = 1.0 atm × 2 .0 L 300 K × mol-K 0 .08206 L-atm × 6.022 × 10 23 molecules mol × 0 .0004 = 1.957 × 10 19 = 2 × 10 19 CO 2 molecules

    10.39 Analyze/Plan .  Follow the strategy for calculations involving many variables. Solve .

    1. P = nRT V ; n = 0 .29 kg O 2 × 1000 g 1 kg × 1 mol O 2 32.00 g O 2 = 9.0625 = 9.1 mol; V = 2 .3 L;

      T = 273 + 9 C = 282 K

      P = 9 .0625 mol 2.3 L × 0.08206 L-atm mol-K × 282 K = 91 atm

    2. V = nRT P ; = 9.0625 mol 0.95 atm × 0.08206 L-atm mol-K × 299 K = 2 .3 × 10 2 L

    10.40

    1. V = 0 .250 L, T = 23 ° C = 296  K, n = 2 .30 g C 3 H 8 × 1 mol C 3 H 8 44.1 g C 3 H 8 = 0.052154 = 0.0522 mol

      P = nRT V = 0.052154 mol × 0.08206 L-atm mol-K × 296 K 0 .250 L = 5.07 atm

    2. STP = 1.00 atm, 273 K

      V = nRT P = 0.052154 mol × 0.08206 L-atm mol-K × 273 K 1 .00 atm = 1.1684 = 1.17 L

    3. C = 5/9 ( F – 32 ); K = C + 273.15 = 5/9 (130 F – 32 ) + 273.15 = 327.59 = 328 K

      P = nRT V = 0.052154 mol × 0.08206 L-atm mol-K × 327.59 K 0 .250 L = 5.608 = 5 .61 atm

    10.41 Analyze/Plan .  Follow the strategy for calculations involving many variables. Solve .

    P = nRT V ; n = 35 .1 g CO 2 × 1 mol CO 2 44.01 g CO 2 = 0.7975 = 0.798 mol; V = 4 .0 L; T = 298 K

    P = 0 .7975 mol 4.0 L × 0.08206 L-atm mol-K × 298 K = 4.876 = 4.9 atm


    10.42  Calculate the mass of He that will produce a pressure of 75 atm in the cylinder, then subtract that mass from 5.225 g He to calculate the mass of He that must be released.

    PV = g MM RT; g = MM × PV RT

    g = 4.0026 g He 1 mol He × mol-K 0 .08206 L-atm × 75 atm × 0.334 L 2 96 K = 4 .1279 = 4 .1 g He remain

    5.225 g He initial – 4.1279 g He remain = 1.0971 = 1.1 g He must be released

    10.43 Analyze/Plan .  Follow the strategy for calculations involving many variables. Solve .

    V = 8 .70 L, T = 24 °C = 297 K, P = 895 torr × 1 atm 760 torr = 1.1776 = 1.18 atm

    1. g = MM × PV RT ; g = 70.91 g Cl 2 1 mol Cl 2 × mol-K 0 .08206 L-atm × 1.1776 atm 297 K × 8.70 L = 29 .8g Cl 2
    2. V 2 = P 1 V 1 T 2 T 1 P 2 = 895 torr × 8 .70 L × 273 K 297 K × 760 torr = 9.42 L
    3. T 2 = P 2 V 2 T 1 P 1 V 1 = 876 torr × 15.00 L × 297 K 895 torr × 8 .70 L = 501 K
    4. P 2 = P 1 V 1 T 2 V 2 T 1 = 895 torr × 8 .70 L × 331 K 5.00 L × 297 K = 1.73 × 10 3 torr = 2 .28 atm

    10.44 T = 23 ° C = 296 K,  P = 16,500 kPa × 1 atm 101 .325 kPa = 162.84 = 163 atm

    V = 5 5.0 gal × 3 .7854 L 1 gal = 208.20 = 208 L

    1. g = MM × PV RT ; g = 32 .0 g O 2 1 mol O 2 × mol-K 0 .08206 L-atm × 162.84 atm 296 K × 2 08 .20 L = 4.4665 × 10 4 g O 2 = 44.7 kg O 2
    2. V 2 = P 1 V 1 T 2 T 1 P 2 = 16 , 500 kPa × 208 .20 L × 273 K 296 K × 101 .325 kPa = 3.13 × 10 4 L
    3. T 2 = P 2 T 1 P 1 = 150 .0 atm × 296 K 16 , 500 kPa × 101.325 kPa 1 atm = 272.7 = 273 K
    4. P 2 = P 1 V 1 T 2 V 2 T 1 = 16 , 500 kPa × 208 .20 L × 297 K 55.0 L × 296 K = 62 , 671 = 6.27 × 10 4 kPa

    10.45 Analyze .  Given: mass of cockroach, rate of O 2 consumption, temperature, percent O 2 in air, volume of air. Find: mol O 2 consumed per hour; mol O 2 in 1 quart of air; mol O 2 consumed in 48 hr.

    1. Plan/Solve .  V of O 2 consumed = rate of consumption × mass × time. n = PV/RT.

      5.2 g × 1 hr × 0 .8 mL O 2 1 g-hr = 4.16 = 4 mL O 2 consumed

      n = PV RT = 1 atm × mol-K 0 .08206 L-atm × 0.00416 L 297 K = 1.71 × 10 4 = 2 × 10 4 mol O 2


    1. Plan/Solve .  qt air → L air → L O 2 available. mol O 2 available = PV/RT.

      mol O 2 /hr (from part (a)) → total mol O 2 consumed. Compare O 2 available and O 2 consumed.

      1 qt air × 0 .946 L 1 qt × 0.21 O 2 in air = 0 .199 L O 2 available

      n = 1 atm × mol-K 0 .08206 L-atm × 0.199 L 297 K = 8.17 × 10 3 = 8 × 10 3 mol O 2 available

      cockroach uses 1 .71 × 10 4 mol 1 hr × 48 hr = 8 .21 × 10 3 = 8 × 10 3 mol O 2 consumed

      Not only does the cockroach use 20% of the available O 2 , it needs all the O 2 in the jar.

    10.46  Change mass to kg; 1 hr = 60 min; pay attention to units.

    1. 185 lb × 1 kg 2. 2046 lb × 4 7 .5 mL O 2 kg-min × 60 min = 2.39 × 10 5 mL
    2. 165 lb × 1 kg 2. 2046 lb × 6 5 .0 mL O 2 kg-min × 60 min = 2.92 × 10 5 mL

    Further Applications of the Ideal-Gas Equation (Section 10.5)

    10.47 Analyze/Plan. At the same temperature and pressure, the density of a gas increases with increasing molar mass.

    HF (20 g/mol) < CO (28 g/mol) < N 2 O (44 g/mol) < Cl 2 (71 g/mol)

    10.48  CO 2 < SO 2 < HBr. For gases at the same conditions, density is directly proportional to molar mass. The order of increasing molar mass is the order of increasing density.

    CO 2 , 44 g/mol < SO 2 , 64 g/mol < HBr, 81 g/mol.

    10.49  (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The balloon thus weighs less than the air displaced by its volume.

    10.50  (b) Xe atoms have a higher mass than N 2 molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser.

    10.51 Analyze/Plan .  Conditions (P, V, T) and amounts of gases are given. Rearrange the relationship PV × MM = gRT to obtain the desired quantity, paying attention (as always!) to units. Solve .

    1. d = MM × P RT ; MM = 46.0 g/mol; P = 0 .970 atm, T = 35 ° C = 308 K

      d = 46 .0 g NO 2 1 mol × mol-K 0 .08206 L-atm × 0.970 atm 308 K = 1.77 g/L

    2. MM = gRT PV = 2.50 g 0 .875 L × 0 .08206 L-atm mol-K × 308 K 685 torr × 760 torr 1 atm = 80.1 g/mol

    10.52

    1. d = MM × P RT ; MM = 146.1 g/mol, T = 21 ° C = 294 K, P = 707 torr

      d = 146 .1 g 1 mol × mol-K 0 .08206 L-atm × 707 torr 294 K × 1 atm 760 torr = 5.63 g/L

    2. MM = dRT P = 7.135 g 1 L × 0 .08206 L-atm mol-K × 285 K 743 torr × 760 torr 1 atm = 171 g/mol

    10.53 Analyze/Plan .  Given: mass, conditions (P, V, T) of unknown gas. Find: molar mass. MM = gRT/PV. Solve .

    MM = gRT PV = 1.012 g 0 .354 L × 0 .08206 L-atm mol-K × 372 K 742 torr × 760 torr 1 atm = 89.4 g/mol

    10.54 MM = gRT PV = 0.846 g 0 .354 L × 0 .08206 L-atm mol-K × 373 K 752 torr × 760 torr 1 atm = 73.9 g/mol

    10.55 Analyze/Plan .  Follow the logic in Sample Exercise 10.9. Solve .

    mol O 2 = PV RT = 3.5 × 10 6 torr × 1 atm 760 torr × mol-K 0 .08206 L-atm × 0.452 L 300 K = 8.456 × 10 11 = 8.5 × 10 –11 mol O 2

    8. 456 × 10 11 mol O 2 × 2 mol Mg 1 mol O 2 × 24.3 g Mg 1 mol Mg = 4.1 × 10 9 g Mg (4 .1 ng Mg)

    10.56 n H 2 = P H 2 V RT ; P = 825 torr × 1 atm 760 torr = 1.0855 = 1.09 atm; T = 273 + 21 ° C = 294 K

    n H 2 = 1.0855 atm × mol-K 0 .08206 L-atm × 1 45 L 294 K = 6.5242 = 6.52 mol H 2

    6. 5242 mol H 2 × 1 mol CaH 2 2 mol H 2 × 42.10 g CaH 2 1 mol CaH 2 = 137.34 = 137 g CaH 2

    10.57

    1. Analyze/Plan .  g glucose → mol glucose → mol CO 2 → V CO 2 Solve .

      24 .5 g × 1 mol glucose 180.1 g × 6 mol CO 2 1 mol glucose = 0.8162 = 0.816 mol CO 2

      V = nRT P = 0.8162 mol × 0 .08206 L-atm mol-K × 310 K 0 .970 atm = 21.4 L CO 2

    2. Analyze/Plan .  g glucose → mol glucose → mol O 2 → V O 2 Solve .

      50 .0 g × 1 mol glucose 180.1 g × 6 mol O 2 1 mol glucose = 1.6657 = 1.67 mol O 2

      V = nRT P = 1.6657 mol × 0 .08206 L-atm mol-K × 2 98 K 1 atm = 40.7 L O 2

    10.58  Follow the logic in Sample Exercise 10.9. The H 2 (g) will be used in a balloon, which operates at atmospheric pressure. Because atmospheric pressure is not explicitly given, assume 1 atm (infinite sig figs).

    n = PV RT = 1 atm × 3 .1150 × 10 4 L 295 K × mol-K 0.08206 L-atm = 1.28678 × 10 3 = 1.29 × 10 3 mol H 2


    From the balanced equation, 1 mol of Fe produces 1 mol of H 2 , so 1.29 × 10 3 mol Fe are required.

    1 .28678 × 10 3 mol Fe × 55.845 g Fe mol Fe × 1 kg 1000 g = 71.86 = 71.9 kg Fe

    10.59 Analyze/Plan .  The gas sample is a mixture of H 2 (g) and H 2 O(g). Find the partial pressure of H 2 (g) and then the moles of H 2 (g) and Zn(s). Solve .

    P t = 738 torr = P H 2 + P H 2 O

    From Appendix B, the vapor pressure of H 2 O at 24 C = 22.38 torr

    P H 2 = ( 738 torr 22 .38 torr) × 1 atm 760 torr = 0.9416 = 0.942 atm

    n H 2 = P H 2 V RT = 0.9416 atm × mol-K 0 .08206 L-atm × 0.159 L 297 K = 0.006143 = 0.00614 mol H 2

    0.006143 mol H 2 × 1 mol Zn 1 mol H 2 × 65.39 g Zn 1 mol Zn = 0.402 g Zn

    10.60  The gas sample is a mixture of C 2 H 2 (g) and H 2 O(g). Find the partial pressure of C 2 H 2 , then moles CaC 2 and C 2 H 2 .

    P t = 753 torr = P C 2 H 2 + P H 2 O P H 2 O at  23 ° C = 21 .07 torr

    P C 2 H 2 = ( 753 torr 21 .07 torr) × 1 atm 760 torr = 0.96307 = 0.963 atm

    1. 524 g CaC 2 × 1 mol CaC 2 64.10 g × 1 mol C 2 H 2 1 mol CaC 2 = 0.023775 = 0.02378 mol C 2 H 2

    V = 0 .023775 mol × 0 .08206 L-atm mol-K × 296 K 0 .96307 atm = 0.600 L C 2 H 2

    Partial Pressures (Section 10.6)

    10.61

    1. When the stopcock is opened, the volume occupied by N 2 (g) increases from 2.0 to 5.0 L. At constant T, P 1 V 1 = P 2 V 2 . 1.0 atm × 2.0 L = P 2 × 5.0 L; P 2 = 0.40 atm
    2. When the gases mix, the volume of O 2 (g) increases from 3.0 to 5.0 L. At constant T, P 1 V 1 = P 2 V 2 . 2.0 atm × 3.0 L = P 2 × 5.0 L; P 2 = 1.2 atm
    3. P t = P N 2 + P O 2 = 0.40 atm + 1 .2 atm = 1 .6 atm

    10.62

    1. The partial pressure of gas A is not affected by the addition of gas C. The partial pressure of A depends only on moles of A, volume of container, and conditions; none of these factors changes when gas C is added.
    2. The total pressure in the vessel increases when gas C is added, because the total number of moles of gas increases.
    3. The mole fraction of gas B decreases when gas C is added. The moles of gas B stay the same, but the total moles increase, so the mole fraction of B (nB/nt) decreases.

    10.63 Analyze .  Given: amount, V, T of three gases. Find: P of each gas, total P.

    Plan . P = nRT/V; P t = P 1 + P 2 + P 3 + Solve .

    1. P He = nRT V = 0.765 mol × 0 .08206 L-atm mol-K × 298 K 10 .00 L = 1.871 = 1.87 atm

      P Ne = nRT V = 0.330 mol × 0 .08206 L-atm mol-K × 298 K 10 .00 L = 0.8070 = 0.807 atm

      P Ar = nRT V = 0.110 mol × 0 .08206 L-atm mol-K × 298 K 10 .00 L = 0.2690 = 0.269 atm

    2. P t = 1.871 atm + 0.8070 atm + 0.2690 atm = 2.9470 = 2.95 atm

    10.64  Given mass, V and T of O 2 and He, find the partial pressure of each gas. Sum to find the total pressure in the tank.

    V = 10.0 L; T = 19 C; 19 + 273 = 292 K

    n O 2 = 51 .2 g O 2 × 1 mol O 2 31 .999 g O 2 = 1 .600 = 1.60 mol O 2

    n He = 32 .6 g He × 1 mol He 4 .0026 g He = 8 .1447 = 8.14 mol He

    P O 2 = 1 .600 mol × 0 .08206 L-atm mol-K × 292 K 10 .0 L = 3 .8338 = 3.84 atm

    P He = 8 .1447 mol × 0 .08206 L-atm mol-K × 292 K 10 .0 L = 19 .5159 = 19.5 atm

    P t = 3.8338 + 19.5159 = 23.3497 = 23.3 atm

    10.65 Analyze . Given 407 ppm CO 2 in the atmosphere; 407 L CO 2 in 10 6 total L air. Find: the mole fraction of CO 2 in the atmosphere. Plan . Avogadro’s law deals with the relationship between volume and moles of a gas.

    Solve . Avogadro’s law states that volume of a gas at constant temperature and pressure is directly proportional to moles of the gas. Using volume fraction to express concentration assumes that the 407 L CO 2 and 10 6 total L air are at the same temperature and pressure. That is, 407 L is the volume that the number of moles of CO 2 present in 10 6 L air would occupy at atmospheric temperature and pressure. The mole fraction of CO 2 in the atmosphere is then just the volume fraction from the concentration by volume.

    X CO 2 = 407 L CO 2 10 6 L air = 0.000407

    10.66 X Xe = 4/100 = 0.04; X Ne = X He = (1 – 0.04)/2 = 0.48

    V t = 0.900 mm × 0.300 mm × 1 0 .0 mm × 1 cm 3 10 3 mm 3 × 1 L 1000 cm 3 = 2.70 × 10 6 L

    P t = 500 torr × 1 atm 760 torr = 0.657895 = 0.658 atm

    n t = PV RT = 0.657895 atm × mol-K 0 .08206 L-atm × 2. 70 × 10 6 L 298 K × 6 .022 × 10 23 atoms mol = 4.3743 × 10 16 = 4.37 × 10 16 total atoms


    Xe atoms = X Xe × total atoms = 0.04(4.3743 × 10 16 ) = 1.75 × 10 15 = 2 × 10 15 Xe atoms

    Ne atoms = He atoms = 0.48(4.3743 × 10 16 ) = 2.10 × 10 16 = 2.1 × 10 16 Ne and He atoms

    Assumptions: To calculate total moles of gas and total atoms, we assumed a reasonable room temperature. Because ‘4% Xe’ was not defined, we conveniently assumed mole percent. The 1:1 relationship of Ne to He is assumed to be by volume and not by mass.

    10.67 Analyze .  Given: mass CO 2 at V, T; pressure of air at same V, T. Find: partial pressure of CO 2 at these conditions, total pressure of gases at V, T.

    Plan . g CO 2 mol CO 2 P CO 2 (via P = nRT/V) ; P t = P CO 2 + P air Solve .

    5.50 g CO 2 × 1 mol CO 2 44.01 g  CO 2 = 0.12497 = 0 .125 mol CO 2 ; T = 273 + 24 °C = 297 K

    P CO 2 = 0.12497 mol × 297 K 10 .0 L × 0 .08206 L-atm mol-K = 0.30458 = 0.305 atm

    P air = 705 torr × 1 atm 760 torr = 0.92763 = 0.928 atm

    P t = P CO 2 + P air = 0.30458 + 0.92763 = 1.23221 = 1.232 atm

    (Result has 3 decimal places and 4 sig figs.)

    10.68 V C 2 H 5 OC 2 H 5 (l) mass C 2 H 5 OC 2 H 5 mol C 2 H 5 OC 2 H 5 P C 2 H 5 OC 2 H 5 (at V, T)

    P t = P N 2 + P O 2 + P C 2 H 5 OC 2 H 5 ; T = 273.15 + 35.0 °C = 308.15 = 308.2 K

    1. 5.00 mL C 2 H 5 OC 2 H 5 × 0 .7134 g C 2 H 5 OC 2 H 5 mL × 1 mol C 2 H 5 OC 2 H 5 74.12 g C 2 H 5 OC 2 H 5 = 0.048125 = 0.0481 mol C 2 H 5 OC 2 H 5

      P = nRT V = 0.048125 mol × 308 .15 6 .00 L × 0 .08206 L-atm mol-K = 0 .20282 = 0 .203 atm

    2. P t = P N 2 + P O 2 + P C 2 H 5 OC 2 H 5 = 0.751 atm + 0 .208 atm + 0 .203 atm = 1 .162 atm

    10.69 Analyze/Plan . When the sample is cooled, the water vapor condenses and all the gas pressure is because of CO 2 (g). The partial pressure CO 2 at 200 °C is equal to the mole fraction of CO 2 times the total pressure of the mixture. Apply Amonton’s law (see Solution 10.27) to the CO 2 pressures at the two temperatures. Solve .

    For a 3:1 mole ratio of CO 2 to H 2 O

    X CO 2 = 3 4 = 0.75 ; P CO 2 = 0.75 × 2. 00 atm = 1.50 atm

    T 1 = 200 °C + 273 = 473 K; T 2 = 10 °C + 273 = 283 K

    P 1 T 1 = P 2 T 2 or  P 2 = P 1 T 2 T 1 = 1.50 atm × 283 K 473 K = 0.8975 = 0.9 atm

    [The result has 1 sig fig if you consider the mole ratio to have 1 sig fig. If you think of the mole ratio as exact (experimentally unlikely), the result will have 3 sig figs.]


    10.70  T = 320 °C + 273 = 593 K; P t (593 K ) = P N 2 + P O 2

    1. Use Amonton’s law (see Solution 10.27) to calculate P N 2 at 593 K.
    2. Use stoichiometry to calculate mol O 2 produced by decomposing 5.15 g Ag 2 O.
    3. Use the ideal-gas law to calculate P O 2 at 593 K. Sum the pressures.
    1. P 1 = 760 torr = 1 atm; T 1 = 32 °C + 273 = 305 K; T 2 = 320 °C + 273 = 593 K

      P 1 T 1 = P 2 T 2 or  P 2 = P 1 T 2 T 1 = 1.00 atm × 593 K 305 K = 1.9443 = 1.94 atm

    2. 2 Ag 2 O(s) → 4 Ag(s) + O 2 (g)

      5 .15 g Ag 2 O × 1 mol Ag 2 O 231.74 g Ag 2 O × 1 mol O 2 2 mol Ag 2 O = 0.01111 = 0.0111 mol O 2

    3. V = 0.0750 L, T = 593 K, n = 0.0111

      P = nRT V ; P = 0 .01111 mol 0.0750 L × 0.08206 L-atm mol-K × 593 K = 4.876 = 7.209 = 7.21 atm

      P t (593 K ) = P N 2 + P O 2 = 1 .94 atm + 7 .21 atm = 9 .15 atm

    10.71 Analyze/Plan .  Mole fraction = pressure fraction. Find the desired mole fraction of O 2 and change to mole percent. Solve .

    X O 2 = P O 2 P t = 0.21 atm 8 .38 atm = 0.025 ; mole % = 0 .025 × 100 = 2 .5%

    10.72

    1. n O 2 = 15.08 g O 2 × 1 mol 31 .999 g = 0.4713 mol;  n N 2 = 8.17 g N 2 × 1 mol 28 .02 g = 0.292 mol

      n H 2 = 2.64 g H 2 × 1 mol 2 .016 g = 1.31 mol; n t = 0.4713 + 0.292 + 1.31 = 2.07 mol

      X O 2 = n O 2 n t = 0.4713 2.07 = 0.228 ; X N 2 = n N 2 n t = 0.292 2.07 = 0.141

      X H 2 = 1.31 2.07 = 0.633

    2. P O 2 = n × RT V ; P O 2 = 0.4713 mol × 0 .08206 L-atm mol-K × 288.15 K 15 .50 L = 0.7190 atm

      P N 2 = 0.292 mol × 0 .08206 L-atm mol-K × 288. 15 K 15 .50 L = 0.445 atm

      P H 2 = 1.31 mol × 0 .08206 L-atm mol-K × 288.15 K 15 .50 L = 2.00 atm

    10.73 Analyze/Plan .  N 2 (g) and O 2 (g) undergo changes of conditions and are mixed. Calculate the new pressure of each gas and add them to obtain the total pressure of the mixture.

    P 2 = P 1 V 1 T 2 /V 2 T 1 ; P t = P N 2 + P O 2 . Solve .

    P N 2 = P 1 V 1 T 2 V 2 T 1 = 5. 25 atm × 1 .00 L × 293 K 12.5 L × 299 K = 0.41157 = 0.412 atm

    P O 2 = P 1 V 1 T 2 V 2 T 1 = 5.25 atm × 5 .00 L × 293 K 12.5 L × 299 K = 2.05786 = 2.06 atm

    P t = 0.41157 atm + 2.05786 atm = 2.46943 = 2.47 atm


    10.74  Calculate the pressure of the gas in the second vessel directly from mass and conditions using the ideal-gas equation.

    1. P SO 2 = gRT M V = 3.00 g SO 2 64.07 g SO 2 / mol × 0 .08206 L-atm mol-K × 299 K 10 .0 L = 0.11489 = 0.115 atm
    2. P N 2 = gRT M V = 2.35 g N 2 28.01 g N 2 / mol × 0 .08206 L-atm mol-K × 299 K 10 .0 L = 0.20585 = 0.206 atm
    3. P t = P SO 2 + P N 2 = 0.11489 atm + 0 .20585 atm = 0 .321 atm

    Kinetic-Molecular Theory of Gases; Effusion and Diffusion (Sections 10.7 and 10.8)

    10.75

    1. Decrease. Increasing the container volume increases the distance between collisions and decreases the number of collisions per unit time.
    2. Increase. Increasing temperature increases the rms speed of the gas molecules, which increases the number of collisions per unit time.
    3. Decrease. Increasing the molar mass of a gas decreases the rms speed of the molecules, which decreases the number of collisions per unit time.

    10.76

    1. False. The average kinetic energy per molecule in a collection of gas molecules is the same for all gases at the same temperature.
    2. True.
    3. False. The molecules in a gas sample at a given temperature exhibit a distribution of kinetic energies.
    4. True.
    5. False. Gas molecules at the same temperature exhibit a distribution of speeds.

    10.77 Analyze/Plan . Given two gases, compare their rms speeds. Use Equation 10.23. Solve .

    u rms (He) u rms ( WF 6 ) = MM(WF 6 ) MM(He) = 297.9 g/mol 4 .003 g/mol = 8.63

    The rms speed of WF 6 is approximately 9 times slower than that of He.

    10.78  The gas undergoes a chemical reaction that has fewer gas particles in products than in reactants. Mass is conserved when a chemical reaction occurs, so the mass of (flask + contents) remains constant. Pressure is directly proportional to number of particles, so pressure decreases as the number of gaseous particles decreases. One simple example of such a reaction is the dimerization of NO 2 : 2 NO 2 (g) → N 2 O 4 .

    10.79 Analyze/Plan . Apply the concepts of the Kinetic-Molecular Theory (KMT) to the situation where a gas is heated at constant volume. Determine how the quantities in (a)–(d) are affected by this change. Solve .

    1. Average kinetic energy is proportional to temperature (K), so average kinetic energy of the molecules increases.
    2. The average kinetic energy of a gas is 1/2 m u rms 2 . Molecular mass doesn’t change as T increases; average kinetic energy increases so rms speed (u) increases. (Also, u rms = (3RT/MM) 1/2 , so u rms is directly related to T.)

    1. As T and thus rms molecular speed increase, molecular momentum (mu) increases and the strength of an average impact with the container wall increases.
    1. As T and rms molecular speed increase, the molecules collide more frequently with the container walls, and the total number of collisions per second increases.

    10.80

    1. They have the same number of molecules (equal volumes of gases at the same temperature and pressure contain equal numbers of molecules).
    2. N 2 is more dense because it has the larger molar mass. Because the volumes of the samples and the number of molecules are equal, the gas with the larger molar mass will have the greater density.
    3. The average kinetic energies are equal (statement 5, section 10.7).
    4. CH 4 will effuse faster. The lighter the gas molecules, the faster they will effuse (Graham’s law).

    10.81

    1. Plan .  The larger the molar mass, the slower the average speed (at constant temperature).

      Solve .  In order of increasing speed (and decreasing molar mass):

      HBr < NF 3 < SO 2 < CO < Ne

    2. Plan .  Follow the logic of Sample Exercise 10.13. Solve.

      u rms = 3 RT MM = ( 3 × 8.314 kg-m 2 / s 2 -mol-K × 298 K 71.0 × 10 3 kg/mol ) 1 / 2 = 324 m/s

    3. Plan .  Use Equation 10.21 to calculate the most probable speed, u mp . MM of O 3 = 48.0 g/mol; T = 270 K. Solve.

      u mp = 2 RT MM = ( 2 × 8.314 kg-m 2 / s 2 -mol-K × 270 K 48.0 × 10 3 kg/mol ) 1 / 2 = 306 m/s

    10.82

    1. Plan .  The greater the molecular (and molar) mass, the smaller the rms and average speeds of the molecules. Calculate the molar mass of each gas, and place them in decreasing order of mass and increasing order of rms and average speed.

      Solve .  CO = 28 g/mol; SF 6 = 146 g/mol; H 2 S = 34 g/mol; Cl 2 = 71 g/mol;

      HBr = 81 g/mol. In order of increasing speed (and decreasing molar mass):

      SF 6 < HBr < Cl 2 < H 2 S < CO

    2. Plan .  Follow the logic of Sample Exercise 10.13. Solve .

      CO : u rms = 3 RT MM = ( 3 × 8.314 kg-m 2 / s 2 -mol-K × 300 K 28.0 × 10 3 kg/mol ) 1 / 2 = 5.17 × 10 2 m/s

      Cl 2 : u rms = ( 3 × 8.314 kg-m 2 / s 2 -mol-K × 300 K 70.9 × 10 3 kg/mol ) 1 / 2 = 3.25 × 10 2 m/s

      As expected, the lighter molecule moves at the greater speed.

    3. Plan .  From Equations 10.20 and 10.21, we see that the ratio of most probable speed to rms speed is (2/3) 1/2 . Use this ratio and the results from part (b) to calculate most probable speeds. Solve .

    1. CO : u mp = (2/3) 1/2 (5.17 × 10 2 m/s) = 422 m/s

    2. Cl 2 : u mp = (2/3) 1/2 (3.25 × 10 2 m/s) = 265 m/s

    3. The lighter molecule, CO, has the greater most probable speed. Note that the most probable speed is less than the rms speed, as shown on Figure 10.12(b).

    10.83  Statements (a) and (d) are true. Statement (b) is false because effusion is the escape of gas molecules through a tiny hole, while diffusion is the distribution of a gas throughout space or throughout another substance. Statement (c) is false because perfume molecules travel to your nose by the process of diffusion, not effusion.

    10.84  Write each proportionality relationship as an equation, then combine them to obtain a formula for mean free path.

    The operational symbols and units are: mean free path, λ, meters (m); temperature, T, kelvins (K); pressure, P, atmospheres (atm); diameter of a gas molecule, d, meters (m), constant, R mfp .

    λ = constant × T; λ = constant/P; λ = constant/d 2

    Combining : λ = R mfp × T P × d 2

    The units of R mfp are chosen and arranged so that they cancel the units of measurement, leaving an appropriate length unit for λ.

    With the units defined above, R mfp will have units of atm-m 3 K .

    (Note that 1 m 3 = 10 3 dm 3 = 1000 L. Substituting, R mfp would have units of atm-L K , with the factor of 1000 incorporated into the value of R mfp .)

    10.85 Plan .  The heavier the molecule, the slower the rate of effusion. Thus, the order for increasing rate of effusion is in the order of decreasing mass. Solve .

    rate 2 H 37 Cl < rate 1 H 37 Cl < rate 2 H 35 Cl < rate 1 H 35 Cl

    10.86 rate 235 U rate 238 U = 238.05 235.04 = 1.0128 = 1.0064

    There is a slightly greater rate enhancement for 235 U(g) atoms than 235 UF 6 (g) molecules (1.0043), because 235 U is a greater percentage (100%) of the mass of the diffusing particles than in 235 UF 6 molecules. The masses of the isotopes were taken from The Handbook of Chemistry and Physics .

    10.87 Analyze .  Given: relative effusion rates of two gases at same temperature. Find: molecular formula of one of the gases. Plan . Use Graham’s law to calculate the formula weight of arsenic(III) sulfide, and thus the molecular formula. Solve .

    rate (sulfide) rate (Ar) = [ 39.9 MM ( sulfide ) ] 1 / 2 = 0.28

    MM (sulfide) = 39.9 / 0.28 2 = 509 g/mol (two significant figures)

    The empirical formula of arsenic(III) sulfide is As 2 S 3 , which has a formula mass of 246.1. Twice this is 490 g/mol, close to the value estimated from the effusion experiment. Thus, the formula of the gas phase molecule is As 4 S 6 .


    10.88  The time required is proportional to the reciprocal of the effusion rate.

    rate (X) rate (O 2 ) = 105 s 31 s = [ 32 g O 2 MM x ] 1 / 2 ; MM x = 32 g O 2 × [ 105 31 ] 2 = 370 g/mol (two sig figs)

    Nonideal-Gas Behavior (Section 10.9)

    10.89

    1. Nonideal-gas behavior is observed at very high pressures and/or low temperatures.
    2. The real volumes of gas molecules and attractive intermolecular forces between molecules cause gases to behave nonideally.

    10.90  Ideal-gas behavior is most likely to occur at high temperature and low pressure, so the atmosphere on Mercury is more likely to obey the ideal-gas law. The higher temperature on Mercury means that the kinetic energies of the molecules will be larger relative to intermolecular attractive forces. Further, the gravitational attractive forces on Mercury are lower because the planet has a much smaller mass. This means that for the same column mass of gas (Figure 10.1), atmospheric pressure on Mercury will be lower.

    10.91  Statement (b) is true. The constants a and b are characteristic of a particular gas and are independent of pressure and temperature.

    10.92 Plan .  The constants a and b are part of the correction terms in the van der Waals equation. The smaller the values of a and b , the smaller the corrections and the more ideal the gas. Solve .

    Ar ( a = 1.34, b = 0.0322) will behave more like an ideal gas than CO 2 ( a = 3.59, b = 0.0427) at high pressures.

    10.93 Analyze/Plan . Follow the logic in Sample Exercise 10.15. Use the ideal-gas equation to calculate pressure in (a), the van der Waals equation in (b). n = 1.00 mol, V = 5.00 L, T = 25 C = 298 K; a = 6.49 L 2 -atm/mol 2 , b = 0.0562 L/mol.

    1. P = nRT V = 1.00 mol × 298 K 5.00 L × 0.08206 L-atm mol-K = 4.89 atm
    2. P = nRT V n b n 2 a V 2 ;

      P = (1 .00 mol)(298 K)(0 .08206 L-atm/mol-K) 5 .00 L (1 .00 mol)(0 .0562 L/mol) ( 1.00 mol ) 2 ( 6.49 L 2 -atm/mol 2 ) ( 5.00 L ) 2

      P = 4.9463 atm – 0.2596 atm = 4.6868 = 4.69 atm

    3. From Sample Exercise 10.15, the difference at 22.41 L between the ideal and van der Waals results is (1.00 – 0.990) = 0.010 atm. At 5.00 L, the difference is (4.89 – 4.69) = 0.20 atm. The effects of both molecular attractions, the a correction, and molecular volume, the b correction, increase with decreasing volume. For the a correction, V 2 appears in the denominator, so the correction increases exponentially as V decreases. For the b correction, n b is a larger portion of the total volume as V decreases. That is, 0.0562 L is 1.1% of 5.0 L, but only 0.25% of 22.41 L. Qualitatively, molecular attractions are more important as the amount of free space decreases and the number of molecular collisions increase. Molecular volume is a larger part of the total volume as the container volume decreases.

    10.94 Analyze .  Conditions and amount of CCl 4 (g) are given. Plan . Use ideal-gas equation and van der Waals equation to calculate pressure of gas at these conditions. Solve .

    1. P = 1 .00 mol × 0 .08206 L-atm mol-K × 353 K 3 3.3 L = 0.870 atm
    2. P = nRT V nb an 2 V 2 = 1.00 × 0.08206 × 353 33.3 ( 1.00 × 0.1383 ) 20.4 ( 1.00 ) 2 ( 33.3 ) 2 = 0.855 atm

      Check .  The van der Waals result indicates that the real pressure will be less than the ideal pressure. That is, intermolecular forces reduce the effective number of particles and the real pressure. This is reasonable for 1 mole of gas at relatively low temperature and pressure.

    3. According to Table 10.3, CCl 4 has larger a and b values. That is, CCl 4 experiences stronger intermolecular attractions and has a larger molecular volume than Cl 2 does. CCl 4 will deviate more from ideal behavior at these conditions than Cl 2 will.

    10.95 Analyze . Given the b value of Xe, 0.0510 L/mol, calculate the radius of a Xe atom.

    Plan . Use Avogadro’s number to change L/mol to L/atom. Use the volume formula,

    V = 4/3 πr 3 and units conversion to obtain the radius in Å. 1 L = 1 dm 3 . Solve .

    0.0510 L 1 mol X e × 1 mol Xe 6 .022 × 10 23 Xe atoms × 1 dm 3 1 L = 8.4689 × 10 26 = 8.47 × 10 26 dm 3

    V = 4/3 π r 3 ; r 3 = 3 V/ 4 π ; r = ( 3 V/ 4 π ) 1 / 3

    r = ( 3 × 8.4689 × 10 26 dm 3 4 × 3.14159 ) 1 / 3 = 2.7243 × 10 9 = 2.72 × 10 9 dm

    2.72 × 10 9 dm × 1 m 10 dm × 1 Å 1 × 10 10 m = 2.72 Å

    The calculated value is the nonbonding radius. From Figure 7.7 in Section 7.3, the bonding atomic radius of Xe is 1.40 Å. We expect the nonbonding radius of an atom to be larger than the bonding radius, but our calculated value is nearly twice as large.

    10.96  The van der Waals radius we calculate from the b parameter in Table 10.3 is more closely associated with the nonbonding atomic radius of an atom. From section 7.3, the nonbonding or van der Waals radius is half of the shortest internuclear distance when two nonbonding atoms collide. So, radii calculated from the van der Waals equation are nonbonding radii. According to the kinetic-molecular theory, ideal-gas particles undergo perfectly elastic, billiard-ball collisions, in keeping with the definition of nonbonding radii.

    Also, from the results of Exercise 10.95, the atomic radius calculated from the van der Waals b value is twice as large as the bonding atomic radius from Figure 7.7. Nonbonding radii are larger than bonding radii because no lasting penetration of electron clouds occurs during a nonbonding collision.


    Additional Exercises

    10.97 Analyze . The height of an Hg column is 760 mm at a pressure of 1.01 × 10 5 Pa. Plan . Develop a relationship between pressure, height of a column of liquid, and density of the liquid. Relationships that might prove useful: P = F/A; F = m × g; m = d × V; V = A × height Solve .

    P = F A = m × g A = d × V × g A = d × A × h × g A = d × h × g;  d = P h × g

    g = 9.81 m/s 2 ; h = 760 mm = 0.760 m; 1 Pa = 1 kg/m-s 2

    d = P h × g = 1 . 0 1 × 1 0 5 kg/m-s 2 0.760 m × 9 .81 m/s 2 = 1.3547 × 10 4 kg m 3 = 1.35 × 10 4 kg m 3

    d = 1.3547 × 10 4 kg m 3 × 1000 g kg × 1 m 3 ( 100 ) 3 cm 3 × 1 cm 3 1 mL = 13.5 g/mL

    10.98  P 1 V 1 = P 2 V 2 ; V 2 = P 1 V 1 /P 2

    V 2 = 3.0 atm × 1 .0 mm 3 730 torr × 760 torr 1 atm = 3.1 mm 3

    10.99  PV = nRT, n = PV/RT. Because RT is constant, n is proportional to PV.

    Total available n = (15.0 L × 1.00 × 10 2 atm) – (15.0 L × 1.00 atm) = 1485 = 1.49 × 10 3 L-atm

    Each balloon holds 2.00 L × 1.00 atm = 2.00 L-atm

    1485 L-atm available 2.00 L-atm/balloon = 742.5 = 742 balloons

    (Only 742 balloons can be filled completely, with a bit of He left over.)

    10.100 P = nRT V ; n = 1.4 × 10 5 mol, V = 0.600 L, T = 23 ° C = 296 K

    P = 1 .4 × 10 5 mol × 0 .08206 L-atm mol-K × 296 K 0 .600 L = 5.7 × 10 4 atm = 0 .43 mm Hg

    10.101

    1. Change mass CO 2 to mol CO 2 . P = 1.00 atm, T = 27 C = 300 K.

      6 × 10 6 tons CO 2 × 2000 lb ton × 453 .6 g lb × 1 mol CO 2 44 .01 g CO 2 = 1.237 × 10 11 = 1 × 10 11 mol

      V = nRT P = 1 .237 × 10 11 mol × 300 K 1 .00 atm × 0 .08206 L-atm mol-K = 3.045 × 10 12 = 3 × 10 12 L

    2. 1.237 × 10 11 mol CO 2 × 44 .01 g CO 2 mol CO 2 × 1 cm 3 1 .2 g × 1 L 1000 cm 3 = 4.536 × 10 9 = 5 × 10 9 L
    3. n = 1.237 × 10 11 mol, P = 70 atm, T = 30 C = 303 K

      V = nRT P = 1 .237 × 10 11 mol × 303 K 70 atm × 0 .08206 L-atm mol-K = 4 .3939 × 10 10 = 4 × 10 10 L


    10.102

    1. n = PV RT = 3 .00 atm × mol-K 0 .08206 L-atm × 110 L 300 K = 13 .4 mol C 3 H 8 (g)
    2. 0 .590 g C 3 H 8 (l) 1 mL × 110 × 10 3 mL × 1 mol C 3 H 8 44.094 g = 1.47 × 10 3 mol C 3 H 8 (l)
    3. Using C 3 H 8 in a 110-L container as an example, the ratio of moles liquid to moles gas that can be stored in a certain volume is 1.47 × 10 3 mol liquid 13.4 mol gas = 110.

      A container with a fixed volume holds many more moles (molecules) of C 3 H 8 (l) because in the liquid phase the molecules are touching. In the gas phase, the molecules are far apart (statement 2, Section 10.7), and many fewer molecules will fit in the container.

    10.103 Vol of room = 12 ft × 20 ft × 9 ft × 12 3 in . 3 1 ft 3 × 2.54 3 cm 3 1 3 in . 3 × 1 L 1000 cm 3 = 61 , 164 = 6 × 10 4 L

    Calculate the total moles of gas in the laboratory at the conditions given.

    n t = PV RT = 1.00 atm × mol-K 0 .08206 L-atm × 61 , 164 L 297 K = 2510 = 3 × 10 3 mol gas

    A Ni(CO) 4 concentration of 1 part in 10 9 means 1 mol Ni(CO) 4 in 1 × 10 9 total moles of gas.

    x mol Ni(CO) 4 2. 510 × 10 3 mol gas = 1 1 × 10 9 = 2.510 × 10 6 = 3 × 10 6 mol Ni(CO) 4

    2. 510 × 10 6 mol Ni(CO) 4 × 170 .74 g Ni(CO) 4 1 mol Ni(CO) 4 = 4.286 × 10 3 = 4 × 10 3 g = 4 mg Ni(CO) 4

    10.104

    1. mol = g/MM; assume mol Ar = mol X;

      g Ar 39.948 g/mol = g X MM X ; 3 .224 g Ar 39.948 g/mol = 8.102 g X MM X

      MM X = (8 .102 g X ) ( 39.948 g/mol ) 3.224 g Ar = 100.39 = 100.4 g/mol

    2. Assume mol Ar = mol X. For gases, PV= nRT and n = PV/RT. For moles of the two gases to be equal, the implied assumption is that P, V, and T are constant. Because we use the same container for both gas samples, constant V is a good assumption. The values of P and T are not explicitly stated.

      We also assume that the gases behave ideally. At ambient conditions, this is a reasonable assumption.

    10.105   It is simplest to calculate the partial pressure of each gas as it expands into the total volume, then sum the partial pressures.

    P 2 = P 1 V 1 /V 2 ; P N 2 = 265 torr (1 .0 L/2 .5 L) = 106 = 1 .1 × 10 2 torr

    P Ne = 800 torr (1.0 L/2.5 L) = 320 = 3.2 × 10 2 torr

    P H 2 = 532 torr (0 .5 L/2 .5 L) = 106 = 1.1 × 10 2 torr

    P t = P N 2 + P Ne + P H 2 = ( 106 + 320 + 106 ) torr = 532 = 5 .3 × 10 2 torr


    10.106

    1. n = PV RT = 0.980 atm × mol-K 0.08206 L-atm × 0.524 L 347 K = 0.018034 = 0.0180 mol air

      mol O 2 = 0.018034 mol air × 0 .2095 mol O 2 1 mol air = 0.003778 = 0.00378 mol O 2

    2. C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g)

      (The H 2 O produced in an automobile engine is in the gaseous state.)

      0.003778 mol O 2 × 1 mol C 8 H 18 12.5 mol O 2 × 114.2 g C 8 H 18 1 mol C 8 H 18 = 0.0345 g C 8 H 18

    10.107

    1. Pressure percent = mole percent. Change pressure/mole percents to mole fraction.

      Partial pressure of each gas is mole fraction ( X ) times total pressure. P x = X x P t

      P N 2 = 0.748(0.985 atm) = 0.737 atm; P O 2 = 0.153(0.985 atm) = 0.151 atm

      P CO 2 = 0.037(0.985 atm) = 0.03645 = 0.036 atm

      P H 2 O = 0.062(0.985 atm) = 0.06107 = 0.061 atm

    2. PV = nRT, n = PV/RT; P = 0.036 atm, V = 0.455 L, T = 37 C = 310 K

      n = 0.03645 atm × 0.455 L 310 K × mol-K 0.08206 L-atm = 6.520 × 10 –4 = 6.5 × 10 –4 mol

    3. C 6 H 12 O 6 + 6 O 2 → 6 CO 2 + 6 H 2 O

      6 .520 × 10 4 mol CO 2 × 1 mol C 6 H 12 O 6 6 mol CO 2 × 180 .15 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 = 0.01958 = 0.020 g C 6 H 12 O 6

    10.108   V and T are the same for He and O 2 .

    P He V = n He RT,    P He /n He = RT/V;   P O 2 /n O 2 = RT/V

    P He n He = P O 2 n O 2 = n O 2 × P O 2 × n He P He ; n He = 1.42 g He × 1 mol He 4.003 g He = 0.3547 = 0.355 mol He

    n O 2 = 158 torr 42.5 torr × 0.355 mol = 1.3188 = 1.32 mol O 2 ; 1.3188 mol O 2 × 32.00 g O 2 1 mol O 2 = 42.2 g O 2

    10.109   At constant temperature, an ideal gas at a certain pressure and volume, P 1 V 1 , expands into a larger volume and lower pressure, P 2 V 2 . This is a Boyle’s law problem.

    Let V 1 = x L, V 2 = 0.800 L + x. Change 1.50 atm to torr, so the pressure units cancel.

    1.50 atm × 760 torr/atm = 1140 torr

    P 1 V 1 = P 2 V 2 . 1140 torr (x L) = 695 torr [(0.800 + x) L]. Units cancel.

    1140 x = 556 + 695 x; 445 x = 556; x = 1.25 L

    Check. 1140 (1.25) = 695 (2.05); 1425 = 1425. The algebra is correct.


    10.110

    1. The quantity d/P = MM/RT should be a constant at all pressures for an ideal gas. It is not, however, because of nonideal behavior. If we graph d/P versus P, the ratio should approach ideal behavior at low P. At P = 0, d/P = 2.2525. Using this value in the formula MM = d/P × RT, MM = 50.46 g/mol.
      image
    2. The ratio d/P varies with pressure because of the finite volumes of gas molecules and attractive intermolecular forces.

    10.111   Calculate the number of moles of Ar in the vessel:

    n = (339.854 – 337.428)/39.948 = 0.060729 = 0.06073 mol

    The total number of moles of the mixed gas is the same (Avogadro’s law). Thus, the average atomic weight is (339.076 – 337.428)/0.060729 = 27.137 = 27.14. Let the mole fraction of Ne be x. Then,

    x (20.183) + (1 – x ) (39.948) = 27.137; 12.811 = 19.765 x; x = 0.6482

    Neon is thus 64.82 mole percent of the mixture.

    10.112 u rms = (3RT/MM) 1/2 ; u rms2 = 2 u rms1 ; T 1 = −33 °C = 240 K

    u rms1 = (3RT 1 /MM) 1/2 ; u rms1 2 = 3(240)R/MM = 720R/MM

    u rms1 = (720R/MM) 1/2 ; u rms2 = 2 u rms1 = (2)(720R/MM) 1/2

    (2)(720R/MM) 1/2 = (3RT 2 /MM) 1/2

    (2) 2 (720R/MM) = 3RT 2 /MM; (2) 2 (720) = 3T 2

    T 2 = (4)(720)/3 = 960 K = 687 C

    Increasing the rms speed (u) by a factor of 2 requires heating to 960 K (or 687 C), increasing the temperature by a factor of 4.

    10.113

    1. Assumption 3 states that attractive and repulsive forces between molecules are negligible. All gases in the list are nonpolar. The largest and most structurally complex molecule, SF 6 , is most likely to depart from this assumption.
    2. The monatomic gas Ne is smallest and least structurally complex, so it will behave most like an ideal gas.
    3. Root-mean-square speed is inversely related to molecular mass. The lightest gas, CH 4 , has the highest rms speed.

    1. The heaviest and most structurally complex is SF 6 . Also, S and F have larger atomic radii than C and H; this means that S–F bonds will be longer than C–H bonds and the volume of SF 6 will be greater than that of CH 4 . It is reasonable to assume that SF 6 will occupy the greatest molecular volume relative to total volume. A quantitative measure is the b value in Table 10.3, with units of L/mol. Unfortunately, SF 6 does not appear in Table 10.3.
    1. Average kinetic energy is only related to absolute (K) temperature. At the same temperature, they all have the same average kinetic-molecular energy.
    1. Rate of effusion is inversely related to molecular mass. The lighter the molecule, the faster it effuses. Ne and CH 4 have smaller molecular masses and effuse faster than N 2 .
    1. If SF 6 occupies the greatest molecular volume [see part (d)], we expect it to have the largest van der Waals b parameter.

    10.114

    1. The effect of intermolecular attraction becomes more significant as a gas is compressed to a smaller volume at constant temperature. This compression causes the pressure, and thus the number of intermolecular collisions, to increase. Intermolecular attraction causes some of these collisions to be inelastic, which amplifies the deviation from ideal behavior.
    2. The effect of intermolecular attraction becomes less significant as the temperature of a gas is increased at constant volume. When the temperature of a gas is increased at constant volume, the pressure of the gas, the number of intermolecular collisions, and the average kinetic energy of the gas particles increase. This higher average kinetic energy means that a larger fraction of the molecules has sufficient kinetic energy to overcome intermolecular attractions, even though there are more total collisions. This increases the fraction of elastic collisions, and the gas more closely obeys the ideal-gas equation.

    10.115   The larger and heavier the particle, in this case a single atom, the more likely it is to deviate from ideal behavior. Other than Rn, Xe is the largest (atomic radius = 1.40 Å), heaviest (molar mass = 131.3 g/mol) and most dense (5.90 g/L) noble gas. Its sus-ceptibility to intermolecular interactions is also demonstrated by its ability to form compounds like XeF 4 .

    10.116

    1. At STP, the molar volume = 1 mol × 0 .08206 L-atm mol-K × 273 K 1 atm = 22.4 L

      Dividing the value for b , 0.0322 L/mol, by 4, we obtain 0.00805 L. Thus, the volume of the Ar atoms is (0.00805/22.4)100 = 0.0359% of the total volume.

    2. At 200 atm pressure (and 0 C, standard temperature) the molar volume is 0.112 L, and the volume of the Ar atoms is 7.19% of the total volume.

    10.117

    1. 120.00 kg N 2 ( g ) × 1000 g 1 kg × 1 mol N 2 28.0135 g N = 4283.6 mol N 2

      P = nRT V = 4283.6 mol × 0 .08206 L-atm mol-K × 553 K 1100 .0 L = 176.72 = 177 atm


    1. Rearranging Equation [10.25] to isolate P, P = nRT V n b n 2 a V 2

      P = ( 4283.6 mol ) ( 0.08206 L-atm/mol-K ) ( 553 K ) 1100.0 L ( 4283 .6 mol ) ( 0.0391 L/mol ) ( 4283.6 mol ) 2 ( 1.39 L 2 -atm/mol 2 ) ( 1100.0 L ) 2 P = 194 , 388 L-atm 1100.0 L- 167.5 L 21.1 atm = 208.5 atm 21.1 atm = 187.4 atm

    1. The pressure corrected for the real volume of the N 2 molecules is 208.5 atm, 31.8 atm higher than the ideal pressure of 176.7 atm. The 21.1 atm correction for intermolecular forces reduces the calculated pressure somewhat, but the “real” pressure is still higher than the ideal pressure. The correction for the real volume of molecules dominates. Even though the value of b is small, the number of moles of N 2 is large enough so that the molecular volume correction is larger than the attractive forces correction.

    Integrative Exercises

    10.118

    1. MM = gRT VP = 1 .56 g 1 .00 L × 0.08206 L-atm mol-K × 323 K 0 .984 atm = 42.0 g/mol

      Assume 100 g cyclopropane

      100 g × 0 .857 C = 85.7 g C × 1 mol C 12 .01 g = 7.136 mol C 7.136 = 1 mol C

      100 g × 0 .143 H = 14.3 g H × 1 mol H 1 .008 g = 14.19 mol H 7.136 = 2 mol H

      The empirical formula of cyclopropane is CH 2 and the empirical formula weight is 12 + 2 = 14 g. The ratio of molar mass to empirical formula weight, 42.0 g/14 g, is 3; therefore, there are three empirical formula units in one cyclopropane molecule. The molecular formula is 3 × (CH 2 ) = C 3 H 6 .

    2. Ar is a monoatomic gas. Cyclopropane molecules are larger and more structurally complex, even though the molar masses of Ar and C 3 H 6 are similar. If both gases are at the same relatively low temperature, they have approximately the same average kinetic energy, and the same ability to overcome intermolecular attractions. We expect intermolecular attractions to be more significant for the more complex C 3 H 6 molecules, and that C 3 H 6 will deviate more from ideal behavior at the conditions listed. This conclusion is supported by the a values in Table 10.3. The a values for CH 4 and CO 2 , more complex molecules than Ar atoms, are larger than the value for Ar. If the pressure is high enough for the volume correction in the van der Waals equation to dominate behavior, the larger C 3 H 6 molecules definitely deviate more than Ar atoms from ideal behavior.
    3. Cyclopropane, C 3 H 6 , MM = 42.0 g/mol; methane, CH 4 MM = 16.0. Rate of effusion through a pinhole is inversely related to molar mass. Cyclopropane would effuse through a pinhole slower than methane, because it has the greater molar mass.

    10.119 Plan .  Write the balanced equation for the combustion of methanol. Because amounts of both reactants are given, determine the limiting reactant. Use mole ratios to calculate moles H 2 O produced, based on the amount of limiting reactant. Change moles to grams H 2 O, then use density to calculate volume of H 2 O(l) produced. Assume the condensed H 2 O(l) is at 25 o C, where density = 0.99707 g/mol. Solve .

    methanol = CH 3 OH(l).  2 CH 3 OH(l) + 3 O 2 (g) → 2 CO 2 (g) + 4 H 2 O(g)

    25.0 mL CH 3 OH × 0 .850 g CH 3 OH mL × 1 mol CH 3 OH 32.04 g = 0.6632 = 0.663 mol CH 3 OH

    mol O 2 = n = PV RT = 1.00 atm × 12 .5 L 273 K × mol-K 0 .08206 L-atm = 0.5580 = 0.558 mol O 2

    0.558 mol O 2 × 2 mol CH 3 OH 3 mol O 2 = 0.372 mol CH 3 OH

    0.558 mol O 2 can react with only 0.372 mol CH 3 OH, so O 2 is the limiting reactant. Note that a large volume of O 2 (g) is required to completely react with a relatively small volume of CH 3 OH(l).

    0 .558 mol O 2 × 4 mol H 2 O 3 mol O 2 × 1 8.02 g H 2 O 1 mol H 2 O × 1 mL H 2 O 0.99707 g H 2 O = 13.446 = 13.4 mL H 2 O

    10.120

    1. Get g C from mL CO 2 ; get g H from mL H 2 O. Also calculate mol C and H, to use in part (b). Get g N by subtraction. Calculate % composition.

      n = PV/RT. At STP, P = 1 atm, T = 273 K. (STP implies an infinite number of sig figs.)

      n C O 2 = 0.08316 L × 1 atm 273 K × mol-K 0 .08206 L-atm = 0.003712 mol CO 2

      0.003712 mol CO 2 × 1 mol C 1 mol CO 2 × 12.0107 g C mol C = 0.044585 = 0.04458 g CO 2

      n H 2 O = 0.07330 L × 1 atm 273 K × mol-K 0 .08206 L-atm = 3.2720 × 10 3 = 3.272 × 10 –3 mol H 2 O

      3.2720 × 10 3 mol H 2 O × 2 mol H 1 mol H 2 O × 1.00794 g H mol H = 6.5959 × 10 3 = 6.596 × 10 –3 g H

      mass % X = mass X sample mass × 100; sample mass = 100.0 mg = 0.1000 g

      % C = 0 .044585 g 0 .1000 g × 100 = 44.585 = 44.58 % C

      % H = 6 .5959 × 10 3 g H 0 .1000 g × 100 = 6.5959 = 6.596 % H

      % Cl = 0 .01644 g Cl 0 .1000 g × 100 = 16.44 % Cl

      % N = 100 – 44.58 – 6.596 – 16.44 = 32.38% N

    2. 0.003712 mol C; 2(3.272 × 10 –3 ) = 6.544 × 10 –3 mol H

      0.01644 g Cl × 1 mol C 35 .453 g Cl = 4.637 × 10 –4 mol Cl

      0.1000 g sample × 0.3238 mass fraction N = 0.03238 g N

      0.03238 g N × 1 mol N 14 .0067 g N = 0.0023118 = 0.002312 mol N


    1. Divide by the smallest number of mol to find the simplest ratio of moles.

    2. 0 .003712 mol C 4 .637 × 10 4 = 8.005 C

    3. 6 .544 × 10 3 mol H 4 .637 × 10 4 = 14.11 H

    4. 4 .637 × 10 4 mol Cl 4 .637 × 10 4 = 1.000 Cl

    5. 0 .002312 mol N 4 .637 × 10 4 = 4.985 N

    6. If we assume 14.11 is “close” to 14 (a reasonable assumption), the empirical formula is C 8 H 14 N 5 Cl.

    1. Molar mass of the compound is required to determine molecular formula when the empirical formula is known.

    10.121

    1. Plan .  Use the ideal-gas law to calculate the moles CO 2 that react.

      Solve. P(reacted) = P(initial) – P(final), at constant V, T. Because both CaO and BaO react with CO 2 in a 1:1 mole ratio, mol CaO + mol BaO = mol CO 2 . Use molar masses to calculate % CaO in sample.

      P(reacted) = 730 torr 150 torr = 580 torr; 580 torr × 1 atm 760 torr = 0.76316 = 0.763 atm

      n = PV RT = 0.76316 atm × 1 .0 L 298 K × mol-K 0 .08206 L-atm = 0.03121 = 0.0312 mol CO 2

    2. Plan. Use the stoichiometry of the reaction and definition of moles to calculate the mass and Mass % of CaO.

      Solve . CaO(s) + CO 2 (s) → CaCO 3 (s). BaO(s) + CO 2 (g) → BaCO 3 (s)

      mol CO 2 reacted = mol CaO + mol BaO

      Let x = g CaO, 4.00 – x = g BaO

      0.03121 = x 56 .08 + 4.00 x 153 .3

      0.03121(56.08)(153.3) = 153.3x + 56.08(4.00 – x)

      268.3 = (153.3x – 56.08x) + 224.3

      43.98 = 97.22x, x = 0.452 = 0.45 g CaO

      0.452 g CaO 4.00 g sample × 100 = 11.3 = 11 % CaO

      (By strict sig fig rules, the result has 2 sig figs, because 268 – 224 = 44 has 0 decimal places and 2 sig figs.)

    10.122

    1. 5.00 g HCl × 1 mol HCl 36.46 g HCl = 0.1371 = 0.137 mol HCl

      5.00 g NH 3 × 1 mol NH 3 17.03 g NH 3 = 0.2936 = 0.294 mol NH 3


    1. The gases react in a 1:1 mole ratio, HCl is the limiting reactant and is completely consumed. (0.2936 mol – 0.1371 mol) = 0.1565 = 0.157 mol NH 3 remain in the system. NH 3 (g) is the only gas remaining after reaction.

    1. V t = 4.00 L. P = nRT V = 0.1565 mol × 0 .08206 L-atm mol-K × 298 K 4 .00 L = 0.957 atm
    1. 0.137 mol HCl × 1 mol NH 4 Cl 1 mol HCl × 5 3.49 g NH 4 Cl 1 mol NH 4 Cl = 7.3284 = 7.33 g NH 4 Cl

    10.123 n = PV RT = 1.00 atm × mol-K 0 .08206 L-atm × 2.7 × 10 12 L 273 K = 1.205 × 10 11 = 1.2 × 10 11 mol CH 4

    CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)      ΔH = –890.4 kJ

    (At STP, H 2 O is in the liquid state.)

    Δ H rxn o = Δ H f o CO 2 (g) + 2 Δ H f o H 2 O(l) Δ H f o CH 4 ( g) 2 Δ H f o O 2 (g)

    Δ H rxn o = 393.5 kJ + 2 ( 285.83 kJ) ( 74.8 kJ) 0 = 890.4 kJ

    890.4 kJ 1 mol CH 4 × 1.205 × 10 11 mol CH 4 = 1.073 × 10 14 = 1.1 × 10 14 kJ

    The negative sign indicates heat evolved by the combustion reaction.

    10.124

    1. 19 e , 9.5 e pairs
      image

      Resonance structures can be drawn with the odd electron on O, but electronegativity considerations predict that it will be on Cl for most of the time.

    2. ClO 2 is very reactive because it is an odd-electron molecule. Adding an electron (reduction) both pairs the odd electron and completes the octet of Cl. Thus, ClO 2 has a strong tendency to gain an electron and be reduced.
    3. ClO 2 , 20 e , 10 e pairs
      image
    4. 4 e– domains around Cl, O–Cl–O bond angle ~107 (<109 owing to repulsion by nonbonding domains)
    5. Calculate mol Cl 2 from ideal-gas equation; determine limiting reactant; mass ClO 2 via mol ratios.

      mol Cl 2 = PV RT = 1.50 atm × 2 .00 L 294 K × mol-K 0 .08206 L-atm = 0.1243 = 0.124 mol Cl 2

      15.0 g NaClO 2 × 1 mol NaClO 2 90.44 g = 0.1659 = 0.166 mol NaClO 2

      2 mol NaClO 2 are required for 1 mol Cl 2 , so NaClO 2 is the limiting reactant.

      For every 2 mol NaClO 2 reacted, 2 mol ClO 2 are produced, so mol ClO 2 = mol NaClO 2 .

      0.1659 mol ClO 2 × 67 .45 g ClO 2 mol = 11.2 g ClO 2


    10.125

    1. ft 3 CH 4 → L CH 4 → mol CH 4 → mol CH 3 OH → g CH 3 OH → L CH 3 OH

      10.7 × 10 9 ft 3 CH 4 × 1 yd 3 3 3 ft 3 × 1 m 3 ( 1.0936 ) 3 yd 3 × 1 L 1 × 10 3 m 3 = 3.03001 × 10 11 = 3.03 × 10 11 L CH 4

      n = PV RT = 3.03 × 10 11 L × 1 .00 atm 298 K × mol-K 0 .08206 L-atm = 1.2391 × 10 10 = 1.24 × 10 10 mol CH 4

      1 mol CH 4 = 1 mol CH 3 OH

      1.2391 × 10 10 mol CH 3 OH × 32 .04 g CH 3 OH mol CH 3 OH × 1 mL CH 3 OH 0.791 g × 1 L 1000 mL = 5 .0189 × 10 8 = 5.02 × 10 8 L CH 3 OH

    2. CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)   ΔH = –890.4 kJ

      Δ H o = Δ H f o CO 2 (g) + 2 Δ H f o H 2 O(l) Δ H f o CH 4 ( g) 2 Δ H f o O 2 (g)

      Δ H o = 393.5 kJ + 2 ( 285.83 kJ) ( 74.8 kJ) 0 = 890.4 kJ

      1.2391 × 10 10 mol CH 4 × 890.4 kJ 1 mol CH 4 = 1.10 × 10 13 kJ

      CH 3 OH(l) + 3/2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)  ΔH = –726.6 kJ

      Δ H o = Δ H f o CO 2 (g) + 2 Δ H f o H 2 O(l) Δ H f o CH 3 OH(l) 3 / 2 Δ H f o O 2 (g) = –393.5 kJ + 2(–285.83 kJ) – (–238.6 kJ) – 0 = –726.6 kJ

      1.2391 × 10 10 mol CH 3 OH × 726.6 kJ 1 mol CH 3 OH = 9.00 × 10 12 kJ

    3. Assume a volume of 1.00 L of each liquid.

      1.00 L CH 4 (l) × 466 g 1 L × 1 mol 16 .04 g × 890.4 kJ mol CH 4 = 2.59 × 10 4 kJ/L CH 4

      1.00 L CH 3 OH × 791 g 1 L × 1 mol 32 .04 g × 726.6 kJ mol CH 3 OH = 1.79 × 10 4 kJ/L CH 3 OH

      Clearly CH 4 (l) has the higher enthalpy of combustion per unit volume.

    10.126  After reaction, the flask contains IF 5 (g) and whichever reactant is in excess. Determine the limiting reactant, which regulates the moles of IF 5 produced and moles of excess reactant.

    I 2 (s) + 5F 2 (g) → 2 IF 5 (g)

    10.0 g I 2 × 1 mol I 2 253.8 g I 2 × 5 mol F 2 1 mol I 2 = 0.1970 = 0.197 mol F 2

    10.0 g F 2 × 1 mol F 2 38.00 g F 2 = 0.2632 = 0.263 mol F 2 available

    I 2 is the limiting reactant; F 2 is in excess.

    0.263 mol F 2 available – 0.197 mol F 2 reacted = 0.066 mol F 2 remain.

    10.0 g I 2 × 1 mol I 2 253.8 g I 2 × 2 mol IF 5 1 mol I 2 = 0.0788 mol IF 5 produced


    1. P IF 5 = nRT V = 0.0788 mol × 0 .08206 L-atm mol-K × 398 K 5 .00 L = 0.515 atm
    2. χ IF 5 = mol IF 5 mol IF 5 + mol F 2 = 0.0788 0.0788 + 0.066 = 0.544
    3. 42 valence e , 21 e pairs
      image
    4. 0.0788 mol IF 5 × 221.90 g IF 5 mol IF 5 = 17.4857 = 17.5 g IF 5 produced

      0.066 mol F 2 × 38.00 g F 2 mol F 2 = 2.508 = 2.5 g F 2 remain

      Total mass in flask = 17.5 g IF 5 + 2.5 g F 2 = 20.00 g; mass is conserved.

    10.127

    1. MgCO 3 (s) + 2HCl(aq) → MgCl 2 (aq) + H 2 O(l) + CO 2 (g)

      CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2 O(l) + CO 2 (g)

    2. n = PV RT = 743 torr × 1 atm 760 torr × mol-K 0 .08206 L-atm × 1.72 L 301 K

      = 0.06808 = 0.0681 mol CO 2

    3. x = g MgCO 3 , y = g CaCO 3 , x + y = 6.53 g

      mol MgCO 3 + mol CaCO 3 = mol CO 2 total

      x 84 .32 + y 100 .09 = 0.06808 ; y = 6.53 x

      x 84 .32 + 6 .53 x 100 .09 = 0.06808

      100.09x – 84.32x + 84.32(6.53) = 0.06808 (84.32)(100.09)

      15.77x + 550.610 = 574.549; x = 1.52 g MgCO 3

      mass % MgCO 3 = 1 .52 g MgCO 3 6.53 g sample × 100 = 23.3 %

      [By strict sig fig rules, the answer has 2 sig figs: 15.77x + 551 (3 digits from 6.53) = 575; 575 – 551 = 24 (no decimal places, 2 sig figs) leads to 1.5 g MgCO 3 and 23% MgCO 3 ]


     

    11 Liquids and Intermolecular Forces

    Visualizing Concepts

    In this chapter we will use the temperature units °C and K interchangeably when designating specific heats and changes in temperature.

    11.1

    1. The diagram best describes a liquid.
    2. In the diagram, the particles are close together, mostly touching but there is no regular arrangement or order. This rules out a gaseous sample, where the particles are far apart, and a crystalline solid, which has a regular repeating structure in all three directions.

    11.2

      1. Hydrogen bonding; H–F interactions qualify for this narrowly defined interaction.
      2. London dispersion forces, the only intermolecular forces between nonpolar F 2 molecules.
      3. Ion-dipole forces between Na + cation and the negative end of a polar covalent water molecule310.
      4. Dipole-dipole forces between oppositely charged portions of two polar covalent SO 2 molecules.
    1. London dispersion forces in (ii) are probably the weakest.

    11.3

    1. The viscosity of glycerol will be greater than that of 1-propanol.
    2. Viscosity is the resistance of a substance to flow. The stronger the intermolecular forces in a liquid, the greater its viscosity. Hydrogen bonding is the predominant force for both molecules. Glycerol has three times as many –OH groups and many more hydrogen-bonding interactions than 1-propanol, so it experiences stronger intermolecular forces and greater viscosity. (Both molecules have the same carbon-chain length, so dispersion forces are similar.)

    11.4 Analyze . When heat is added to a liquid, the temperature of the liquid rises. If enough heat is added to reach the boiling point (bp), any excess heat is used to vaporize the liquid. If heat is still available when all the liquid is converted to gas, the temperature of the gas rises.

    Plan . Use the specific heat of CH 4 (l) to calculate the amount of heat required to raise the temperature of 32.0 g of CH 4 (l) from –170 °C to –161.5 °C. If this is less than 42 kJ, use ΔH vap to calculate the energy required to vaporize the liquid, and so on, until exactly 42.0 kJ has been used to increase the temperature and/or change the state of CH 4 .


    Solve . Heat the liquid to its boiling point: ΔT = [–161.5 °C – (–170 °C)] = 8.5 °C = 8.5 K (Note that the magnitude of one degree is the same in Kelvins and Celsius.)

    3. 48 J g K × 3 2 .0 g CH 4 × 8 .5 K = 9 46.56 = 9.5 × 10 2 J = 0 .95 kJ

    Heating CH 4 (l) to its boiling point requires only 0.95 kJ. We have added 42 kJ, so there is definitely enough heat to vaporize the liquid. ΔH vap for CH 4 (l) is 8.20 kJ/mol. The 32.0 g sample is 2.00 mol CH 4 (l), so the energy required to vaporize the sample at –161.5 °C is (2 × 8.20 kJ/mol = ) 16.4 kJ. The energy used to heat the sample to –161.5 °C and vaporize it at this temperature is (0.947 kJ + 16.4 kJ) = 17.347 = 17.3 kJ. We have (42.0 kJ – 17.347 kJ) = 24.653 = 24.7 kJ left to heat the gas.

    Δ T = 24.65 kJ × g K 2 . 2 2 J × 1000 J kJ × 1 3 2 .0 g CH 4 = 346.99 = 347 K = 347 C o

    The final temperature of the methane gas, CH 4 (g), is then (–161.5 °C + 346.99 °C) = 185.49 = 185 °C.

    11.5

    1. 385 torr. Find 30 °C on the horizontal axis, and follow a vertical line from this point to its intersection with the red vapor pressure curve. Follow a horizontal line from the intersection to the vertical axis and read the vapor pressure.
    2. 22 °C. Reverse the procedure outlined in part (a). Find 300 torr on the vertical axis, follow it to the curve and down to the value on the horizontal axis.
    3. 47 °C. The normal boiling point of a liquid is the temperature at which its vapor pressure is 1 atm, or 760 mm Hg. A vapor pressure of 1 atm is very near the top of this diagram, at approximately 47 °C.

    11.6 Analyze/Plan . We are given the structures of two molecules with the same molecular formula and asked questions about their physical properties. Because the molecules have the same molecular formula, their van der Waals forces are similar. Consider any differences in intermolecular forces experienced by the two molecules. Solve .

    1. Propanol, the molecule on the left, has an O–H bond and experiences hydrogen bonding, while ethyl methyl ether does not.
    2. We expect propanol to have a larger dipole moment. Both molecules are somewhat polar, but propanol has hydrogen bonding.
    3. Propanol boils at 97.2 °C, while ethyl methyl ether boils at 10.8 °C. Molecules in liquid propanol are attracted to each other by hydrogen bonding. More kinetic energy and thus a higher temperature is required to separate the molecules and produce a gas.

    11.7

    1. 360 K, the normal boiling point; 260 K, normal freezing point. The left-most line is the freezing/melting curve, the right-most line is the condensation/boiling curve. The normal boiling and freezing points are the temperatures of boiling and freezing at 1 atm pressure.
    2. The material is solid in the left-most green (or pale blue) zone, liquid in the blue zone, and gas in the tan zone. (i) gas; (ii) solid; (iii) liquid.
    3. The triple point, where all three phases are in equilibrium, is the point where the three lines on the phase diagram meet. For this substance, the triple point is approximately 185 K at 0.45 atm.

    11.8

    1. The substance is in a liquid crystalline state at temperatures T 1 and T 2 . At T 1 , the molecules are aligned in layers and the long molecular axes are perpendicular to the layer planes; this describes a smectic A phase. At T 2 , the long molecular axes are aligned but the ends are not aligned; this describes a nematic phase.
    2. T 3 is the highest temperature. The molecular arrangement in this phase has the least order, so it represents the highest temperature. (The molecules are closely packed, but not aligned in any way; this describes an ordinary liquid phase.)

    Molecular Comparisons of Gases, Liquids, and Solids (Section 11.1)

    11.9

    1. solid < liquid < gas
    2. gas < liquid < solid
    3. Matter in the gaseous state is most easily compressed, because particles are far apart and there is much empty space.

    11.10

    1. In solids, particles are in essentially fixed positions relative to each other, so the average energy of attraction is stronger than average kinetic energy. In liquids, particles are close together but moving relative to each other. The average attractive energy and average kinetic energy are approximately balanced. In gases, particles are far apart and in constant, random motion. Average kinetic energy is much greater than average energy of attraction.
    2. As the temperature of a substance is increased, the average kinetic energy of the particles increases. In a collection of particles (molecules), the state is determined by the strength of interparticle forces relative to the average kinetic energy of the particles. As the average kinetic energy increases, more particles are able to overcome intermolecular attractive forces and move to a less ordered state, from solid to liquid to gas.
    3. If a gas is placed under very high pressure, the particles undergo many collisions with the container and with each other. The large number of particle-particle collisions increases the likelihood that intermolecular attractions will cause the molecules to coalesce (liquefy).

    11.11

    1. It increases. Kinetic energy is the energy of motion. As melting occurs, the motion of atoms relative to each other increases, which increases kinetic energy. As the kinetic energy of individual atoms increases, the overall average kinetic energy of the sample increases.
    2. It increases. As the atoms move relative to one another, the average distance between them increases. The physical property that corroborates this is density. The density of a solid is usually greater than the density of its liquid, indicating a greater sample volume for the liquid. This greater sample volume is the result of greater distance between atoms in three dimensions.

    11.12

    1. Ar < CCl 4 < Si. Intermolecular energy of attraction increases from gas to liquid to solid. (See Solution 11.10.)
    2. Ar < CCl 4 < Si. Boiling point increases as intermolecular energy of attraction increases. The greater the intermolecular attractive energy among particles, the greater the kinetic energy and temperature required to overcome this attractive energy and produce the less ordered gaseous state.

    11.13

    1. At standard temperature and pressure, the molar volumes of Cl 2 and NH 3 are nearly the same because they are both gases. In the gas phase, molecules are far apart and most of the volume occupied by the substance is empty space. Differences in molecular characteristics such as weight, shape, and dipole moment have little bearing on the molar volume of a gas.

      The ideal-gas law states that one mole of any gas at STP will occupy a fixed volume. The slight difference in molar volumes of the two gases is predicted by the van der Waals correction, which quantifies deviation from ideal behavior.

    2. On cooling to 160 K, both compounds condense from the gas phase to the solid state. Condensation, as the word implies, eliminates most of the empty space between molecules, so we expect a significant decrease in the molar volume.
    3. 1 cm 3 2 . 0 2 g Cl 2 × 7 0 .096 g Cl 2 1 mol Cl 2 = 35.1 cm 3 / mol Cl 2 = 0.0351 L / mol Cl 2

      1 cm 3 0. 8 4 g NH 3 × 1 7 .031 g NH 3 1 mol NH 3 = 20.3 cm 3 / mol NH 3 = 0.0203 L / mol NH 3

    4. Solid state molar volumes are not as similar as those in the gaseous state. In the solid state, most of the empty space is gone, so molecular characteristics do influence molar volumes. Cl 2 is heavier than NH 3 and the Cl–Cl bond distance is almost double the N H bond distance (Figure 7.7). Intermolecular attractive forces among polar NH 3 molecules bind them more tightly than forces among nonpolar Cl 2 molecules. These factors all contribute to a molar volume for Cl 2 (s) that is almost twice that of NH 3 (s).
    5. Like solids, liquids are condensed phases. That is, there is little empty space between molecules in the liquid state. We expect the molar volumes of the liquids to be closer to those in the solid state than those in the gaseous state.

    11.14

    1. The average distance between molecules is greater in the liquid state. Density is the ratio of the mass of a substance to the volume it occupies. For the same substance in different states, mass will be the same. The smaller the density, the greater the volume occupied, and the greater the distance between molecules. The liquid at 130° has the lower density (1.08 g/cm 3 ), so the average distance between molecules is greater.
    2. Less. For the same mass of compound, the sample with the higher density will occupy the smaller volume.

    Intermolecular Forces (Section 11.2)

    11.15

    1. London dispersion forces
    2. dipole-dipole forces
    3. hydrogen-bonding forces

    11.16

    1. Intra molecular interactions are generally stronger than intermolecular interactions. That is, interactions within molecules, such as chemical bonds, are stronger than interactions between molecules.
    2. Intermolecular interactions are broken when a liquid is converted to a gas.

    11.17

    1. SO 2 is a polar covalent molecule, so dipole-dipole and London dispersion forces must be overcome to convert the liquid to a gas.
    2. CH 3 COOH is a polar covalent molecule that experiences London dispersion, dipole-dipole, and hydrogen-bonding (O–H bonds) forces. All of these forces must be overcome to convert the liquid to a gas.
    3. H 2 S is a polar covalent molecule that experiences London dispersion and dipole-dipole forces, so these must be overcome to change the liquid into a gas. (H–S bonds do not lead to hydrogen-bonding interactions.)

    11.18

    1. CH 3 OH experiences hydrogen bonding, but CH 3 SH does not.
    2. Both gases are influenced by London dispersion forces. The heavier the gas particles, the stronger the London dispersion forces. The heavier Xe is a liquid at the specified conditions, whereas the lighter Ar is a gas.
    3. Both gases are influenced by London dispersion forces. The larger, diatomic Cl 2 molecules are more polarizable, experience stronger dispersion forces, and have the higher boiling point.
    4. Acetone and 2-methylpropane are molecules with similar molar masses and London dispersion forces. Acetone also experiences dipole-dipole forces and has the higher boiling point.

    11.19

    1. Polarizability increases as molecular size (and thus molecular weight) increases. In order of increasing polarizability: CH 4 < SiH 4 < SiCl 4 < GeCl 4 < GeBr 4 .
    2. The magnitude of London dispersion forces and thus the boiling points of molecules increase as polarizability increases. The order of increasing boiling points is the order of increasing polarizability:

      CH 4 < SiH 4 < SiCl 4 < GeCl 4 < GeBr 4

    11.20

    1. True. A more polarizable molecule can develop a larger transient dipole, increasing the strength of electrostatic attractions and dispersion forces among molecules.
    2. False. The noble gases are all monoatomic. Going down the family, the atomic radius and the size of the electron cloud increase. The larger the electron cloud, the more polarizable the atom, the stronger the London dispersion forces, and the higher the boiling point. (In general, strength of forces and boiling point vary in the same direction, not opposite directions.)
    3. False. Generally, dipole-dipole forces are stronger than dispersion forces for molecules of similar size and mass. The size of the molecule and the magnitude of its dipole moment (if there is one) determine the relative magnitudes of dispersion and dipole-dipole forces.
    4. True. For molecules with similar molecular weights and elemental composition, linear molecules have the possibility for greater contact along and around their surfaces than spherical molecules. Their electron clouds are thus more polarizable, and dispersion forces are greater.
    5. True. The larger the electron cloud, the more easily it can be distorted by transient electronic interactions.

    11.21 Analyze/Plan .  For molecules with similar structures, the strength of dispersion forces increases with molecular size (molecular weight and number of electrons in the molecule).

    Solve:

    1. H 2 S
    2. CO 2
    3. GeH 4

    11.22  For molecules with similar structures, the strength of dispersion forces increases with molecular size (molecular weight and number of electrons in the molecule).

    1. Br 2
    2. CH 3 CH 2 CH 2 CH 2 CH 2 SH
    3. CH 3 CH 2 CH 2 Cl. These two molecules have the same molecular formula and molecular weight (C 3 H 7 Cl, molecular weight = 78.5 amu), so the shapes of the molecules determine which has the stronger dispersion forces. According to Figure 11.6, the cylindrical (not branched) molecule will have stronger dispersion forces.

    11.23  Both hydrocarbons experience dispersion forces. Rod-like butane molecules can contact each other over the length of the molecule, whereas spherical 2-methylpropane molecules can only touch tangentially. The larger contact surface of butane facilitates stronger forces and produces a higher boiling point.

    11.24  Both molecules experience hydrogen bonding through their –OH groups and dispersion forces between their hydrocarbon portions. The position of the –OH group in isopropyl alcohol shields it somewhat from approach by other molecules and slightly decreases the extent of hydrogen bonding. Also, isopropyl alcohol is less rod-like (it has a shorter chain) than propyl alcohol, so dispersion forces are weaker. Because hydrogen bonding and dispersion forces are weaker in isopropyl alcohol, it has the lower boiling point.

    11.25

    1. A molecule must contain H atoms bound to either N, O, or F atoms to participate in hydrogen bonding with like molecules.
    2. CH 3 NH 2 and CH 3 OH have N–H and O–H bonds, respectively; they will form hydrogen bonds with other molecules of the same kind. (CH 3 F has C–F and C–H bonds, but no H–F bonds.)

    11.26

    1. HF has the higher boiling point because hydrogen bonding is stronger than dipole-dipole forces.
    2. CHBr 3 has the higher boiling point because it has the higher molar mass, which leads to greater polarizability and stronger dispersion forces.
    3. ICl has the higher boiling point because it is a polar molecule. For molecules with similar structures and molar masses, dipole-dipole forces are stronger than dispersion forces.

    11.27

    1. Replacing a hydroxyl hydrogen with a CH 3 group eliminates hydrogen bonding in that part of the molecule. This reduces the strength of intermolecular forces and leads to a (much) lower boiling point.
    2. CH 3 OCH 2 CH 2 OCH 3 is a larger, more polarizable molecule with stronger London dispersion forces and thus a higher boiling point.

    11.28

    1. C 4 H 10 . Both molecules experience dispersion forces. C 4 H 10 has the higher boiling point due to greater molar mass and similar strength of forces.
    2. CH 3 CH 2 CH 2 CH 2 OH has the higher boiling point due to the influence of hydrogen bonding.
    3. SO 3 . This is a tough call; SO 2 has dipole-dipole forces; SO 3 has dispersion forces but a larger molecular weight. The relative strength of dispersion and dipole-dipole forces depends on the mass and shape of the molecules. SO 3 molecules have greater molecular weight and are planar, so alignment is facile and dispersion forces are strong; SO 3 has the higher boiling point (confirmed by CRC Handbook of Chemistry and Physics ).
    4. Cl 2 CO has the higher boiling point due to greater molecular weight and stronger dispersion forces. (Note that H 2 CO does not have hydrogen bonding, because the H atoms are bound to C, not to O.)

    11.29

    Physical Property H 2 O H 2 S
    Normal Boiling Point, °C 100.00 −60.7
    Normal Melting Point, °C 0.00 −85.5

    Based on its much higher normal melting and boiling point, H 2 O has stronger intermolecular forces.

    H 2 O, with H bound to O, has hydrogen bonding. H 2 S, with H bound to S, has dipole-dipole forces. (The electronegativities of H and S, 2.1 and 2.5, respectively, are similar. The H−S bond dipoles in H 2 S are not large, but S does have two nonbonded electron pairs. The molecule has medium polarity.) Both molecules have London dispersion forces.

    11.30  Statement (c) is the best explanation. Chloroform is somewhat polar and has a larger dipole moment than carbon tetrachloride. The C–H bond in chloroform is not, however, a candidate for hydrogen bonding. The only explanation is that carbon tetrachloride is more polarizable, by virtue of its larger molar mass, than chloroform.

    11.31  SO 4 2– has a greater negative charge than BF 4 , so ion-ion electrostatic attractions are greater in sulfate salts. These strong forces limit the ion mobility required for the formation of an ionic liquid. (This is called an electronic effect.)

    11.32  The longer the alkyl side chain of the 1-alkyl-3-methylimidazolium cation, the more irregular the shape of the cation. Particles with irregular shapes are more difficult to pack into solids, so the melting point of the salt decreases as the length of the alkyl group and irregularity increases. (This is called a steric effect.)

    Select Properties of Liquids (Section 11.3)

    11.33

    1. As temperature increases, surface tension decreases; they are inversely related.
    2. As temperature increases, viscosity decreases; they are inversely related.

    1. Surface tension and viscosity are both directly related to the strength of intermolecular attractive forces. The same attractive forces that cause surface molecules to be difficult to separate cause molecules elsewhere in the sample to resist movement relative to one another. Liquids with high surface tension have intermolecular attractive forces sufficient to produce a high viscosity as well.

    11.34  The order of increasing strength of intermolecular forces is also the order of increasing viscosity and surface tension

    1. CH 3 CH 2 CH 3 < CH 2 Cl 2 < CH 3 CH 2 OH
    2. CH 3 CH 2 CH 3 < CH 2 Cl 2 < CH 3 CH 2 OH
    3. CH 3 CH 2 CH 3 < CH 2 Cl 2 < CH 3 CH 2 OH

    11.35

    1. Diagram (ii) shows stronger adhesive forces between the surface and the liquid. These adhesive forces attract the liquid to the surface and flatten the drop.
    2. Diagram (i) represents water on a nonpolar surface. The stronger hydrogen bonding cohesive forces among water molecules in the liquid prevent the drop from spreading.
    3. Diagram (ii) represents water on a polar surface. Adhesive dipole-dipole interactions between water molecules and the surface compete successfully with cohesive hydrogen bonding forces in the liquid and the drop spreads.

    11.36

    1. image
    2. All have bonds (N H or O–H, respectively) capable of forming hydrogen bonds. Hydrogen bonding is the strongest intermolecular interaction between neutral molecules and leads to very strong cohesive forces in liquids. The stronger the cohesive forces in a liquid, the greater the surface tension.

    11.37

    1. The three molecules have similar structures and all experience hydrogen- bonding, dipole-dipole, and dispersion forces. The main difference in the series is the increase in the number of carbon atoms in the alkyl chain, with a corresponding increase in chain length, molecular weight, and strength of dispersion forces. The boiling points, surface tension, and viscosities all increase because the strength of dispersion forces increases.
    2. Ethylene glycol has an OH group at both ends of the molecule. This greatly increases the possibilities for hydrogen bonding, so the overall intermolecular attractive forces are greater and the viscosity of ethylene glycol is much greater.
    3. Water has the highest surface tension but lowest viscosity because it is the smallest molecule in the series. Because water molecules are small, they approach each other closely and form many strong hydrogen bonds. There is no hydrocarbon chain to disrupt hydrogen bond formation or to inhibit their attraction to molecules in the interior of the drop. Water molecules at the surface of a drop are missing a few hydrogen bonds and are strongly pulled into the center of the drop, resulting in high surface tension. The absence of an alkyl chain also means the molecules can move around each other easily, resulting in the low viscosity.

    11.38

    1. For molecules with similar shapes, viscosity usually decreases with decreasing molecular weight. Because n-pentane has one fewer carbon atom and a shorter chain than n-hexane, the molecules are slightly more free to move around each other and n-pentane will have the smaller viscosity.
    2. According to Figure 11.6, neopentane is roughly spherical, whereas n-pentane is cylindrical or rod shaped. The spherical neopentane has weaker dispersion forces and the molecules are more free to tumble, so it will have the smaller viscosity.

    Phase Changes (Section 11.4)

    11.39

    1. melting, endothermic
    2. evaporation or vaporization, endothermic
    3. deposition, exothermic
    4. condensation, exothermic

    11.40

    1. condensation, exothermic
    2. sublimation, endothermic
    3. vaporization (evaporation), endothermic
    4. freezing, exothermic

    11.41

    1. Melting, (s) → (l)
    2. Endothermic. Energy is always required to overcome intermolecular forces that organize molecules into a solid.
    3. Heat of vaporization is usually larger than heat of fusion. Vaporization requires enough energy to separate molecules by large distances. Melting or fusion only requires that molecules move relative to each other, not that they are separated.

    11.42

    1. Liquid ethyl chloride at room temperature is far above its boiling point. When the liquid contacts the room temperature surface, heat sufficient to vaporize the liquid is transferred from the surface to the ethyl chloride, and the heat content of the molecules increases. At constant atmospheric pressure, ΔH = q, so the heat content and the enthalpy content of C 2 H 5 Cl(g) are higher than that of C 2 H 5 Cl(l). This indicates that the specific heat of the gas is less than that of the liquid, because the heat content of the gas starts at a higher level.
    2. Liquid C 2 H 5 Cl is vaporized (boiled), C 2 H 5 Cl(g) is warmed to the final temperature, and the solid surface is cooled to the final temperature. The enthalpy of vaporization (ΔH vap ) of C 2 H 5 Cl(l), the specific heat of C 2 H 5 Cl(g), and the specific heat of the solid surface must be considered.

    11.43 Analyze .  The heat required to vaporize 60 g of H 2 O equals the heat lost by the cooled water.

    Plan .  Using the enthalpy of vaporization, calculate the heat required to vaporize 60 g of H 2 O in this temperature range. Using the specific heat capacity of water, calculate the mass of water than can be cooled 15 °C if this much heat is lost.


    Solve .  Evaporation of 60 g of water requires:

    60 g H 2 O × 2 .4 kJ 1 g H 2 O = 1.44 × 10 2 kJ = 1 .4 × 10 5 J

    Cooling a certain amount of water by 15 °C:

    1.44 × 10 5 J × 1 g K 4.184 J × 1 15 o C = 2294 = 2.3 × 10 3 g H 2 O

    Check .  The units are correct. A surprisingly large mass of water (2300 g ≈ 2.3 L) can be cooled by this method.

    11.44  Energy released when 200 g of H 2 O is cooled from 15 °C to 0 °C:

    4.184 J g K × 200 g H 2 O × 15 o C = 12 .55 × 10 3 J = 13 kJ

    Energy released when 200 g of H 2 O is frozen (there is no change in temperature during a change of state):

    334 J g × 200 g H 2 O = 6 .68 × 10 4 J = 66 .8 kJ

    Total energy released = 12.55 kJ + 66.8 kJ = 79.35 = 79.4 kJ

    Mass of freon that will absorb 79.4 kJ when vaporized:

    79.35 kJ × 1 × 10 3 J 1 kJ × 1 g CCl 2 F 2 289 J = 275 g CCl 2 F 2

    11.45 Analyze/Plan .  Follow the logic in Sample Exercise 11.3. Solve . Physical data for ethanol, C 2 H 5 OH, is: mp = –114 °C; ΔH fus = 5.02 kJ/mol; C s(solid) = 0.97 J/g-K; bp = 78 °C; ΔH vap = 38.56 kJ/mol; C s(liquid) = 2.3 J/g-K. Solve .

    1. Heat the liquid from 35 °C to 78 °C, ΔT = 43 °C = 43 K.

      42.0 g C 2 H 5 OH × 2 .3 J g K × 43 K × 1 kJ 1000 J = 4.1538 = 4.2 kJ

      Vaporize (boil) the liquid at 78 °C, using ΔH vap .

      42.0 g C 2 H 5 OH × 1 mol C 2 H 5 OH 46 .07 g × 38.56 kJ mol = 35.1535 = 35.2 kJ

      Total energy required is 4.1538 kJ + 35.1535 kJ = 39.3073 = 39.3 kJ.

    2. Heat the solid from –155 °C to –114 °C, ΔT = 41 °C = 41 K.

      42.0 g C 2 H 5 OH × 0 .97 J g K × 41 K × 1 kJ 1000 J = 1.6703 = 1.7 kJ

      Melt the solid at –114 °C, using ΔH fus .

      42.0 g C 2 H 5 OH × 1 mol C 2 H 5 OH 46 .07 g × 5.02 kJ mol = 4.5765 = 4.58 kJ


      Heat the liquid from –114 °C to 78 °C, ΔT = 192 °C = 192 K.

      42.0 g C 2 H 5 OH × 2 .3 J g K × 192 K × 1 kJ 1000 J = 18.5472 = 19 kJ

      From part (a), vaporizing (boiling) 42.0 g of C 2 H 5 OH liquid at 78 °C requires 35.1535 kJ = 35.2 kJ.

      Total energy required = 1.6703 kJ + 4.5765 kJ + 18.5472 kJ + 35.1535 kJ = 59.9476 = 60 kJ.

      Check .  The relative energies of the various steps are reasonable; vaporization is the largest. The sum has no decimal places because (19 kJ) has no decimal places.

    11.46  Consider the process in steps, using the appropriate thermochemical constant.

    Heat the liquid from 10.00 °C to 47.6 °C, ΔT = 37.6 °C = 37.6 K, using the specific heat of the liquid.

    35.0 g C 2 Cl 3 F 3 × 0.91 J g K × 3 7 .6 K × 1 kJ 1000 J = 1.1976 = 1.2 kJ

    Boil the liquid at 47.6 °C (320.6 K), using the enthalpy of vaporization.

    35.0 g C 2 Cl 3 F 3 × 1 mol C 2 Cl 3 F 3 187.4 g C 2 Cl 3 F 3 × 27.49 kJ mol = 5.1342 = 5.13 kJ

    Heat the gas from 47.6 °C to 105.00 °C, ΔT = 57.4 °C = 57.4 K, using the specific heat of the gas.

    35.0 g C 2 Cl 3 F 3 × 0.67 J g K × 57.4 K × 1 kJ 1000 J = 1.3460 = 1.3 kJ

    The total energy required is 1.1967 kJ + 5.1342 kJ + 1.3460 kJ = 7.6778 = 7.7 kJ.

    11.47

    1. False. The critical pressure is the pressure required to cause liquefaction at the critical temperature.
    2. True.
    3. False. In general, the higher the critical temperature, the higher the critical pressure.
    4. True. The more intermolecular forces in a substance, the greater the kinetic energy required to overcome them. This greater kinetic energy translates to higher critical temperatures and pressures (as well as melting and boiling points).

    11.48

    1. CCl 3 F, CCl 2 F 2 , and CClF 3 are polar molecules that experience dipole-dipole and London dispersion forces with like molecules. CF 4 is a nonpolar compound that experiences only dispersion forces.
    2. According to Solution 11.47(b), the higher the critical temperature, the stronger the intermolecular attractive forces of a substance. Therefore, the strength of intermolecular attraction increases moving from right to left across the series and as molecular weight increases. CF 4 < CClF 3 < CCl 2 F 2 < CCl 3 F.

    1. The increasing intermolecular attraction with increasing molecular weight indicates that the critical temperature and pressure of CCl 4 will be greater than that of CCl 3 F. Looking at the numerical values in the series, an increase of 88 K in critical temperature and 3.1 atm in critical pressure to the corresponding values for CCl 3 F seem reasonable.
      Physical Property CCl 3 F CCl 4 (predicted) CCl 4 (CRC)
      Critical Temperature (K) 471 557 556.6
      Critical Pressure (atm) 43.5 46.6 44.6

      The predicted values for CCl 4 are in very good agreement with literature values. The key concept is that dispersion, not dipole-dipole, forces dominate the physical properties in the series.

    Vapor Pressure (Section 11.5)

    11.49  Properties (c) intermolecular attractive forces, (d) temperature, and (e) density of the liquid affect vapor pressure of a liquid.

    11.50  A normal boiling point of 56 °C places the liquid-vapor curve for acetone between the curves for diethyl ether and ethanol on Figure 11.25. Following a vertical line of increasing vapor pressure at 25 °C, we first cross the ethanol curve, then the (virtual) acetone curve. This means that, at 25 °C, the vapor pressure of acetone is higher than the vapor pressure of ethanol. (The lower boiling point of acetone is a strong indicator that it will have a higher vapor pressure than ethanol at a given temperature.)

    11.51

    1. Analyze/Plan .  Given the molecular formulae of several substances, determine the kind of intermolecular forces present, and rank the strength of these forces. The weaker the forces, the more volatile the substance. Solve .

      CBr 4 < CHBr 3 < CH 2 Br 2 < CH 2 Cl 2 < CH 3 Cl < CH 4

    2. CH 4 < CH 3 Cl < CH 2 Cl 2 < CH 2 Br 2 < CHBr 3 < CBr 4
    3. Boiling point increases as the strength of intermolecular forces increases. By analogy to attractive forces in HCl (Section 11.2), the trend will be dominated by dispersion forces, even though four of the molecules (CHBr 3 , CH 2 Br 2 , CH 2 Cl 2 , and CH 3 Cl) are polar. Thus, the order of increasing boiling point is the order of increasing molar mass and increasing strength of dispersion forces.

    11.52

    1. False. The heavier (and larger) CBr 4 has stronger dispersion forces, a higher boiling point, lower vapor pressure, and is less volatile.
    2. True.
    3. False.
    4. False.

    11.53

    1. The water in the two pans is at the same temperature, the boiling point of water at the atmospheric pressure of the room. During a phase change, the temperature of a system is constant. All energy gained from the surroundings is used to accomplish the transition, in this case to vaporize the liquid water. The pan of water that is boiling vigorously is gaining more energy and the liquid is being vaporized more quickly than in the other pan, but the temperature of the phase change is the same.

    1. Vapor pressure does not depend on either volume or surface area of the liquid. As long as the containers are at the same temperature, the vapor pressures of water in the two containers are the same.

    11.54  Statement (c) is the best explanation for the cool tea. Statements (a) and (b) are somewhat true, but the effects would not be rapid. That the boiling point is lower at lower pressure is more significant. Statement (d) is false.

    11.55 Analyze/Plan . Follow the logic in Sample Exercise 11.4. The boiling point is the temperature at which the vapor pressure of a liquid equals atmospheric pressure. Solve .

    1. The boiling point of ethanol at 200 torr is ~48 °C.
    2. The vapor pressure of ethanol at 60 °C is approximately 340 torr. Thus, at 60 °C ethyl alcohol would boil at an external pressure of 340 torr.
    3. The boiling point of diethyl ether at 400 torr is ~17 °C.
    4. 40 °C is above the normal boiling point of diethyl ether, so the pressure at which 40 °C is the boiling point is greater than 760 torr. According to Figure 11.25, a boiling point of 40 °C requires an external pressure of 1000 torr. (At these conditions, the vapor pressure of diethyl ether is 1000 torr.)

    11.56

    1. image

      A plot of vapor pressure versus temperature data for H 2 O from Appendix B is shown here. The vapor pressure of water at body temperature, 37 °C, is approximately 50 torr.

    2. The data point at 760.0 torr, 100 °C is the normal boiling point of H 2 O. This is the temperature at which the vapor pressure of H 2 O is equal to a pressure of 1 atm or 760 torr.
    3. At an external (atmospheric) pressure of 633 torr, the boiling point of H 2 O is approximately 96 °C.
    4. At an external pressure of 774 torr, the boiling point of water is approximately 100.5 °C.

    Phase Diagrams (Section 11.6)

    11.57

    1. The critical point is the temperature and pressure beyond which the gas and liquid phases are indistinguishable.
    2. The gas/liquid line ends at the critical point because at conditions beyond the critical temperature and pressure, there is no distinction between gas and liquid. In experimental terms, a gas cannot be liquefied at temperatures higher than the critical temperature, regardless of pressure.

      (At conditions beyond the critical point, a substance is known as a supercritical fluid. Supercritical fluids have many practical applications, such as decaffeination of green coffee beans and dry cleaning.)

    11.58

    1. The triple point on a phase diagram represents the temperature and pressure at which the gas, liquid, and solid phases are in equilibrium.
    2. No. A phase diagram represents a closed system, one where no matter can escape and no substance other than the one under consideration is present; air cannot be present in the system. Even if air is excluded, at 1 atm of external pressure, the triple point of water is inaccessible, regardless of temperature [see Figure 11.28].

    11.59

    1. The water vapor would deposit to form a solid at a pressure of around 4 torr. At higher pressure, perhaps 5 atm or so, the solid would melt to form liquid water. This occurs because the melting point of ice, which is 0 °C at 1 atm, decreases with increasing pressure.
    2. In thinking about this exercise, keep in mind that the total pressure is being maintained at a constant 0.50 atm. That pressure is composed of water vapor pressure and some other pressure, which could come from an inert gas. At 100 °C and 0.50 atm, water is in the vapor phase. As it cools, the water vapor will condense to the liquid at the temperature where the vapor pressure of liquid water is 0.50 atm. From Appendix B, we see that condensation occurs at approximately 82 °C. Further cooling of the liquid water results in freezing to the solid at approximately 0 °C. The freezing point of water increases with decreasing pressure, so at 0.50 atm, the freezing temperature is very slightly above 0 °C.

    11.60

    1. Solid CO 2 sublimes to form CO 2 (g) at a temperature of about –60 °C.
    2. Solid CO 2 melts to form CO 2 (l) at a temperature of about –55 °C. The CO 2 (l) boils when the temperature reaches approximately –45 °C.

    11.61 Analyze/Plan . Follow the logic in Sample Exercise 11.5, using the phase diagram for neon. Solve .

    1. The normal melting point is the temperature where solid becomes liquid at 1 atm pressure. Following a horizontal line at 1 atm to the solid-liquid line, the normal melting point is approximately 24 K.
    2. Neon sublimes, changes directly from solid to gas, at pressures less than the triple point pressure, approximately 0.5 atm.
    3. Room temperature is 298 K, in the region where neon is a supercritical fluid. Neon cannot be liquefied at any temperature above the critical temperature, approximately 45 K, regardless of pressure.

    11.62

    1. The normal boiling point is the temperature where liquid becomes gas at 1 atm pressure. Moving vertically down from 1 atm on the liquid-gas line to the temperature axis, the normal boiling point is approximately 27 to 28 K, or –246 °C to –245 °C.
    2. The much higher critical temperature and pressure of Ar (150.9 K, 48 atm) compared with those of Ne (25 K, 0.43 atm), indicate that Ar experiences much stronger intermolecular forces than Ne.

    11.63 Analyze/Plan . Follow the logic in Sample Exercise 11.5, using the phase diagram for methane in Figure 11.30. Solve .

    1. According to Sample Exercise 11.5, the triple point of methane (CH 4 ) is approximately (–180 °C, 0.1 atm). The solid-liquid line in the phase diagram is essentially vertical in the pressure range 0.1-100 atm. This means that conditions at the surface of Titan (–178 °C, 1.6 atm) are very close to the solid-liquid line. Methane on the surface of titan is likely to exist in both solid and liquid forms.
    2. Methane is a liquid at –178 °C and 1.6 atm. Moving upward through Titan’s atmosphere at a constant temperature of –178 °C, pressure decreases. At a pressure slightly greater than 0.1 atm, we expect to see vaporization to gaseous methane. If we begin with solid methane at 1.6 atm and a temperature slightly below –180 °C, we expect sublimation to gaseous methane at a pressure slightly less than 0.1 atm.

    11.64  The density of Ga(s), 5.91 g/cm 3 , is less than the density of Ga(l), 6.1 g/cm 3 , just above the melting temperature. “Typically” the density of a solid is greater than the density of its liquid. Gallium is then an atypical substance, like water, where the solid state is denser and more compact than the liquid. This results in a backward sloping solid-liquid line on the phase diagram for water, and we also expect to see this unusual feature on the diagram for gallium.

    Liquid Crystals (Section 11.7)

    11.65  In a nematic liquid crystalline phase, molecules are aligned along their long axes, but the molecular ends are not aligned. In an ordinary liquid, molecules have no orderly arrangement; they are randomly oriented, or amorphous. Both an ordinary liquid and a nematic liquid crystal phase are fluids; molecules are free to move relative to one another. In an ordinary liquid, molecules can move in any direction. In a nematic phase, molecules are free to translate in all dimensions. Molecules cannot tumble or rotate out of the molecular plane, or the order of the nematic phase is lost and the sample becomes an ordinary liquid.

    11.66  Reinitzer observed that cholesteryl benzoate has a phase that exhibits properties intermediate between those of the solid and liquid phases. This “liquid-crystalline” phase, formed by melting at 145 °C, is viscous and opaque; its viscosity decreases on heating and it becomes clear at 179 °C.


    11.67

    1. True.
    2. False. Liquid crystalline molecules are often rod-like with some rigidity in the long direction.
    3. True. Liquid crystalline is a phase of matter with distinct phase-change temperatures.
    4. False. If no significant intermolecular forces were present, there would be no driving force for the intermediate ordering typical of liquid crystals.
    5. False. Molecules containing only carbon and hydrogen do no exhibit the significant intermolecular forces required for liquid crystal formation.
    6. True.

    11.68

    1. Graph B applies to a liquid crystalline material.
    2. Melting solid to liquid. Graph A applies to a ”regular” material. The constant temperature of the 2–3 segment represents the first phase change, melting solid.
    3. Melting solid to liquid crystal. Segment 2–3 is the first constant temperature process. For a liquid crystalline material, this is melting solid to liquid crystal.
    4. Heating liquid. In a regular material, the second heating change is heating the liquid.
    5. Heating the liquid crystal. In a liquid crystalline material the second heating seqment is heating the liquid crystal.

    11.69  Because order is maintained in at least one dimension, the molecules in a liquid-crystalline phase are not totally free to change orientation. This makes the liquid-crystalline phase more resistant to flow, more viscous, than the isotropic liquid.

    11.70  A nematic phase is composed of sheets of molecules aligned along their lengths, but with no additional order within the sheet or between sheets. A cholesteric phase also contains this kind of sheet, but with some ordering between sheets. In a cholesteric phase, there is a characteristic angle between molecules in one sheet and those in an adjacent sheet. That is, one sheet of molecules is twisted at some characteristic angle relative to the next, producing a “screw” axis perpendicular to the sheets.

    11.71  As the temperature of a substance increases, the average kinetic energy of the molecules increases. More molecules have sufficient kinetic energy to overcome intermolecular attractive forces, so overall ordering of the molecules decreases as temperature increases. Melting provides kinetic energy sufficient to disrupt alignment in one dimension in the solid, producing a smectic phase with ordering in two dimensions. Additional heating of the smectic phase provides kinetic energy sufficient to disrupt alignment in another dimension, producing a nematic phase with one-dimensional order.

    11.72  In the nematic phase, molecules are aligned in one dimension, the long dimension of the molecule. In a smectic phase (A or C), molecules are aligned in two dimensions. Not only are the long directions of the molecules aligned, but the ends are also aligned. The molecules are organized into layers; the height of the layer is related to the length of the molecule.


    Additional Exercises

    11.73

    1. decrease
    2. increase
    3. increase
    4. increase
    5. increase
    6. increase
    7. increase

    11.74

    1. 0 K – 54 K. A substance is a solid until its temperature reaches the melting point.
    2. 54 K – 95 K. The density of a substance decreases significantly when it goes from solid to liquid. The data shows that this happens between 90 and 100 K for O 2 . It is reasonable to assume the boiling point is in the middle of this range. (A quick Internet search indicates that the boiling point of O 2 is 90.1 K.)
    3. 95 K – 140 K. For temperatures in the table, O 2 is a gas when the density is low.
    4. 95 K. As noted in part (b), O 2 is a liquid at 90 K and a gas at 100 K. The boiling point is somewhere in the range 90 – 100 K (actually 90.1 K)
    5. Dispersion forces only

    11.75

    1. Correct.
    2. The lower boiling liquid must experience less total intermolecular forces.
    3. If both liquids are structurally similar nonpolar molecules, the lower boiling liquid has a lower molecular weight than the higher boiling liquid.
    4. Correct.
    5. At their boiling points, both liquids have vapor pressures of 760 mm Hg.

    11.76

    1. The cis isomer has stronger dipole-dipole forces; the trans isomer is nonpolar. (Because both molecules have the same molecular weight, we can say that the dipole-dipole and dispersion forces of the cis isomer are stronger than the dispersion-only forces of the trans isomer.)
    2. The molecule with the stronger intermolecular interactions will have the higher boiling point. The cis isomer boils at 60.3 °C and the trans isomer boils at 47.5 °C.

    11.77  Statement (b) best explains the data, although statement (d) is also true, because the molecular structures of the compounds are similar. Statements (a) and (c) are false.

    11.78

    1. Four, all of them. All covalent compounds exhibit dispersion interactions.
    2. Three. Benzene is nonpolar and exhibits only dispersion forces. When Cl, Br, or OH is substituted for H in benzene, the molecule becomes polar and the molecules exhibit dipole-dipole interactions.
    3. One. The O–H bond in phenol participates in hydrogen bonding.
    4. Bromine is larger and more polarizable than chlorine, so the dispersion forces in bromobenzene are stronger than those in chlorobenzene.
    5. Phenol exhibits hydrogen bonding, which is the strongest intermolecular interaction among covalent molecules.

    11.79  The GC base pair, with more hydrogen bonds, is more stable to heating. To break up a base pair by heating, sufficient thermal energy must be added to break the existing hydrogen bonds. With 50% more hydrogen bonds, the GC pair is definitely more stable (harder to break apart) than the AT pair.

    11.80  The two O−H groups in ethylene glycol are involved in many hydrogen-bonding interactions. Pentane is nonpolar and experiences only dispersion forces.

    1. Ethylene glycol is more viscous because of its strong hydrogen bonding.
    2. Pentane has the lower boiling point because it experiences weaker intermolecular forces.
    3. Ethylene glycol is used as antifreeze because it is viscous and doesn’t boil off in the very hot radiator.
    4. Pentane is used as the “blowing agent” because it has a low boiling point and is volatile.

    11.81  A plot of the number of carbon atoms versus boiling point follows. For eight C atoms, C 8 H 18 , the boiling point is approximately 130 °C. The more carbon atoms in the hydrocarbon, the longer the chain, the more polarizable the electron cloud, the stronger the London dispersion forces, and the higher the boiling point.

    image

    11.82  Ionic liquids are the liquid phase of ionic compounds. Upon melting, the ions are free to move relative to one another. The ion-ion interparticle attractive forces at work in an ionic liquid are extremely strong relative to dispersion, dipole-dipole, and even hydrogen-bonding forces operating in most molecular solvents. These powerful ion-ion forces must be broken for an ion to escape to the vapor phase. In the distribution of particle energies at room temperature, very few ions have sufficient kinetic energy to escape these interactions and move to the vapor phase. With few particles in the vapor phase, the vapor pressures of ionic liquids are extremely low.

    11.83

    1. Sweat, or salt water, on the surface of the body vaporizes to establish its typical vapor pressure at atmospheric pressure. Because the atmosphere is a totally open system, saturated vapor pressure is never reached, and the sweat evaporates continuously. Evaporation is an endothermic process. The heat required to vaporize sweat is absorbed from your body, helping to keep it cool.

    1. The vacuum pump reduces the pressure of the atmosphere (air + water vapor) above the water. Eventually, atmospheric pressure equals the vapor pressure of water and the water boils. Boiling is an endothermic process, and the temperature drops if the system is not able to absorb heat from the surroundings fast enough. As the temperature of the water decreases, the water freezes. (On a molecular level, the evaporation of water removes the molecules with the highest kinetic energies from the liquid. This decrease in average kinetic energy is what we experience as a temperature decrease.)

    11.84

    1. If the Clausius-Clapeyron equation is obeyed, a graph of ln P versus 1/T (K −1 ) should be linear. Here are the data in a form for graphing.

      T (K)

      1/T

      P (torr)

      ln P

      image

      280.0

      3.571 × 10 −3

      32.42

      3.479

      300.0

      3.333 × 10 −3

      92.47

      4.527

      320.0

      3.125 × 10 −3

      225.1

      5.417

      330.0

      3.030 × 10 −3

      334.4

      5.812

      340.0

      2.941 × 10 −3

      482.9

      6.180

      According to the graph, the Clausius-Clapeyron equation is obeyed, to a first approximation.

      Δ H vap = slope × R;  slope = 3 .479 - 6.180 ( 3.571 - 2.941 ) × 10 - 3 = - 2.701 0.630 × 10 - 3 = - 4.29 × 10 3 Δ H vap = - ( - 4.29 × 10 3 ) × 8.314 J / mol K = 35 .7 kJ / mol

    2. The normal boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure, 760 torr. From the graph, ln 760 = 6.63, 1/T for this vapor pressure = 2.828 × 10 −3 ; T = 353.6 K

    11.85

    1. The Clausius-Clapeyron equation is ln P = - Δ H vap RT + C .

      For two vapor pressures, P 1 and P 2 , measured at corresponding temperatures T 1 and T 2 , the relationship is

      ln P 1 - ln P 2 = ( - Δ H vap RT 1 + C ) - ( - Δ H vap RT 2 + C ) ln P 1 - ln P 2 = - Δ H vap R ( 1 T 1 - 1 T 2 ) + C - C ; ln P 1 P 2 = - Δ H vap R ( 1 T 1 - 1 T 2 )

    2. P 1 = 13.95 torr, T 1 = 298 K; P 2 =144.78 torr, T 2 = 348 K

      ln 13 .95 144 .78 = - Δ H vap 8 .314 J/mol - K ( 1 298 - 1 348 ) - 2.33974 (8 .314 J / mol - K ) = - Δ H vap (4 .821 × 10 - 4 / K )

      ΔH vap = 4.035 × 10 4 = 4.0 × 10 4 J/mol = 40 kJ/mol

      [(1/T 1 ) – (1/T 2 )] has 2 sig figs and so does the result.


    1. The normal boiling point of a liquid is the temperature at which the vapor pressure of the liquid is 760 torr.

      P 1 = 144.78 torr, T 1 = 348 K; P 2 = 760 torr, T 2 = bp of octane

      l n ( 144.78 760.0 ) = - 4.035 × 10 4 J / mol 8.314 J / mol K ( 1 348 K - 1 T 2 )

      - 1.6581 - 4.8533 × 10 3 = 2.874 × 10 - 3 - 1 T 2 ; 1 T 2 = 2.874 × 10 - 3 - 3.416 × 10 - 4

      1 / T 2 = 2.532 × 10 - 3 = 2.53 × 10 - 3 ; T 2 = 395 K (122 o C)

      From the plot of boiling point versus number of carbon atoms in Solution 11.81, we read an approximate boiling point for octane of 130 °C. These two temperatures are close, but do differ by more than 5%. Considering experimental uncertainties in the vapor pressure (vp) data, and the empirical nature of the plot, the two values are surprisingly close. The literature boiling point of octane, 126 °C, is exactly midway between our two estimates.

    1. P 1 = vp of octane at –30 °C, T 1 = 243 K; P 2 = 144.78 torr, T 2 = 348 K

      ln P 1 144.7 8 torr = - 4.035 × 10 4 J / mol 8.314 J / mol - K ( 1 243 - 1 348 )

      ln P 1 144. 78 t orr = - 4.035 × 10 4 J / mol 8.314 J/mol-K × 1. 242 × 10 - 3 = - 6.026 = - 6.03 P 1 144.78 torr = e - 6.026 ; P 1 = 0.002415 (144 .78) = 0.3496 = 0.35 torr

      [This result has 2 sig figs because (ln = −6.03) has 2 decimal places. In a ln or log, the places left of the decimal show order of magnitude, and places right of the decimal show sig figs in the real number.] The result, 0.35 torr at –30 °C, is reasonable, because we expect vapor pressure to decrease as temperature decreases, and we are approaching the freezing point of octane, –57 °C.

    11.86  Physical data for the two compounds from the Handbook of Chemistry and Physics :

    MM dipole moment boiling point
    CH 2 CI 2 85 g/mol 1.60 D 40.0 °C
    CH 3 I 142 g/mol 1.62 D 42.4 °C
    1. The two substances have very similar molecular structures; each is an unsymmetrical tetrahedron with a single central carbon atom and no hydrogen bonding. Because the structures are very similar, the magnitudes of the dipole-dipole forces should be similar. This is verified by their very similar dipole moments. The heavier compound, CH 3 I, will have slightly stronger London dispersion forces. Because the nature and magnitude of the intermolecular forces in the two compounds are nearly the same, it is very difficult to predict which will be more volatile [or which will have the higher boiling point as in part (b)].

    1. Given the structural similarities discussed in part (a), one would expect the boiling points to be very similar, and they are. Based on its larger molar mass (and dipole-dipole forces being essentially equal) one might predict that CH 3 I would have a slightly higher boiling point; this is verified by the known boiling points.
    1. According to Equation [11.1], ln P = - Δ H vap RT + C

      A plot of ln P versus 1/T for each compound is linear. Because the order of volatility changes with temperature for the two compounds, the two lines must cross at some temperature; the slopes of the two lines, ΔH vap for the two compounds, and the y-intercepts, C, must be different.

    1. CH 2 Cl 2

      CH 3 I

      ln P T (K) 1/T ln P T (K) 1/T
      2.303 229.9 4.351 × 10 −3 2.303 227.4 4.398 × 10 −3
      3.689 250.9 3.986 × 10 −3 3.689 249.0 4.016 × 10 −3
      4.605 266.9 3.747 × 10 −3 4.605 266.2 3.757 × 10 −3
      5.991 297.3 3.364 × 10 −3 5.991 298.5 3.350 × 10 −3
      image

      For CH 2 Cl 2 , - Δ H vap / R = slope = ( 5.991 - 2.303 ) ( 3.364 × 10 - 3 - 4.350 × 10 - 3 ) = - 3.688 0.987 × 10 - 3 = - 3.74 × 10 3 = - Δ H vap / R

      ΔH vap = 8.314 (3.74 × 10 3 ) = 3.107 × 10 4 J/mol = 31.1 kJ/mol

      For CH 3 I , - Δ H vap / R = slope = (5 .991 - 2.303 ) ( 3.350 × 10 - 3 - 4.398 × 10 - 3 ) = - 3.688 1.048 × 10 - 3 = - 3.519 × 10 3 = - Δ H vap / R

      ΔH vap = 8.314 (3.519 × 10 3 ) = 2.926 × 10 4 J/mol = 29.3 kJ/mol


    11.87  The normal melting point (nmp) and normal boiling point (nbp) are at a pressure of 1 atm. The triple point (tp) occurs at 1000 Pa. Change this pressure to atm.

    1 000 × 1 atm 1 .01325 × 10 5 Pa = 0.009869 = 9.87 × 10 - 3 = 1 × 10 - 2 atm

    image

    With no information about the critical point of naphthalene, we cannot denote it or the supercritical fluid region.

    11.88  When voltage is applied to a liquid crystal display, the molecules align with the voltage and the appearance of the display changes. At low Antarctic temperatures, the liquid crystalline phase is closer to its freezing point. The molecules have less kinetic energy due to temperature and the applied voltage may not be sufficient to overcome orienting forces among the molecules. If some or all of the molecules do not rotate when the voltage is applied, the display will not function properly.

    11.89

    image

    Integrative Exercises

    11.90

    1. Isooctane will have a lower viscosity than octane. Isooctane molecules are more spherical and cannot become as entangled as the flexible chains of octane molecules.
    2. In order of increasing boiling point: hexane < heptane < octane < nonane < decane Because the molecules have similar structures, the strength of dispersion forces increases with increasing chain length and molar mass. The stronger the dispersion forces the higher the boiling point.

    1. Statement (i) is the likely explanation. Statement (ii) is false and (iii) is unlikely.
    1. Statement (ii) is the explanation. n-octyl alcohol exhibits hydrogen bonding. Statement (i) is false.

    11.91

    1. 24 valence e , 12 e pairs
      image

      The geometry around the central C atom is trigonal planar, and around the two terminal C atoms, tetrahedral.

    2. Polar. The C=O bond is quite polar and the dipoles in the trigonal plane around the central C atom do not cancel.
    3. Dipole-dipole and London dispersion forces
    4. Because the molecular weights of acetone and 1-propanol are similar, the strength of the London dispersion forces in the two compounds is also similar. The big difference is that 1-propanol has hydrogen bonding, whereas acetone does not. These relatively strong attractive forces lead to the higher boiling point for 1-propanol.

    11.92

    image

    It is useful to draw the structural formulas because intermolecular forces are determined by the size and shape (structure) of molecules.

    1. Molar mass :  compounds (i) and (ii) have similar rod-like structures; (ii) has a longer rod. The longer chain leads to greater molar mass, stronger London dispersion forces and higher heat of vaporization.
    2. Molecular shape :  compounds (iii) and (v) have the same chemical formula and molar mass but different molecular shapes (they are structural isomers). The more rod-like shape of (v) leads to more contact between molecules, stronger dispersion forces, and higher heat of vaporization.
    3. Molecular polarity :  rod-like hydrocarbons (i) and (ii) are essentially nonpolar, owing to free rotation about C–C σ bonds, whereas (iv) is quite polar, owing to the C=O group. (iv) has a smaller molar mass than (ii) but a larger heat of vaporization, which must be due to the presence of dipole-dipole forces in (iv). [Note that (iii) and (iv), with similar shape and molecular polarity, have very similar heats of vaporization.]

    1. Hydrogen-bonding interactions :  molecules (v) and (vi) have similar structures, but (vi) has hydrogen bonding and (v) does not. Even though molar mass and thus dispersion forces are larger for (v), (vi) has the higher heat of vaporization. This must be due to hydrogen-bonding interactions.

    11.93  Ethanol will evaporate until its vapor fills the flask at a pressure of 40.0 torr. Calculate the mass of 2.00 L of ethanol vapor at 19 °C and a pressure of 40.0 torr. Subtract this mass from the original 1.00 g to find the mass of liquid ethanol remaining.

    T = 19 °C + 273.15 = 292 K; MM = 46.07 g/mol;

    40.0 torr × 1 atm 760 torr = 0.052632 = 0.0526 atm

    g = MM × PV RT ; g = 46 .07 g ethanol 1 mol ethanol × mol K 0 .08206 L - atm × 0.052632 atm 292 K × 2. 0 L = 0 .202 g vapor

    g ethanol liquid = 1.00 g – 0.202 g vapor = 0.798 = 0.80 g

    11.94

    1. For butane to be stored as a liquid at temperatures above its boiling point (–5 °C), the pressure in the tank must be greater than atmospheric pressure. In terms of the phase diagram of butane, the pressure must be high enough so that, at tank conditions, the butane is “above” the gas-liquid line and in the liquid region of the diagram.

      The pressure of a gas is described by the ideal-gas law as P = nRT/V; pressure is directly proportional to moles of gas. The more moles of gas present in the tank the greater the pressure, until sufficient pressure is achieved for the gas to liquefy. At the point where liquid and gas are in equilibrium and temperature is constant, liquid will vaporize or condense to maintain the equilibrium vapor pressure. That is, as long as some liquid is present, the gas pressure in the tank will be constant.

    2. If butane gas escapes the tank, butane liquid will vaporize (evaporate) to maintain the equilibrium vapor pressure. Vaporization is an endothermic process, so the butane will absorb heat from the surroundings. The temperature of the tank and the liquid butane will decrease.
    3. 250 g C 4 H 10 × 1 mol C 4 H 10 58.12 g C 4 H 10 × 21.3 kJ mol = 91.6 kJ

      V = nRT P = 250 g × 1 mol 58 .12 g × 0.08206 L-atm mol K × 308 K 755 torr × 760 torr 1 atm = 109.44 = 109 L

    11.95 Plan .

    1. Using thermochemical data from Appendix B, calculate the energy (enthalpy) required to melt and heat the H 2 O.
    2. Using Hess’s Law, calculate the enthalpy of combustion, ΔH comb , for C 3 H 8 .
    3. Solve the stoichiometry problem.

    Solve .

    1. Heat H 2 O(s) from - 20 o C to 0 .0 o C; 5 .50 × 10 3 g H 2 O × 2 .092 J g- o C × 20 o C = 2.301 × 10 2 = 2.3 × 10 2 kJ

      Melt H 2 O ( s ) ; 5 .50 × 10 3 g H 2 O × 6 .008 kJ mol H 2 O × 1 mol H 2 O 18.02 g H 2 O = 1834 = 1 .83 × 10 3 kJ


    1. Heat H 2 O(l) from 0 o C to 75 o C;  5 .50 × 10 3 g H 2 O × 4 .184 J g- o C × 75 o C = 1726 = 1.7 × 10 3 kJ

      Total energy = 230.1 kJ + 1834 kJ + 1726 kJ = 3790 = 3.8 × 10 3 kJ

      (The result is significant to 100 kJ, limited by 1.7 × 10 3 kJ)

    2. C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(l)

      Assume that one product is H 2 O(l), because this leads to a more negative ΔH comb and fewer grams of C 3 H 8 (g) required.

      Δ H comb = 3 Δ H f CO 2 ( g ) + 4 Δ H f H 2 O(l) - Δ H f C 3 H 8 ( g ) - 5 Δ H f O 2 ( g ) = 3 ( - 393.5 kJ ) + 4 ( - 285.83 kJ ) - ( - 103.85 kJ ) - 5 ( 0 ) = - 2219.97 = - 2220 kJ
    3. 3.790 × 10 3 kJ required × 1 mol C 3 H 8 2219.97 kJ × 44.096 g C 3 H 8 1 mol C 3 H 8 = 75 g C 3 H 8

      (3.8 × 10 3 kJ required has 2 sig figs and so does the result)

    11.96 P = nRT V = g RT M V ; T = 273 .15 + 26 .0 o C = 299 .15 = 299 .2 K;  V = 5 .00 L

    g C 6 H 6 (g) = 7.2146 – 5.1493 = 2.0653 g C 6 H 6 (g)

    P (vapor) = 2 .0653 g 78 .11 g / mol × 299.15 K 5 .00 L × 0.08206 L-atm mol K × 760 torr 1 atm = 98.660 = 98.7 torr

    11.97 Plan .  Relative humidity and vp of H 2 O at give T → P H 2 O → ideal-gas law → mol H 2 O(g) → H 2 O molecules. Change °F → °C, volume of room from ft 3 → L.

    Solve . °C = 5/9 (°F – 32); °C = 5/9 (68 °F – 32) = 20 °C;

    rh = (P H 2 O in air / vp of H 2 O) × 100

    From Appendix B, vp of H 2 O at 20 °C = 17.54 torr

    P H 2 O in air = rh × vp of H 2 O/100 = 58 × 17.54 torr/100 = 10.173 = 10 torr

    V = 12 ft × 10 ft × 8 ft × 12 3 in 3 ft 3 × 2.54 3 cm 3 in 3 × 1 L 1000 cm 3 = 2.718 × 10 4 = 3 × 10 4 L

    (The result has 1 sig fig, as does the measurement 8 ft.)

    PV = nRT; n = PV/RT

    n = 10.173 torr × 1 atm 760 torr × mol K 0.08206 L-atm × 2.718 × 10 4 L 293 K = 15.13 = 2 × 10 1 mol H 2 O

    15 .13 mol H 2 O × 6 .022 × 10 23 molecules 1 mol = 9.112 × 10 24 = 9 × 10 24 H 2 O molecules


     

    12 Solids and Modern Materials

    Visualizing Concepts

    12.1   The red-orange compound is more likely to be a semiconductor and the white one an insulator. The red-orange compound absorbs light in the visible spectrum (red-orange is reflected, so blue-green is absorbed), whereas the white compound does not. This indicates that the red-orange compound has a lower energy electron transition than the white one. Semiconductors have lower energy electron transitions than insulators.

    12.2   When choosing a unit cell, remember that the environment of each lattice point must be identical and that unit cells must tile to generate the complete two-dimensional lattice. For a given structure, there are often several ways to draw a unit cell. We will select the unit cell with higher symmetry (more 90° or 120° angles) and smaller area ( a × b ).

    image

    12.3  Sketch (a) represents ductility. Metals can be drawn into wires.

    Sketch (b) represents malleability. Metals deform under pressure; they can be pounded into sheets.

    12.4   Arrangement (i) represents close packing. Columns of atoms are offset relative to one another to minimize the amount of empty space in the two-dimensional lattice.

    12.5

    1. Clearly, the structure is close packed. The question is: cubic or hexagonal? This is a side view of a close-packed array, like the one in Figure 12.13. The key is the arrangement of the third row relative to the first. Looking at any three rows of cannon balls, there is a ball in the third row directly above (at the same horizontal position as) one in the first row. This is an ABABAB pattern and the structure is hexagonal close packed.
    2. CN = 12, regardless of whether the structure is hexagonal or cubic close packed.

    1. CN(1) = 9, CN(2) = 6. The coordination numbers of these two balls are less than 12 because they are on the “surface” of the structure. Of the 12 maximum closest positions, several are unoccupied. Ball 1 is missing two balls that would be in front of it in its own layer, and one ball of the triad in the layer above, for a total of 3 missing and 9 occupied positions. Ball 2 is missing these same three balls, along will all 3 balls in the triad of the layer that would be below it. Ball 2 has six missing and six occupied nearest neighbors.

    12.6   Arrangement (b) is more stable. In (a), the cation is so small that its neighboring anions are nearly touching. Very close contacts among like-charged particles produce strong electrostatic repulsions and an unstable arrangement.

    12.7   Fragment (b) is more likely to give rise to electrical conductivity. Arrangement (b) has a delocalized system of π electrons, in which electrons are free to move. Mobile electrons are required for electrical conductivity.

    12.8

    1. Band A is the valence band.
    2. Band B is the conduction band.
    3. Band A (the valence band) consists of bonding molecular orbitals (MOs).
    4. This is the electronic structure of a p-type doped semiconductor. The electronic structure shows a few empty MOs or “positive” holes in the valence band. This fits the description of a p-type doped semiconductor.
    5. The dopant is Ga. Of the three elements listed, only Ga has fewer valence electrons than Ge, the requirement for a p-type dopant.

    12.9   Polymer (a) is more crystalline and has the higher melting point. The polymer chains in cartoon (a) are linear and have two ordered regions shown in the upper left and lower right. The greater the degree of order, the more crystalline the material. The ordered regions of polymer (a) indicate that there are stronger intermolecular forces attracting the chains to each other. Stronger intermolecular forces mean that polymer (a) will have the higher melting point.

    12.10  The smaller the nanocrystals, the greater the band gap, E g , and the shorter the wavelength of emitted light. According to Figure 6.4, the shortest visible wavelengths appear violet or purple and the longest appear red.

    1. The 4.0 nm nanocrystals will have the smallest E g , and emit the longest wavelength. That describes vial 2, the red one.
    2. The 2.8 nm nanocrystals will have the largest E g , and emit the shortest wavelength. That describes vial 1, the green one.
    3. The band gap of the 100 nm CdTe crystals is 1.5 eV.

      λ = hc E = 6.626 × 10 - 34 J- s × 2 .998 × 10 8 m s × 1 1.5 eV × 1 eV 1.602 × 10 - 19 J = 8.27 × 10 - 7 m Υ = c λ = 2 .998 × 10 8 m/s 8.27 × 10 - 7 m = 3.63 × 10 14 s - 1

      The visible portion of the electromagnetic spectrum has wavelengths up to 750 nm (7.50 × 10 −7 m). The 100 nm CdTe crystals emit a wavelength longer than this, 827 nm (8.27 × 10 −7 m). This light is in the IR portion of the spectrum and is not visible to the human eye.


    Classification of Solids (Section 12.1)

    12.11  Statement (b) best explains the difference. In molecular solids, relatively weak intermolecular forces (hydrogen-bonding, dipole-dipole, dispersion) bind the molecules in the lattice, so relatively little energy is required to disrupt these forces. In covalent-network solids, covalent bonds join atoms into an extended network. Melting or deforming a covalent-network solid means breaking these covalent bonds, which requires a large amount of energy.

    12.12

    1. Covalent-network solid. According to Figure 12.1, diamond and silicon are covalent-network solids. Individual atoms are bound into a three-dimensional network by strong covalent (single) bonds.
    2. Covalent-network solid. The physical properties could describe a covalent-network or ionic solid. Silicon and oxygen are both nonmetals, so their bonding is likely to be covalent.

    12.13

    1. hydrogen-bonding forces, dipole-dipole forces, London dispersion forces
    2. covalent chemical bonds (mainly)
    3. ionic bonds, the Coulombic forces between anions and cations (mainly)
    4. metallic bonds

    12.14

    1. metallic
    2. molecular or metallic (physical properties of metals vary widely)
    3. covalent-network or ionic
    4. covalent-network

    12.15

    1. ionic
    2. metallic
    3. covalent-network (Based on melting point, it is probably a network solid. Transition metals in high oxidation states often form bonds with nonmetals that have significant covalent character. It could also be characterized as ionic with some covalent character to the bonds.)
    4. molecular
    5. molecular
    6. molecular

    12.16

    1. InAs—covalent-network (InAs is a compound semiconductor, with an average of four electrons per atom and somewhat polar covalent bonds.)
    2. MgO—ionic crystal (metal and nonmetal)
    3. HgS—ionic crystal (metal and nonmetal)
    4. In—metallic (metal, Figure 7.13)
    5. HBr—molecular (two nonmetals)

    12.17  Metallic. The melting point eliminates molecular and covalent-network solids. Because the solid conducts electricity and is insoluble in water, it is metallic. Ionic solids are insulators that are often soluble in water.

    12.18  Molecular. The substance is low-melting, which strongly suggests it is molecular. Many molecular solids, such as sucrose, are soluble in water. That an aqueous solution of the white solid does not conduct electricity confirms that this is a molecular solid.


    Structures of Solids (Section 12.2)

    12.19 Analyze/Plan . Crystalline solids have a regular repeat in all three directions. Amorphous solids have no regular repeating structure. Draw diagrams that reflect these definitions. Solve .

    1. image
    2. image

    12.20  Statement (e) is the best explanation for the difference in density. In amorphous silica (SiO 2 ) the regular structure of quartz is disrupted; the loose, disordered structure, has many vacant “pockets” throughout. There are fewer SiO 2 groups per volume in the amorphous solid; the packing is less efficient and less dense.

    12.21 Analyze . Given two two-dimensional structures, draw and describe the unit cells and lattice vectors. Plan . When choosing a unit cell, the environment of each lattice point must be identical and the unit cells must tile to generate the complete two-dimensional lattice. For a given structure, there are often several ways to draw a unit cell. The radii of A and B are equal. Solve.

    image

    12.22

    image

    12.23 Plan . Refer to Figure 12.6 to find geometric characteristics of the seven three- dimensional primitive lattices.


    Solve . Tetragonal, a = b ≠ c , α = β = γ = 90°. In the new lattice, one of the edge lengths is longer than (not the same as) the other two, and all angles remain 90°.

    12.24  Rhombohedral, a = b = ≠ c , α = β = γ 90°. In the new lattice, the angles between the unit cell edges remain equal, but have some value less than 90°. The unit cell edge lengths remain equal.

    12.25  Choice (e), both rhombohedral and triclinic. According to Figure 12.6, if no lattice vectors are perpendicular to each other, none of the unit cell angles (α, β, γ) are 90°. This is characteristic of two of the three- dimensional primitive lattices: triclinic and rhombohedral.

    12.26  Choice (c), rhombohedral. If all three lattice vectors have the same length, a = b = c . This is characteristic of the rhombohedral as well as the cubic lattice.

    12.27  Choice (b), 2. A body-centered cubic lattice is composed of body-centered cubic unit cells. A unit cell contains the minimum number of atoms when it has atoms only at the lattice points. A body-centered cubic unit cell like this is shown in Figure 12.12(b). There is one atom totally inside the cell (1 × 1) and one at each corner (8 × 1/8) for a total of 2 atoms in the unit cell. (Only metallic elements have body-centered cubic lattices and unit cells.)

    12.28  Choice (d), 4. A face-centered cubic lattice is composed of face-centered cubic unit cells. A unit cell contains the minimum number of atoms when it has atoms only at the lattice points. A face-centered cubic unit cell like this is shown in Figure 12.12(c). There is one atom centered on each face (6 × 1/2) and one at each corner (8 × 1/8) for a total of 4 atoms in the unit cell. (Only metallic elements have face-centered cubic lattices and unit cells.)

    12.29 Analyze . Given a diagram of the unit cell dimensions and contents of nickel arsenide, determine what kind of lattice this crystal possess, and the empirical formula of the compound. Plan . Refer to Figure 12.6 to find geometric characteristics of the seven three-dimensional primitive lattices. Decide where atoms of the two elements are located in the unit cell and use Table 12.1 to help determine the empirical formula.

    1. a = b = 3.57 Å. c = 5.10 Å ≠ a or b . α = β = 90°, γ = 120°. This unit cell is hexagonal. There are no atoms in the exact middle of the cell or on the face centers, so it is a primitive hexagonal unit cell. Nickel arsenide has a primitive hexagonal unit cell and crystal lattice.
    2. There are Ni atoms at each corner of the cell (8 × 1/8) and centered on four of the unit cell edges (4 × 1/4) for a total of 2 Ni atoms. There are 2 As atoms totally inside the cell. The unit cell contains 2 Ni and 2 As atoms; the empirical formula is NiAs.

    12.30

    1. a = b = 3.55 Å. c = 6.18 Å a or b . α = β = γ = 90°. This is a tetragonal unit cell and crystal lattice.
    2. There is one Al atom totally inside the cell and K atoms at each corner (8 × 1/8). There are F atoms centered on four of the unit cell faces (4 × 1/2) and two F atoms totally inside the cell. The unit cell contains 1 K, 1 Al, and 4 F atoms. The empirical formula is KAlF 4 .

    Metallic Solids (Section 12.3)

    12.31 Analyze/Plan .  Consider body-centered and face-centered structures shown in Figures 12.11 and 12.12 to relate the structures and densities of the metals.

    Solve . A body-centered cubic structure has more empty space than a face-centered cubic one. (A faced-centered cubic structure is one of the close-packed structures.) The more empty space, the less dense the solid. We expect potassium, which has the lowest density of the listed elements, to adopt the body-centered cubic structure.

    12.32  Metallic: (b) NiCo alloy and (c) W. The lattices of these substances are composed of neutral metal atoms. Delocalization of valence electrons produces metallic properties.

    Not metallic: (d) Ge is a metalloid, not a metal. (a) TiCl 4 and (e) ScN are ionic compounds; in ionic compounds, electrons are localized on the individual ions, precluding metallic properties.

    12.33 Analyze . Give diagrams of three structure types, find which is most densely packed and which is least densely packed. Plan . Assume that the same element packs in each of the three structures, so that atomic mass and volume are constant. Then we are analyzing the packing efficiency or relative amount of empty space in each structure. Solve .

    Structure type A has a face-centered cubic unit cell with metal atoms only at the lattice points; this corresponds to a cubic close-packed structure. Structure type B has a body- centered cubic unit cell with metal atoms at the lattice points; this is also a body- centered cubic structure. Structure type C has a hexagonal unit cell with two atoms totally inside the cell. Building up many unit cells into a lattice (Figure 12.14) leads to a hexagonal close-packed structure.

    1. In both cubic and hexagonal-close packed structures, any individual atom has twelve nearest neighbor atoms. Both structures are close packed and have equal amounts of empty space. Structure types A and C have equally dense packing and are more densely packed than structure type B.
    2. Structure type B, which is not close packed, has the least dense atom packing.

    12.34

    1. The density of a crystalline solid is (unit cell mass/unit cell volume). Solve for unit cell volume, then use geometry and the properties of a body-centered cubic unit cell to calculate the atomic radius of sodium. There are 2 Na atoms in each body-centered cubic unit cell (Figure 12.12).

      V = unit cell mass ρ = 2 Na atoms × 2 2 .99 g Na 6.022 × 10 23 Na atoms × cm 3 0.97 g × 1 Å 3 ( 10 - 8 cm) 3 = 78.71 = 79 Å 3

      For a cubic unit cell, V = a 3 . We need the relationship between atomic radius and unit cell edge length for a body-centered cubic unit cell. In a body-centered cubic metal structure, the atoms touch along the body diagonal, d 2 . Then, d 2 = 4 r. From the Pythagorean theorem, d 2 = 3 a . (See Solution 12.36 (c).)

      a = ( V ) 1 / 3 = ( 78.71 ) 1 / 3 = 4.2857 = 4.3 Å.

      d 2 = 4 r Na ; d 2 = 3 a ; r Na = 3 a / 4 = 3 × 4. 2857 A 4 = 1.8557 = 1.9 Å


    1. A cubic close-packed metal structure has a face-centered cubic unit cell; there are 4 atoms in each unit cell and atoms touch along the face diagonal, d 1 . Then, d 1 = 4 r. From the Pythagorean theorem, d 1 = 2 a .

      d 1 = 4 r Na ; d 1 = 2 a ; a = 4 r Na / 2 = 4 × 1. 8557 Å 2 = 5.2489 = 5.2 Å = 5 .2 × 10 - 8 cm

      ρ = 4 Na atoms ( 5.2489 × 10 - 8 cm) 3 × 2 2 .99 g Na 6.022 × 10 23 Na atoms = 1.056 = 1 .1 g/cm 3

      Sodium metal with a cubic close-packed structure would not float on water.

    12.35 Analyze .  Given the cubic unit cell edge length and arrangement of Ir atoms, calculate the atomic radius and the density of the metal. Plan . In a face-centered cubic metal structure, there is space between the atoms along the unit cell edge, but they touch along the face diagonal. See Figure 12.12. Use the geometry of the right equilateral triangle to calculate the atomic radius. From the definition of density and paying attention to units, calculate the density of Ir(s). Solve .

    1. The length of the face diagonal of a face-centered cubic unit cell is four times the radius of the atom and 2 times the unit cell dimension or edge length, a for cubic unit cells.

      4 r = 2 a ; r = 2 a / 4 = 2 × 3.833 Å 4 = 1.3552 = 1.355 Å

    2. The density of iridium is the mass of the unit cell contents divided by the unit cell volume. There are 4 Ir atoms in a face-centered cubic unit cell.

      ρ = 4 Ir atoms ( 3.833 × 10 - 8 cm) 3 × 192.22 g Ir 6.022 × 10 23 Ir atoms = 22.67 g/cm 3

      Check .  The units of density are correct. Note that Ir is quite dense.

    12.36

    1. In a body-centered cubic unit cell, there is one atom totally inside the unit cell (1 × 1) and one atom at each of the eight corners (8 × 1/8), for a total of 2 Ca atoms in each unit cell.
    2. Because atoms are only at lattice points in a body-centered metal structure, each metal atom has an equivalent environment. That is, atoms at the corner of the cell and the interior atom must have equivalent environments. Consider the Ca atom at the middle of the unit cell. It has eight nearest neighbors, the eight Ca atoms at the corners of the cell. Interior atoms in adjacent unit cells are farther from the reference Ca atom and are not “nearest.” (For a corner Ca atom, the eight nearest neighbors are the interior atoms in the eight unit cells that include 1/8 of that corner atom.)
    3. In a body-centered cubic metal structure, the atoms touch along the body diagonal, d 2 . Describe the length of the body diagonal in terms of the unit cell dimension a . Define a triangle with sides that are: d 2 , the body diagonal of the cube; d 1 , the face diagonal of the cube; and a , one edge of the cube. The length of the face diagonal is 2 a and the edge length is a . By the Pythagorean theorem,

      d 2 = d 1 2 + a 2 = ( 2 a ) 2 + a 2 = 3 a

      4 r Ca = 3 a ; a = 4 r Ca / 3 = 4 × 1. 97 A ˚ 3 = 4.5495 = 4.55 A ˚


    1. ρ = 2 Ca atoms ( 4.5495 × 10 - 8 cm) 3 × 4 0 .08 g Ca 6.022 × 10 23 Ca atoms = 1.4136 = 1.41 g/cm 3

    12.37

    1. Analyze . We are given a face-centered cubic unit cell with edge length 5.588 Å. Plan . In a face-centered cubic metal structure, the atoms touch along the face diagonal of the unit cell. The length of the face diagonal of a face-centered cubic unit cell is four times the radius of the atom and 2 times the unit cell dimension or edge length, a for cubic unit cells. Solve .

      4 r = 2 a ; r = 2 a / 4 = 2 × 5.588 A ˚ 4 = 1.9757 = 1.976 A ˚

    2. Plan . The density of calcium is the mass of the unit cell contents divided by the unit cell volume. There are 4 Ca atoms in a face-centered cubic unit cell. Solve .

      ρ = 4 Ca atoms ( 5.588 × 10 - 8 cm) 3 × 40.078 g Ca 6.022 × 10 23 Ca atoms = 1.526 g/cm 3

      Check .  The units of density are correct.

    12.38 Analyze/Plan . We are given the radius of atoms in various cubic structures and asked to calculate the volume of each structure. Use Figure 12.12 to determine the relationship between atomic radius and unit cell edge length a. For cubic unit cells, the three edge lengths are equal. The unit cell volume is a 3 . Solve .

    1. In a primitive cubic structure, atoms touch along a unit cell edge, a .

      a = 2 r; V = 8 r 3 . V = 8(1.82) 3 = 48.2285 = 48.2 Å 3

      (Primitive cubic metal structures are relatively rare.)

    2. In a face-centered cubic metal structure, the atoms touch along the face diagonal, d 1 . Then, d 1 = 4 r. From the Pythagorean theorem, the length of the face diagonal equals 2 times the unit cell dimension, a .

      4 r = 2 a ; a = 4 × r 2 ; V = a 3 = ( 4 × r 2 ) 3 = 136.411 = 136 Å 3

    12.39 Analyze .  Given the structure of aluminum metal and the atomic radius of an Al atom, find the number of Al atoms in each unit cell and the coordination number of each Al atom. Calculate (estimate) the length of the unit cell edge and the density of aluminum metal.

    Plan . Use Figure 12.12(c) to count the number of Al atoms in one unit cell and use Figure 12.13 to visualize the coordination number of each Al atom. According to Figure 12.12(c), there is space between the atoms along the unit cell edge, but they touch along the face diagonal. Use the geometry of the right equilateral triangle and the atomic radius to calculate the unit cell edge length. From the definition of density and paying attention to units, calculate the density of aluminum metal. Solve .

    1. 8 corners × 1/8 atom/corner + 6 faces × ½ atom/face = 4 atoms
    2. Each aluminum atom is in contact with 12 nearest neighbors, 6 in one plane, 3 above that plane, and 3 below. Its coordination number is thus 12.

    1. The length of the face diagonal of a face-centered cubic unit cell is four times the radius of the metal and 2 times the unit cell dimension (usually designated a for cubic cells).

      4 × 1.43 Å = 2 × a ; a = 4 × 1.43 Å 2 = 4.0447 = 4.04 Å = 4 .04 × 10 - 8 cm

    1. The density of the metal is the mass of the unit cell contents divided by the volume of the unit cell.

      density = 4 Al atoms ( 4.0447 × 10 - 8 cm) 3 × 26.98 g Al 6.022 × 10 23 Al atoms = 2.71 g/cm 3

    12.40 Analyze .  Given the atomic arrangement, length of the cubic unit cell edge, and density of the solid, calculate the atomic weight of the element. Plan . If we calculate the mass of a single unit cell, and determine the number of atoms in one unit cell, we can calculate the mass of a single atom and of a mole of atoms. Solve .

    The volume of the unit cell is (4.078 × 10 −8 cm) 3 . The mass of the unit cell is:

    19.30 g cm 3 × ( 4.078 × 10 - 8 ) 3 cm 3 unit cell = 1.3089 × 10 - 21 = 1.31 × 10 - 21 g/unit cell

    There are four atoms of the element present in the face-centered cubic unit cell. Thus the atomic weight is:

    1.3089 × 10 - 21 g unit cell × 1 unit cell 4 atoms × 6.0221 × 10 23 atoms 1 mol = 197.06 = 197 g/mol

    Check .  The result is in the range of known atomic weights and the units are correct. The element is most likely gold.

    12.41  Statement (b) is false. Alloys are mixtures, not compounds, that vary in composition. One or more of the components of the alloy can be a nonmetal.

    12.42

    1. False. Substitutional and interstitial alloys are both solution alloys.
    2. True
    3. True

    12.43 Analyze/Plan . Consider the descriptions of various alloy types in Section 12.3. Solve .

    1. Fe 0.97 Si 0.03 ; interstitial alloy. The radii of Fe and Si are substantially different, so Si could fit in “holes” in the Fe lattice. Also, the small amount of Si relative to Fe is characteristic of an interstitial alloy.
    2. Fe 0.60 Ni 0.40 , substitutional alloy. The two metals have very similar atomic radii and are present in similar amounts.
    3. SmCo 5 , intermetallic compound. The two elements are present in stoichiometric amounts.

    12.44

    1. Cu 0.66 Zn 0.34 , substitutional alloy; similar atomic radii, substantial amounts of both components
    2. Ag 3 Sn, intermetallic compound; set stoichiometric ratio of components
    3. Ti 0.99 O 0.01 , interstitial alloy; very different atomic radii, tiny amount of smaller component

    12.45

    1. True
    2. False. Interstitial alloys form between elements with very different bonding atomic radii.
    3. False. Nonmetallic elements are typically found in interstitial alloys.

    12.46

    1. True
    2. True
    3. False. In stainless steel, the chromium atoms replace iron atoms in the structure. (The atomic radii of Fe and Cr are 1.25 Å and 1.27 Å, respectively. Metal atoms with similar radii form substitutional alloys.)

    12.47 Analyze . Given the color of a gold alloy, find the other element(s) in the alloy and the type of alloy formed. Plan . Refer to “Chemistry Put to Work: ALLOYS OF GOLD”. Solve .

    1. White gold, nickel or palladium, substitutional alloy
    2. Rose gold, copper, substitutional alloy
    3. Green gold, silver, substitutional alloy

    12.48  The term expansion implies that unit cell edge lengths will increase. This agrees with the fact that the volume increases. Density is the mass of a unit cell divided by its volume. Since the number of atoms in a unit cell does not change when the metal is heated, but the volume increases, the density will decrease.

    Metallic Bonding (Section 12.4)

    12.49

    1. True
    2. False. See statement (a).
    3. False. Delocalized electrons in metals facilitate the transfer of kinetic energy, which is the basis of thermal conductivity.
    4. False. Metals have large thermal conductivities.

    12.50  The part of the bar sitting in the dark will feel hot. The structure of the metal and its delocalized electrons extend over the whole bar. The high thermal conductivity of the metal facilitates heat transfer from the part of the bar in the sun to the part of the bar in the shade.

    12.51 Plan . By analogy to Figure 12.22, the most bonding, lowest energy MOs have the fewest nodes. As energy increases, the number of nodes increases. When constructing an MO diagram from AOs, total number of orbitals is conserved. The MO diagram for a linear chain of six Li atoms will have six MOs, starting with zero nodes and maximum overlap, and ending with five nodes and minimum overlap. Solve .


    image
    1. Six. Six AOs require six MOs.
    2. Zero nodes in lowest energy orbital
    3. Five nodes in highest energy orbital
    4. Two nodes in the HOMO
    5. Three nodes in the LUMO
    6. The HOMO-LUMO energy gap for the six-atom diagram is smaller than the one for the four-atom diagram. In general, the more atoms in the chain, the smaller the HOMO-LUMO energy gap.

    12.52

    image
    1. 8
    2. 0
    3. 7
    4. 3
    5. 4
    6. smaller

    12.53 Analyze/Plan .  Consider the definition of ductility, as well as the discussion of metallic bonding in Section 12.4. Solve .

    Ductility is the property related to the ease with which a solid can be drawn into a wire. Basically, the softer the solid the more ductile it is. The more rigid the solid, the less ductile it is. For metals, ductility decreases as the strength of metal-metal bonding increases, producing a stiffer lattice less susceptible to distortion.

    1. Ag is more ductile. Mo, with 6 valence electrons, has a filled bonding band, strong metal-metal interactions, and a rigid lattice. This predicts high hardness and low ductility. Ag, with 11 valence electrons, has a nearly filled antibonding band as well as a filled bonding band. Bonding is weaker than in Mo, and Ag is more ductile.
    2. Zn is more ductile. Si is a covalent-network solid with all valence electrons localized in bonds between Si atoms. Covalent-network substances are high-melting, hard, and not particularly ductile.

    12.54  Statement (d) is false and does not follow from the fact that alkali metals have relatively weak metal-metal bonds. We expect strong metal-metal bonds to be present in metals with high melting points (Figure 12.21).

    12.55  The relevant electron configurations are:

    Y: [Kr]5s 2 4d 1 ; Zr: [Kr]5s 2 4d 2 ; Nb: [Kr]5s 2 4d 3 ; Mo: [Kr]5s 1 4d 5 ;

    The order of increasing melting points is Y < Zr < Nb < Mo.

    Moving across the fifth period from Y to Mo, the number of valence electrons increases, from 3 for Y to 6 for Mo. More valence electrons (up to 6) mean increased occupancy of the bonding molecular orbital band, and increased strength of metallic bonding. Melting requires that atoms are moving relative to each other. Stronger metallic bonding requires more energy to break bonds and mobilize atoms, resulting in higher melting points from Y to Mo.

    12.56  In each group, choose the metal that has the number of valence electrons closest to six.

    1. Re
    2. Mo
    3. Ru

    Ionic and Molecular Solids (Sections 12.5 and 12.6)

    12.57

    1. Sr:  Sr atoms occupy the 8 corners of the cube.

      8 corners × 1/8 sphere/corner = 1 Sr atom

      O:  O atoms occupy the centers of the 6 faces of the cube.

      6 faces × 1/2 atom/face = 3 O atoms

      Ti:   There is 1 Ti atom at the body center of the cube.

      Empirical Formula: SrTiO 3

    2. Six. The Ti atom at the center of the cube is coordinated to the six O atoms at the face centers.

    1. Twelve. Each Sr atom occupies one corner of 8 unit cells. Sr is coordinated to 3 oxygen positions in each unit cell for a total of 24 oxygen positions. However, each O position is in the center of a cell face, with half-occupancy in each cell.

      24 oxygen positions × ½ occupancy = 12 oxygen atoms. Each Ti atom is coordinated to 12 oxygen atoms.

    12.58  Co:  8 corners × 1/8 sphere/corner = 1 Co atom

    O:  2 atoms completely inside the unit cell = 2 O atoms

    Formula: CoO 2

    The oxidation number of Co is +4, the oxidation state is Co(IV).

    12.59 Analyze/Plan . The density of MnS is the mass of the unit cell contents divided by the volume of the unit cell. MnS has the same structure as rock salt, or NaCl. Refer to the NaCl structure in Figures 12.25 and 12.26 to determine the unit cell contents. Use the unit cell edge length to find unit cell volume. Calculate density. Solve .

    By analogy to the NaCl structure, with Mn replacing Na and S replacing Cl, there are 8 sulfide ions (green spheres) on the corners of the unit cell, and 6 sulfide ions in the middle of the faces. The number of sulfide ions per unit cell is then [8(1/8) + 6(1/2)] = 4. There is 1 manganese ion (purple sphere almost hidden in figure) completely inside the unit cell and 12 manganese ions (purple spheres) along the unit cell edges. The number of manganese ions is then [1 + 12(1/4)] = 4. This result satisfies charge balance requirements.

    4 Mn 2+ , 4 S 2– , 4 MnS formula units. The mass of 1 MnS formula unit is 87.003 g/ 6.022 × 10 23 NaF units.

    d = 4 MnS units ( 5.223 A ˚ ) 3 × 87.003 g 6.0221 × 10 23 MnS units × ( 1 Å 1 × 10 - 8 cm ) 3 = 4.056 g/cm 3

    Check . The value for the density of alabandite (MnS) reported in the CRC Handbook of Chemistry and Physics, 74th Ed., is 3.99 g/cm 3 . The calculated density is within 2% of the reported value.

    12.60  Calculate the mass of a single unit cell and then use density to find the volume of a single unit cell. The edge length is the cube root of the volume of a cubic cell. From the previous exercise, there are four PbSe units in a NaCl-type unit cell . The unit cell edge length is designated a .

    8.27 g/cm 3 = 4 PbSe units a 3 × 286.2 g 6 .022 × 10 23 PbSe units × ( 1 Å 1 × 10 - 8 cm ) 3

    a 3 = 229.87 Å 3 , a = 6.13 Å

    12.61 Analyze .  Given the atomic arrangement and length of the unit cell side, calculate the density of HgS and HgSe. Qualitatively and quantitatively compare the densities of the two solids. Plan . Calculate the mass and volume of a single unit cell and then use them to find density. The unit cell volume is the cube of edge length. Solve .


    1. According to Figures 12.25 and 12.26, sulfide ions (yellow spheres) occupy the corners and faces of the unit cell (in a face-centered cubic arrangement), for a total of 6(1/2) + 8(1/8) = 4 S 2– ions per unit cell. There are 4 mercury ions (gray spheres) totally in the interior of the cell. This means there are 4 HgS units in a unit cell with the zinc blende structure.

      density = 4 HgS units ( 5.852 Å ) 3 × 232.655 g 6.022 × 10 23 HgS units × ( 1 Å 1 × 10 - 8 cm ) 3 = 7.711 g/cm 3

    2. We expect Se 2– to have a larger ionic radius than S 2– , because Se is below S in the chalcogen family and both ions have the same charge. Thus, HgSe will occupy a larger volume and the unit cell edge will be longer.
    3. For HgSe, also with the zinc blende structure:

      density = 4 HgSe units ( 6.085 Å ) 3 × 279.55 g HgSe 6.022 × 10 23 HgSe units × ( 1 Å 1 × 10 - 8 cm ) 3 = 8.241 g/cm 3

      Even though HgSe has a larger unit cell volume than HgS, it also has a larger molar mass. The mass of Se is more than twice that of S, whereas the radius of Se 2– is only slightly larger than that of S 2– (Figure 7.8). The greater mass of Se accounts for the greater density of HgSe.

    12.62

    1. Each cell edge goes through two half Rb + ions (at the corners) and one full I ion (centered on the edge). The length of an edge, a , is then

      a = 2r Rb + 2r I = 2(1.66 Å) + 2(2.06 Å) = 7.44 Å

    2. density = 4 RbI units ( 7.44 Å ) 3 × 212.37 g 6.022 × 10 23 RbI units × ( 1 Å 1 × 10 - 8 cm ) 3 = 3.43 g/cm 3
    3. In the CsCl-type structure, I anions sit at the corners of the cube and an Rb + cation sits completely inside the unit cell (at or near the body center). Assume that the anions and cations touch along the body diagonal d 2 , so that

      d 2 = 2r Rb + 2r I = 2(1.66 Å) + 2(2.06 Å) = 7.44 Å. (This is the edge length of the NaCl-type structure.)

      From Solution 12.36, the relationship between the body diagonal and edge of a cube is: d 2 = 3 × a ; a = d 2 / 3 . a = 7.44 Å / 3 = 4.2955 = 4.30 Å

    4. There is one RbI unit (8 × 1/8 I anions and 1 Rb + cation) in the CsCl-type unit cell.

      density = 1 RbI unit ( 4.2955 Å ) 3 × 212.37 g 6.022 × 10 23 RbI units × ( 1 Å 1 × 10 - 8 cm ) 3 = 4.45 g/cm 3

      The density of the CsCl-type structure is greater than the density of the NaCl-type structure, owing to the much smaller unit cell volume.

    12.63 Analyze . Given that CuI, CsI, and NaI uniquely adopt one of the structure types pictured in Figure 12.26, match the ionic compound with its structure. Use ionic radii to inform your decision. Plan . Note the relative cation and anion radii in the three structures in Figure 12.26. Match these ratios with those of CuI, CsI, and NaI. Solve .


    1. In the CsCl structure, the anion and cation have about the same radius; in the NaCl structure, the anion is somewhat larger than the cation; in the ZnS structure the anion is much larger than the cation.

      In the three compounds given, Cs + (r = 1.81 Å) and I (r = 2.06 Å) have the most similar radii; CsI will adopt the CsCl-type structure. The radii of Na + (r = 1.16 Å) and I (r = 2.06 Å) are somewhat different; NaI will adopt the NaCl-type structure. The radii of Cu + (r = 0.74 Å) and I (r = 2.06 Å) are very different; CuI has the ZnS-type structure.

    2. As cation size decreases, coordination number of the anion decreases. In CsI, I has a coordination number of eight; in NaI, I has a coordination number of six; in CuI, I has a coordination number of four.

    12.64

    1. In CaF 2 the ionic radii are very similar, Ca 2+ (r = 1.14 Å) and F (r = 1.19 Å). In ZnF 2 the cation radius is smaller and the ionic radii are more different, Zn 2+ (r = 0.88 Å) and F (r = 1.19 Å). Cations in both structures in the exercise are shown with equal radii, so direct inspections does not answer the question. We can, however, refer to the trend that, for compounds with the same cation/anion ratio, as cation size decreases, coordination number of the anion decreases. The anion coordination number (CN) for the rutile structure is 3 and for the fluorite structure is 4. The compound with the smaller cation, ZnF 2 , will adopt the rutile structure and the compound with the larger cation, CaF 2 , will adopt the fluorite structure. (Detailed analysis of coordination numbers follows.)
    2. Rutile (top) structure, cation (blue) coordination number (CN) = 6, anion (green) CN = 3. [The blue cation in the interior or the cell is coordinated to the six (green) anions associated with the same cell; either of the green interior anions is associated with the triangle of blue cations located inside and at the two nearest corners of the cell.]

      Fluorite (bottom) structure, cation (blue) CN = 8, anion (green) CN = 4. [A blue cation at one of the face centers is coordinated to the four nearest green anions inside the cell and to four identical anions in an adjacent cell that also contains the cation. Any of the green interior anions is coordinated to a tetrahedron of blue cations located at one corner and the three nearest face centers.]

    12.65 Analyze . Given three magnesium compounds in which the coordination number (CN) of Mg 2+ is six, determine the coordination number of the anion. Plan . Use Equation 12.1 with the cation/anion ratio of each compound and the Mg 2+ coordination number to calculate the anion coordination number in each compound. Solve .

    1. MgS: 1 cation, 1 anion, cation CN = 6

      number of cations per formula unit number of anions per formula unit = anion coordination number cation coordination number

      anion CN = cation CN × # of cations per formula unit # of anions per formula unit = 6 × 1 1 = 6


    1. MgF 2 : 1 cation, 2 anions, cation CN = 6

      anion CN = cation CN × # of cations per formula unit # of anions per formula unit = 6 × 1 2 = 3

    1. MgO: 1 cation, 1 anion, cation CN = 6, anion CN = 6. The cation/anion ratio and cation CN are the same as in part (a), so the anion CN is the same (6).

    12.66 # of cations per formula unit # of anions per formula unit = anion coordination number cation coordination number

    cation CN = anion CN × # of anions per formula unit # of cations per formula unit

    1. AlF 3 :  1 cation, 3 anions, anion CN = 2; cation CN = (2 × 3)/1 = 6
    2. Al 2 O 3 : 2 cations, 3 anions, anion CN = 6; cation CN = (6 × 3)/2 = 9
    3. AlN:  1 cation, 1 anion, anion CN = 4; cation CN = (4 × 1)/1 = 4

    12.67

    1. False. Although both molecular solids and covalent-network solids have covalent bonds, the melting points of molecular solids are much lower because intermolecular forces among their molecules are much weaker than covalent bonds among atoms in a covalent-network solid.
    2. True. The statement is true if there are no significant differences in polarity or molar mass of the molecules being compared.

    12.68

    1. False. Melting point is a bulk property, not a molecular property.
    2. True. Strengths of intermolecular forces determine properties of the bulk material like melting point.

    Covalent-Network Solids (Section 12.7)

    12.69

    1. Ionic solids are much more likely to dissolve in water. Polar water molecules can disrupt ionic bonds to surround and separate ions and form a solution, but they cannot break the covalent bonds of a covalent-network solid.
    2. Covalent-network solids can become better conductors of electricity via chemical substitution. Most semiconductors are covalent-network solids, and doping or chemical substitution changes their electrical properties.

    12.70  The extended network of localized covalent bonds in a covalent-network solid produces solids that are inflexible, hard, and high-melting. The delocalized nature of metallic bonding leads to flexibility, because atoms can move relative to one another without “breaking” bonds. Metals have a wide range of hardness and melting point, depending on the occupancy of the bands.

    1. Ductility, metallic solid
    2. Hardness, covalent-network solid and metallic solid, depending on the strength of metallic bonding
    3. High melting point, covalent-network solid and metallic solid, depending on the strength of metallic bonding

    12.71 Analyze/Plan . Follow the logic in Sample Exercise 12.3. Solve .

    1. CdS. Both semiconductors contain Cd; S and Te are in the same family, and S is higher.
    2. GaN. Ga is in the same family and higher than In; N is in the same family and higher than P.
    3. GaAs. Both semiconductors contain As; Ga and In are in the same family and Ga is higher.

    12.72

    1. InP. P is in the same family and higher than As.
    2. AlP. Al and P are horizontally separated, resulting in greater bond polarity, and Al and P are in the row above Ge. Band gap values in Table 12.4 confirm this order.
    3. AgI. The four elements are in the same row; Ag and I are farther apart than Cd and Te.

    12.73 Analyze .  Given: GaAs. Find: dopant to make n-type semiconductor.

    Plan .  An n-type semiconductor has extra negative charges. If the dopant replaces a few Ga atoms, it should have more valence electrons than Ga, Group 3A.

    Solve .  The obvious choice is a Group 4A element, either Ge or Si. Ge would be closer to Ga in bonding atomic radius (Figure 7.7).

    12.74  p-type semiconductors have a slight electron deficit. If the dopant replaces As, Group 5A, it should have fewer than five valence electrons. The dopant will be a 4A element, probably Si or Ge. Si would be closer to As in bonding atomic radius.

    12.75

    1. Analyze .  Given: 1.1 eV. Find: wavelength in meters that corresponds to the energy 1.1 eV. Plan .  Use dimensional analysis to find wavelength.

      Solve .  1 eV = 1.602 × 10 −19 J (inside back cover of text); λ = hc/E

      λ = 6.626 × 10 - 34 J-s × 3.00 × 10 8 m s × 1 1.1 eV × 1 eV 1 .602 × 10 - 19 J = 1.128 × 10 - 6 = 1.1 × 10 - 6 m

    2. Si can absorb energies ≥ 1.1 eV, or wavelengths ≤ 1.1 × 10 −6 m. A wavelength of 1.1 × 10 −6 m corresponds to 1100 nm. The range of wavelengths in the visible portion of the spectrum is 300 to 750 nm. Silicon absorbs all wavelengths in the visible portion of the solar spectrum.
    3. Si absorbs wavelengths less than 1100 nm. This corresponds to approximately 80–90% of the total area under the curve.

    12.76

    1. From Table 12.4, the band gap of CdTe is 1.50 eV (or 145 kJ/mol).
    2. λ = hc E = 6.626 × 10 - 34 J- s × 2 .998 × 10 8 m s × 1 1.50 eV × 1 eV 1.602 × 10 - 19 J = 8.267 × 10 - 7 = 8.27 × 10 - 7 m

    (c,d) CdTe can absorb energies ≥ 1.50 eV or wavelengths ≤ 8.27 × 10 −7 m. A wavelength of 8.27 × 10 −7 m corresponds to 827 nm. CdTe absorbs a smaller range of wavelengths and thus a smaller portion of the solar spectrum than Si.


    12.77 Plan/Solve .  Follow the logic in Solution 12.75.

    λ = hc/E = 6.626 × 10 - 34 J-s × 3 .00 × 10 8 m s × 1 1.74 eV × 1 eV 1 .602 × 10 - 19 J = 7.131 × 10 - 7 = 7.13 × 10 - 7 m = 713 nm

    The emitted light with a wavelength of 713 nm is red light in the visible region of the electromagnetic spectrum.

    12.78  The band gap, ΔE, of GaAs is 1.43 eV.

    λ = hc E = 6.626 × 10 - 34 J- s × 2 .998 × 10 8 m s × 1 1.43 eV × 1 eV 1.602 × 10 - 19 J = 8.67 × 10 - 7 m (867 nm)

    The visible portion of the electromagnetic spectrum has wavelengths up to 750 nm (7.50 × 10 −7 m). GaAs emits longer, 867 nm, light in the IR portion of the spectrum.

    12.79 Analyze/Plan . From Table 12.4, E g for GaAs (x = 0) is 1.43 eV and for GaP (x = 1) is 2.26 eV. If E g varies linearly with x, the band gap for x = 0.5 should be approximately the average of the two extreme values.

    Solve . (1.43 + 2.26)/2 = 1.845 = 1.85 eV.

    λ = hc/E = 6.626 × 10 - 34 J-s × 2 .998 × 10 8 m s × 1 1.845 eV × 1 eV 1 .602 × 10 - 19 J = 6.721 × 10 - 7 m = 672 nm

    12.80  Reverse the logic in Solution 12.79. Given λ, calculate E g . Then solve for x assuming the value of E g is a linear combination of the stoichiometric contributions of GaP and GaAs. That is, 2.26 x + 1.43(1 – x) = E g .

    E g = hc λ = 6.626 × 10 - 34 J- s 6 .60 × 10 - 7 m × 2.998 × 10 8 m s × 1 eV 1.602 × 10 - 19 J = 1.8788 = 1.88 eV

    2.26 x + 1.43(1 – x) = 1.8788; 2.26 x – 1.43 x + 1.43 = 1.8788

    0.83 x = 0.4488; x = 0.5407 = 0.54

    Check . From Solution 12.79, an E g of 1.85 eV corresponds to x = 0.5. E g =1.88 eV should have a very similar composition, and x = 0.54 is very close to x = 0.5. The P/As composition is very sensitive to wavelength, and provides a useful mechanism to precisely tune the wavelength of the diode.

    Polymers (Section 12.8)

    12.81

    1. A monomer is a small molecule with low molecular mass that can be joined with other monomers to form a polymer. It is the repeating unit of a polymer.
    2. Ethene (ethylene) is commonly used as a monomer, while ethanol and methane are not. Methane does not have reactive groups that can be chemically linked into polymers. Ethanol is reactive because of its one alcohol functional group, but it cannot form long polymer chains. Alcohol monomers used to form polyesters have two alcohol groups.

    12.82 n -decane does not have a sufficiently high chain length or molecular mass to be considered a polymer.


    12.83  A value of 100 amu is too low to be the molecular weight of a polymer; it would not represent enough monomer units. Often a single monomer has a molecular weight greater than 100 amu. Reasonable values for a polymer’s molecular weight are 10,000 amu (typical for low density polyethylene), 100,000 amu, and 1,000,000 amu (typical for high density polyethylene).

    12.84  True. In an addition polymerization, all atoms present in the monomer are present in the polymer. (We are assuming 100% yield. However, there is usually some unreacted monomer as well as new polymer.)

    12.85 Analyze .  Given two types of reactant molecules, we are asked to write a condensation reaction with an ester product. Plan .  A condensation reaction occurs when two smaller molecules combine to form a larger molecule and a small molecule, often water. Consider the structures of the two reactants and how they could combine to join the larger fragments and split water. Solve .

    A carboxylic acid contains the image functional group; an alcohol contains the –OH functional group. These can be arranged to form the image ester functional group and H2O. Condensation reaction to form an ester:

    image

    If a dicarboxylic acid (two –COOH groups, usually at opposite ends of the molecule) and a dialcohol (two –OH groups, usually at opposite ends of the molecule) are combined, there is the potential for propagation of the polymer chain at both ends of both monomers. Polyethylene terephthalate (Table 12.6) is an example of a polyester formed from the monomers ethylene glycol and terephthalic acid.

    12.86

    image

    12.87 Analyze/Plan .  In an addition polymerization, the monomer contains a multiple bond. We expect the monomer associated with this polymer to have either a double or triple bond.

    Solve . This polymer is similar to polyethylene (Figure 12.35), with two hydrogen atoms replaced by chlorine atoms. The monomer in polyethylene is ethylene or ethene. The monomer for this polymer is dichloroethylene or dichloroethene.

    image

    12.88

    1. By analogy to polyisoprene, Figure 12.42,
      image
    2. image

    12.89 Analyze/Plan . Given the structure of a polymer, identify the monomers that react to form the polymer. Compare the polymer structure to other condensation polymers shown in Table 12.6. Solve .

    Kevlar is a polymer similar to nylon. The connecting unit is image

    According to Figure 12.37, the monomers involved in this type polymer are a diacid, a molecule with two carboxylic acid groups, and a diamine, a molecule with two –NH 2 groups. The spacers in Kevlar are both benzene rings. The monomers used to produce Kevlar are:

    image

    12.90

    1. image
    2. image

      In the peptide shown here, the N terminus amino acid contains R1 and the C terminus amino acid contains R3.

    3. Six (R1–R2–R3, R2–R3–R1, R3–R1–R2, R1–R3–R2, R3–R2–R1, R2–R1–R3)

    12.91

    1. Most of a polymer backbone is composed of σ bonds. The geometry around individual atoms is tetrahedral with bond angles of 109°, so the polymer is not flat, and there is relatively free rotation around the σ bonds. The flexibility of the molecular chains causes flexibility of the bulk material. Flexibility is enhanced by molecular features that inhibit order, such as branching, and diminished by features that encourage order, such as cross-linking or delocalized π electron density.
    2. Less flexible. Cross-linking is the formation of chemical bonds between polymer chains. It reduces flexibility of the molecular chains and increases the hardness of the material. Cross-linked polymers are less chemically reactive because of the links.

    12.92  At the molecular level, the longer, unbranched chains of HDPE fit closer together and have more crystalline (ordered, aligned) regions than the shorter, branched chains of LDPE. Closer packing leads to higher density.

    12.93  Low degree of crystallinity. A good plastic wrap is extremely flexible; the lower the degree of crystallinity, the less rigid and more flexible the polymer.

    12.94

    1. True
    2. True
    3. True

    Nanomaterials (Section 12.9)

    12.95  Continuous energy bands of molecular orbitals require a large number of atoms contributing a large number of atomic orbitals to the molecular orbital scheme. If a solid has dimensions 1-10 nm, nanoscale dimensions, there may not be enough contributing atomic orbitals to produce continuous energy bands of molecular orbitals.

    12.96

    1. Calculate the wavelength of light that corresponds to 2.4 eV, then look at a visible spectrum such as Figure 6.4 to find the color that corresponds to this wavelength.

      λ = hc E = 6.626 × 10 - 34 J-s × 3 .00 × 10 8 m s × 1 2.4 eV × 1 eV 1 .602 × 10 - 19 J = 5.167 × 10 - 7 = 5.2 × 10 - 7 m( 520 nm)

      The emitted 520 nm light is green.

    2. Yes. The wavelength of blue light is shorter than the wavelength of green light. Emitting a shorter wavelength requires a larger band gap. As particle size decreases, band gap increases. Appropriately sized CdS quantum dots, far smaller than a large CdS crystal, would have a band gap greater than 2.4 eV and could emit shorter wavelength blue light.
    3. No. The wavelength of red light is greater than the wavelength of green light. The large CdS crystal has plenty of AOs contributing to the MO scheme to ensure maximum delocalization. A bigger crystal will not reduce the size of the band gap. The 2.4 eV band gap of the large CdS crystal represents the minimum energy and maximum wavelength of light that can be emitted by this material.

    12.97

    1. False. As particle size decreases, the band gap increases. The smaller the particle, the fewer AOs that contribute to the MO scheme, the more localized the bonding and the larger the band gap.
    2. False. The wavelength of emitted light corresponds to the energy of the band gap. As particle size decreases, band gap increases and wavelength decreases (E = hc/λ).

    12.98  True. Blue light has short wavelengths, corresponding to a relatively large band gap. As particle size decreases, band gap increases and wavelength decreases. We could begin with a semiconductor with a smaller band gap and make it a nanoparticle to increase E g and decrease wavelength. (Nanoparticle size becomes one more way to tune the properties of semiconductors.)

    12.99 Analyze .  Given: Au, 4 atoms per unit cell, 4.08 Å cell edge, volume of sphere = 4/3 π r 3 . Find: Au atoms in 20 nm diameter sphere.

    Plan .  Relate the number of Au atoms in the volume of 1 cubic unit cell to the number of Au atoms in a 20 nm diameter sphere. Change units to Å (you could just as well have chosen nm as the common unit), calculate the volumes of the unit cell and sphere, and use a ratio to calculate atoms in the sphere.

    Solve .  vol. of unit cell = (4.08 Å) 3 = 67.9173 = 67.9 Å 3

    20 nm diameter = 10 nm radius; 10 nm × 1 × 10 - 9 m 1 nm × 1 Å 1 × 10 - 10 m = 100 Å radius

    (Note that 1 nm = 10 Å.)

    vol. of sphere = 4/3 × 3.14159 × (100 Å) 3 = 4.18879 × 10 6 = 4.19 × 10 6 Å 3

    4 Au atoms 67.9173 Å 3 = x Au atoms 4.18879 × 10 6 Å 3 ; x = 2.46699 × 10 5 = 2.47 × 10 5 Au atoms

    12.100

    1. There are four InP formula units in each cubic unit cell. Calculate the number of unit cells contained in a 3.00 nm cubic crystal and a 5.00 nm crystal. The volume of one unit cell is

      ( 5.869 ) 3 Å 3 × 1 nm 3 10 3 Å 3 = 0.20216 = 0.2022 nm 3

      The volume of the 3.00 nm crystal is (3.00) 3 nm 3 = 27.0 nm 3

      27 .0 nm 3 / crystal 0 .20216 nm 3 / unit cell × 4 InP units unit cell = 534.23 = 534 InP units

      That is, 534 In atoms and 534 P atoms.

      The volume of the 5.00 nm crystal is (5.00) 3 nm 3 = 125 nm 3

      125 nm 3 / crystal 0 .20216 nm 3 / unit cell × 4 InP units unit cell = 2473.3 = 2 .473 × 10 3 InP units

      That is, 2473 In atoms and 2473 P atoms.

    2. As a quantum dot gets smaller, its band gap gets larger. Larger band gaps correspond to a higher energy and shorter wavelength of emitted light. Blue light has a shorter wavelength than orange light (Figure 6.4). Therefore, the smaller 3.00 nm cube emits blue light and the 5.00 nm cube emits orange light.

    12.101  Statement (b) is correct.

    1. Neither graphite nor graphene is molecular. Both are covalent-network solids with covalent-network bonding.
    2. Neither graphite nor graphene is an insulator or a metal.
    3. Both are pure carbon.
    4. The carbon atoms in both are sp 2 hybridized.

    12.102  Statement (e) is correct. Statements (b) and (c) both support the notion that buckyballs are discrete molecules and not extended materials.

    Additional Exercises

    12.103

    1. NaCl, ionic; MgCl 2 , ionic; PCl 3 , molecular; SCl 2 , molecular. The melting points of the solids are definitive. PCl 3 and SCl 2 are molecular solids because of their low melting points. NaCl and MgCl 2 are ionic solids because of their relatively high melting points. A covalent-network solid has a much higher melting point than either of these compounds.
    2. CaCl 2 has a much higher melting point than SiCl 4 . The ionic solid CaCl 2 is composed of a metal cation and a nonmetal anion. The molecular solid SiCl 4 is composed of a metalloid, Si, and a nonmetal, Cl.

    12.104  According to Figure 12.6, a tetragonal unit cell has a square base, with the third lattice vector perpendicular to the base but with a different length. In other words, a = b ≠ c , α = β = γ = 90 o . To create a face-centered tetragonal unit cell, add a lattice point in the center of each face (both square and rectangular faces). Draw a second face-centered tetragonal unit cell above or below the first one.

    The square base of a new tetragonal unit cell can be drawn by connecting the face centers of the four rectangular faces. Connecting these four face centers with those of the adjacent (old) tetragonal cell creates the new tetragonal unit cell. The lattice point at the center of the old square face becomes the body center of the new unit cell.

    image

    12.105  Orthorhombic, a ≠ b ≠ c , α = β = γ = 90 o . In the new lattice, the edge lengths are all different, and all angles remain 90 o .


    12.106  Qualitatively from Figure 12.12, a face-centered cubic structure has a greater portion of its volume occupied by metal atoms and less “empty” space than a body-centered structure. The face-centered structure will have the greater density.

    Quantitatively, use the atomic radius of iron, 1.32 Å from Figure 7.7, to estimate the unit cell edge length in each of the structures, then calculate the estimated density of both structures. Recall that a face-centered cubic structure has 4 atoms per unit cell, and a body-centered structure has 2.

    Face-centered cubic: 4 r Fe = 2 a ; a = 4 r Fe / 2 = 4 × 1. 32 Å 2 = 3.7335 = 3.73 Å

    ρ = 4 Fe atoms ( 3.7335 × 10 - 8 cm) 3 × 5 5 .845 g Fe 6.022 × 10 23 Fe atoms = 7.1277 = 7.13 g/cm 3

    Body-centered cubic: 4 r Fe = 3 a ; a = 4 r Fe / 3 = 4 × 1. 32 Å 3 = 3.0484 = 3.05 Å

    ρ = 2 Fe atoms ( 3.0484 × 10 - 8 cm) 3 × 5 5 .845 g Fe 6.022 × 10 23 Fe atoms = 6.5472 = 6.55 g/cm 3

    The face-centered cubic structure has the greater density.

    12.107 The metallic properties of malleability, ductility, and high electrical and thermal conductivity are results of the delocalization of valence electrons throughout the lattice. Delocalization occurs because metal atom valence orbitals of nearest-neighbor atoms interact to produce nearly continuous molecular orbital energy bands. When C atoms are introduced into the metal lattice, their valence orbitals do not have the same energies as metal orbitals, and the interaction is different. This causes a discontinuity in the band structure and limits delocalization of electrons. The properties of the carbon-infused metal begin to resemble those of a covalent-network lattice with localized electrons. The substance is harder and less conductive than the pure metal.

    12.108 Density is the mass of the unit cell contents divided by the unit cell volume [(edge length) 3 ]. Refer to Figure 12.12(c) to determining the number of Ni atoms and Figure 12.17 for the number of Ni 3 Al units in each unit cell.

    Ni: There are 4 Ni atoms in each face-centered cubic unit cell (8 × 1/8 at the corners, 6 × 1/2 on the face centers).

    density = 4 Ni atoms ( 3.53 Å ) 3 × 58.6934 g Ni 6.022 × 10 23 Ni atoms × ( Å 1 × 10 - 8 cm ) 3 = 8.86 g / cm 3

    Ni 3 Al: There is 1 Ni 3 Al unit in each cubic unit cell. According to Figure 12.17, Ni is at the face centers ( 6 × 1/2 = 3 Ni atoms) and Al is at the corners (8 × 1/8 = 1 Al atom); the stoichiometry is correct.

    density = 1 Ni 3 Al unit ( 3.56 Å ) 3 × 203.062 g Ni 3 Al 6.022 × 10 23 Ni atoms × ( Å 1 × 10 - 8 cm ) 3 = 7.47 g / cm 3

    The density of the Ni 3 Al alloy (intermetallic compound) is ~85% of the density of pure Ni. The sizes of the two unit cells are very similar. In Ni 3 Al, one out of every four Ni atoms is replaced with an Al atom. The mass of an Al atom is (~27 amu) is about half that of a Ni atom (~59 amu); the mass of the unit cell contents of Ni 3 Al is ~7/8 (87.5%) that of Ni, and the densities show the same relationship.


    12.109 Ni 3 Al: Ni is at the face centers ( 6 × 1/2 = 3 Ni atoms) and Al is at the corners (8 × 1/8 = 1 Al atom). The atom ratio in the structure matches the empirical formula.

    Nb 3 Sn: The top, front, and side faces of the unit cell are clearly visible; each has 2 Nb atoms centered on it, as opposed to totally inside the cell. The three opposite faces are not completely visible in the diagram, but must be the same by translational symmetry. Two Nb atoms centered on each of 6 faces (12 × 1/2 = 6 Nb atoms). One Sn atom completely inside the unit cell and one at each corner (1 + 8 × 1/8 = 2 Sn atoms). The atom ratio is 6 Nb : 2 Sn or 3 Nb : 1 Sn; the atom ratio in the structure matches the empirical formula.

    SmCo 5 : One Sm atom at each corner (8 × 1/8 = 1 Sm atom). One Co atom totally inside the unit cell (1 Co atom), two Co atoms centered on the top and bottom faces (4 × 1/2 = 2 Co atoms), four Co atoms centered on 4 side faces (4 × 1/2 = 2 Co atoms), for a total of (1 + 2 + 2 = 5 Co atoms). The ratio is 1 Sm : 5 Co, which matches the empirical formula.

    12.110

    1. CsCl, primitive cubic lattice (Figure 12.25)
    2. Au, face-centered cubic lattice (Figure 12.18)
    3. NaCl, face-centered cubic lattice (Figure 12.25)
    4. Po, primitive cubic lattice, rare for metals (Section 12.3)
    5. ZnS, face-centered cubic lattice (Figure 12.25)

    12.111 Calculate the wavelength of light that corresponds to an energy of 2.20 eV.

    λ = hc E = 6.626 × 10 - 34 J-s × 2 .998 × 10 8 m s × 1 2.20 eV × 1 eV 1 .602 × 10 - 19 J = 5.636 × 10 - 7 = 5.64 × 10 - 7 m ( 564 nm)

    [The absorbed 564-nm light is yellow-green and cinnabar appears the complementary color, red-violet or vermillion.]

    12.112 Aluminum, density = 2.70 g/mL, is indeed more dense than silicon, density = 2.33 g/mL. This difference is significant, but not as large as one might expect relative to the difference in electrical conductivity. It is true that under very high pressure, about 12 GPa or 120,000 atm, the structure of silicon changes from the diamond structure to a close-packed structure. The conductivity also changes from that of a semiconductor to that of a metal. The structure change can be detected by monitoring the conductivity (or resistivity) of the material.

    In terms of atomic properties, aluminum adopts a metallic (close-packed) structure because it is electron deficient. With only three valence electrons, it is difficult for an aluminum atom to acquire a complete octet by covalent bonding. A close-packed structure (face-centered cubic) provides each aluminum atom with twelve nearest neighbors and the possibility of electron delocalization to satisfy its bonding needs. On the other hand, each silicon atom has four valence electrons and can complete its octet by forming four bonds in a covalent-network structure with localized bonding.

    12.113

    1. Zinc sulfide, ZnS (Figure 12.26).
    2. Covalent. Silicon and carbon are both nonmetals, and their electronegativities are similar.

    1. Silicon carbide is hard and high-melting because it is a covalent-network solid. In SiC, the C atoms form a face-centered cubic array with Si atoms occupying alternate tetrahedral holes in the lattice. This means that the coordination numbers of both Si and C are 4; each Si is bound to four C atoms in a tetrahedral arrangement, and each C is bound to four Si atoms in a tetrahedral arrangement, producing an extended three-dimensional network. SiC is high-melting because a great deal of chemical energy is stored in the covalent SI C bonds. Melting requires breaking covalent Si–C bonds, which takes a huge amount of thermal energy. It is hard because the three-dimensional lattice resists any change that would weaken the Si–C bonding network.

    12.114

    1. V = (edge length) 3 = (0.5 mm) 3 = 0.125 = 0.1 mm 3

      d = m V ; m = d × V = 8.96 g cm 3 × 0.125 mm 3 × 1 cm 3 ( 10 ) 3 mm 3 = 1.12 × 10 - 3 = 1 × 10 - 3 g Cu

      1.12 × 10 - 3 g Cu × 1 mol Cu 63.546 g Cu × 6.022 × 10 23 Cu atoms mol Cu = 1.06 × 10 19 = 1 × 10 19 Cu atoms

    2. 1 × 10 - 19 J range of energy levels 1.06 × 10 19 Cu atoms = 9 × 10 - 39 J / Cu atoms

      This spacing is substantially smaller than the 1 × 10− 18 J separation between energy levels in the hydrogen atom.

    12.115 Semiconductors have a filled valence band and an empty conduction bond, separated by a characteristic difference in energy, the band gap, E g . When a semiconductor is heated, more electrons have sufficient energy to jump the band gap, and conductivity increases. Metals have a partially filled continuous energy band. Heating a metal increases the average kinetic energy of the metal atoms, usually through increased vibrations within the lattice. The greater vibrational energy of the atoms leads to imperfections in the lattice and discontinuities in the energy band. Thermal vibrations create barriers to electron delocalization and reduce the conductivity of the metal.

    12.116

    1. There are 8 Na atoms and 4 O atoms in the unit cell. The red O atoms adopt a face-centered cubic arrangement. There are O atoms at each corner of the cube, and in the center of each face.

      8 corners × 1/8 atom/corner + 6 faces × ½ atom/face = 4 O atoms

      The 8 green Na atoms are completely in the interior of the unit cell.

      Check .  These numbers of atoms agree with the empirical formula of Na 2 O.

    2. Analyze/Plan . Visualize the coordination number and environment of the sodium ion.

      Solve . In this structure, Na atoms sit completely inside the unit cell (but not at the body center). Like Zn atoms in the ZnS structure (Figure 12.26), they have 4 nearest-neighbor anions in a tetrahedral coordination environment.

    image

      [In Na 2 O, the 4 O atoms (dark gray) around a particular Na atom (light gray) are the O atom on the nearest corner and 3 oxygen atoms on adjacent face centers that form a triangular face of the coordination tetrahedron. For the upper, left, front Na atom, this is the O atom on the top, left, front corner and the 3 oxygen atoms on the top, left, and front faces.]

    1. There are 8 Na atoms and 4 O atoms or 4 Na 2 O formula units in the cubic unit cell. Density is unit cell contents divided by unit cell volume.

      density = 4 Na 2 O units ( 5.550 Å ) 3 × 61.979 g Na 2 O 6.0221 × 10 23 Na 2 O units × ( Å 1 × 10 - 8 cm ) 3 = 2.40 8 g/cm 3

    12.117

    1. image
    2. Teflon TM is formed by addition polymerization.

    12.118

    image

    12.119 X-ray diffraction is the phenomenon that enables us to measure interatomic distances in crystals. Diffraction is most efficient when the wavelength of light is similar to the size of the object (e.g., the slit) doing the diffracting. Interatomic distances are on the order of 1-10 Å, and the wavelengths of X-rays are also in this range. Visible light has wavelengths of 400-700 nm, or 4000-7000 Å, too long to be diffracted efficiently by atoms (electrons) in crystals.


    12.120 n λ = 2 d sin θ; n = 1, λ = 1.54 Å, θ = 14.22 °; calculate d.

    d = n λ 2 sin θ = 1 × 1.54 Å 2 sin(14 .22) = 3.1346 = 3.13 Å

    12.121 Germanium is in the same family but below Si on the periodic chart. This means that Ge will probably have bonding characteristics and crystal structure similar to those of Si. Because Ge has a larger bonding atomic radius than Si, we expect a larger unit cell and d -spacing for Ge. In Bragg’s law, n λ = 2 d sin θ, d and sin θ are inversely proportional. That is, the larger the d -spacing, the smaller the value of sinθ and θ. In a diffraction experiment, we expect a Ge crystal to diffract X-rays at a smaller θ-angle than a Si crystal, assuming the X-rays have the same wavelength.

    12.122

    1. Both diamond (d = 3.5 g/cm 3 ) and graphite (d = 2.3 g/cm 3 ) are covalent-network solids with efficient packing arrangements in the solid state; there is relatively little empty space in their respective crystal lattices. Diamond, with bonded C–C distances of 1.54 Å in all directions, is more dense than graphite, with shorter C–C distances within carbon sheets but longer 3.35 Å separations between sheets (Figure 12.29). Buckminsterfullerene has much more empty space, both inside each C 60 “ball” and between balls, than either diamond or graphite, so its density will be considerably less than 2.3 g/cm 3 .
    2. In a face-centered cubic unit cell, there are 4 complete C 60 units.

      4 C 60 units ( 14.2 Å ) 3 × 720.66 g 6 .022 × 10 23 C 60 units × ( 1 Å 1 × 10 - 8 cm ) 3 = 1.67 g/cm 3

      (1.67 g/cm 3 is the smallest density of the three allotropes: diamond, graphite, and buckminsterfullerene.)

    12.123

    (b) Increase. Promotion of an electron from the valence band to the conduction band generates a mobile electron in the conduction band and a mobile hole in the valence band. Both phenomena promote delocalization and increase the conductivity of the semiconductor.

    Integrative Exercises

    12.124 The karat scale is based on mass%, not mol%. In each case, change mol% to mass % Au. Then, karat = mass fraction Au × 24. Determine color using mass% and Figure 12.18.

    1. Assume 0.50 mol Ag and 0.50 mol Au.

      0 .50 mol Ag × 107 .87 g Ag mol Ag = 53.935 = 54 g Ag

      0 .50 mol Au × 196 .97 g Au mol Au = 98.485 = 98 g Au

      mass % Au = 98 .485 g Au (98 .485 + 53 .935) g total × 100 = 64.61 = 65 % Au

      karat = 0.6461 × 24 = 15.5064 = 16 karat Au

      On Figure 12.18, the Au/Ag alloy line is labeled in terms of mass% Ag. For an alloy that is 65% Au and 35% Ag, the color is greenish yellow.


    1. Assume 0.50 mol Cu and 0.50 mol Au.

      0 .50 mol Cu × 63 .546 g Cu mol Cu = 31.773 = 32 g Cu

      0 .50 mol Au = 98.485 = 98 g Au

      mass% Au = 98 .485 g Au (98 .485 + 31 .773) g total × 100 = 75.61 = 76 % Au

      karat = 0.7561 × 24 = 18.15 = 18 karat Au

      On Figure 12.18, for an alloy that is 76% Au and 24% Cu, the color is reddish gold.

    12.125 Analyze .  Given: mass % of Al, Mg, O; density, unit cell edge length. Find: number of each type of atom. Plan . We are not given the type of cubic unit cell, primitive, body centered, face centered. So we must calculate the number of formula units in the unit cell, using density, cell volume, and formula weight. Begin by determining the empirical formula and formula weight from mass % data. Solve . Assume 100 g spinel.

    37.9 g Al × 1 mol Al 26 .98 g Al = 1.405 mol Al; 1.405/0.7036 2 17 .1 g Mg × 1 mol Mg 24.305 g Mg = 0.7036 mol Mg; 0.7036/0.7036 = 1 45 .0 g O × 1 mol O 16 .00 g Al = 2.813 mol O; 2 .813/0 .7036 4

    The empirical formula is Al 2 MgO 4 ; formula weight = 142.3 g/mol

    Calculate the number of formula units per unit cell.

    8.09 Å = 8.09 × 10 −10 m = 8.09 × 10 −8 cm; V = (8.09 × 10 −8 ) 3 cm 3

    3.57 g cm 3 × ( 8.09 × 10 - 8 ) 3 cm 3 × 1 mol 142 .3 g × 6.022 × 10 23 units mol = 7.999 = 8

    There are 8 formula units per unit cell, for a total of 16 Al atoms, 8 Mg atoms, and 32 O atoms.

    [The relationship between density (d), unit cell volume (V), number of formula units (Z), formula weight (FW), and Avogadro’s number (N) is a useful one. It can be rearranged to calculate any single variable, knowing values for the others. For densities in g/cm 3 and unit cell volumes in cm 3 the relationship is Z = (N × d × V)/FW.]

    12.126 Refer to Section 12.7 and Figure 12.29.

    1. In diamond, each C atom is bound to 4 other C atoms. According to VSEPR, the geometry around a central atom with 4 bonding electron pairs is tetrahedral and the C–C–C bond angles are 109 o .
    2. Within a graphite sheet, each C atom is bound to 3 other C atoms in a trigonal planar arrangement. The C–C–C bond angles are 120 o .)
    3. Within a sheet, sp 2 hybrid orbitals are involved in the sigma bonding network framework. This leaves atomic p orbitals to be involved in interactions between sheets. These interactions are responsible for stacking.

    12.127

    1. image

      ΔH = D(C=C) – 2D(C–C) = 614 – 2(348) = –82 kJ/mol C 2 H 4

    2. image

      ΔH = 2D(C–O) + 2D(N–H) – 2D(C–N) – 2D(H–O)

      ΔH = 2(358) + 2(391) – 2(293) – 2(463) = –14 kJ/mol

      (This is –14 kJ/mol of either reactant.)

    3. image

      ΔH = 2D(C–O) + 2D(O–H) – 2D(C–O) – 2D(O–H) = 0 kJ

    12.128

    1. sp 3 hybrid orbitals at C, 109° bond angles around C
    2. image

      Isotactic polypropylene has the highest degree of crystallinity and highest melting point. The regular shape of the polymer backbone allows for close, orderly (almost zipper-like) contact between chains. This maximizes dispersion forces between chains and produces higher order (crystallinity) and melting point. Atactic polypropylene has the least order and the lowest melting point.

    1. image

      Both participate in hydrogen-bonding interactions with H 2 O molecules. These are strong intermolecular forces that hold the “moisture” at the surface of the fabric next to the skin. Polypropylene has no strong interactions with water, and capillary action “wicks” the moisture away from the skin.

    12.129

    1. image
    2. C–Cl bonds are weakest, so they are most likely to break upon heating.
    3. The repeating unit in polyvinyl chloride consists of two C atoms, each in a different environment. Consider the net changes in these two C atoms when the polymer is converted to diamond a high pressure.

      Diamond is a covalent-network structure where each C atom is tetrahedrally bound to four other C atoms [Figure 12.29(a)].

      image

      Assume that there is no net change to the C−C bonds in the structure, even though they may be broken and reformed. The net change to the 2-C vinyl chloride unit is then breaking three C−H bonds and one C−Cl bond, and making four C−C bonds.

      Δ H = 3D(C - H) - D(C - Cl) - 4D(C - C) = 3(413) + 328 - 4(348) = 523 kJ/vinyl chloride unit

    12.130

    1. Follow the logic outlined in Solution 12.99.

      vol. of unit cell = (5.43 Å) 3 = 160.1030 = 1.60 × 10 2 Å 3

      1 cm 3 × ( 1 ) 3 Å 3 ( 1 × 10 - 8 ) 3 cm 3 = 1 × 10 24 Å 3 (volume of material)

      4 Si atoms 160.103 Å 3 = x Si atoms 1 × 10 24 Å 3 ; x = 2.4984 × 10 22 = 2.50 × 10 22 Si atoms

      (To 1 sig fig, the result is 2 × 10 22 Si atoms.)


    1. 1 ppm phosphorus = 1 P atom per 1 × 10 6 Si atoms

      1 P atom 1 × 10 6 Si atoms = x P atoms 2.4984 × 10 22 Si Atoms ; x = 2.4984 × 10 22 = 2.50 × 10 16 P atoms

      2.4984 × 10 16 P atom × 1 mol 6 .022 × 10 23 atoms × 30.97376 g P mol × 1 mg 1 × 10 - 3 g = 1.29 × 10 - 3 mg P (1 .29 μ g)

    12.131 Analyze/Plan . Write a balanced chemical equation for the reaction. Calculate moles FeTiO 3 to be produced, then moles and grams of the two reactants required to produce this amount of product. Solve .

    1. FeO(s) + TiO 2 (s) → FeTiO 3 (s)
    2. The formula weight of FeTiO 3 is 151.71 g/mol.
    3. 2 .500 g FeTiO 3 × 1 mol 151 .710 g FeTiO 3 = 0 .0164788 = 0 .01648 mol FeTiO 3
    4. The coefficients in the chemical equation are all 1, so we need 0.01648 moles of each of the reactants to produce 0.01648 mol FeTiO 3 (s) (assuming the reaction goes to completion). The molar mass of FeO is 71.844 g/mol. The molar mass of TiO 2 is 79.8558 g/mol.

      0 .0164788 mol FeO × 71 .844 g FeO mol FeO = 1 .18391 = 1 .184 g FeO

    5. 0 .0164788 mol TiO 2 × 79 .8558 g TiO 2 mol = 1 .31593 = 1 .316 g TiO 2

    12.132 The bonding atomic radius of a Si atom is 1.11 Å (Figure 7.7), so the diameter is 2.22 Å.

    14 nm × 1 × 10 - 9 m 1 nm × 1 Å 1 × 10 - 10 m × 1 Si atom 2.22 Å = 63 Si atoms .


     

    13 Properties of Solutions

    Visualizing Concepts

    13.1 (a) < (b) < (c). In Section 13.1, entropy is qualitatively defined as randomness or dispersal in space. In container (a), the two kinds of particles are not mixed and the particles are close together, so (a) has the least entropy. In container (b), the particles occupy approximately the same volume as container (a) but the two kinds of particles are homogeneously mixed, so the degree of dispersal and randomness is greater than in (a). In container (c), the two kinds of particles are homogeneously mixed and they occupy a larger volume than in (b), so (c) has the greatest entropy.

    13.2

    1. The oxygen atom of the water molecule is associated with the cation. In the polar water molecule, the partial negative charge is localized on the oxygen atom. The electrostatic attractive interaction is between the positive charge of the cation and the partially negative charge of the oxygen atom.
    2. Statement (c). The smaller ionic radius of lithium ion means that the positive charge is localized over a smaller volume; it is essentially a point charge. This increases the strength of the interaction between lithium cation and each individual water molecule.

    13.3

    1. No.
    2. The ionic solid with the smaller lattice energy will be more soluble in water. Lattice energy is the main component of ΔH solute , the enthalpy required to separate solute particles. The solid with the smaller lattice energy will have the less endothermic ΔH solute . Assuming ΔH solvent and ΔH mix are the same for the two solids, the one with the smaller lattice energy will have a less endothermic ΔH soln and be more soluble in water.

    13.4   Statements (a) and (d) are true. Statement (b) is false because of the average distance between particles. Statement (c) is false because gases, like all pure substances, have characteristic boiling points.

    13.5   Diagram (b) is the best representation of a saturated solution. There is some undissolved solid with particles that are close together and ordered, in contact with a solution containing mobile, separated solute particles. As much solute has dissolved as can dissolve, leaving some undissolved solid in contact with the saturated solution.

    13.6   Statement (b) is the best explanation. Statement (a) is false because gas molecules are in constant random motion. Statement (c) is false because only N, O, and F atoms can formally act as hydrogen bond acceptors. Statement (d) is false because lighter gases are more likely to form saturated solutions with water. The term saturated means that all solute that can dissolve has dissolved. Less soluble solutes, in this case lighter gases, are more likely to form saturated solutions.


    13.7   Vitamin B 6 is more water soluble and vitamin E is more fat soluble. The B 6 molecule contains three –OH groups and the image that can enter into many hydrogen-bonding interactions with water. Its relatively small molecular size indicates that dispersion forces will not play a large role in intermolecular interactions and that hydrogen bonding will dominate. On the other hand, the long, rod-like hydrocarbon chain of vitamin E will lead to stronger dispersion forces among it and the mostly nonpolar fats. Although vitamin E has one –OH and one image group, the long hydrocarbon chain prevents water from surrounding and separating the vitamin E molecules, reducing its water solubility.

    13.8   The second friend is correct. As pressure decreases, gas solubility decreases. The first bubbles were from air gases dissolved at atmospheric pressure but not dissolved at lower pressure. The second batch of bubbles is from the water boiling. As the pressure in the vessel decreases, eventually the vapor pressure of water equals external pressure, and the water boils.

    13.9

    1. Yes, the molarity changes with a change in temperature. Molarity is defined as moles solute per unit volume of solution. If solution volume is different, molarity is different.
    2. No, molality does not change with change in temperature. Molality is defined as moles solute per kilogram of solvent. Even though the volume of solution has changed because of increased kinetic energy, the mass of solute and solvent have not changed, and the molality stays the same.

    13.10

    1. The blue line represents the solution. According to Raoult’s law, the presence of a nonvolatile solute lowers the vapor pressure of a volatile solvent. At any given temperature, the blue line has a lower vapor pressure and represents the solution.
    2. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to atmospheric pressure. Assuming atmospheric pressure of 1.0 atm, the boiling point of the solvent (red line) is approximately 64 C. The boiling point of the solution is approximately 70 C.

    13.11  Ideally, 0.50 L. If the volume outside the balloon is very large compared to 0.25 L, solvent will flow across the semipermeable membrane until the molarities of the inner and outer solutions are equal, 0.1 M . This requires an “inner” solution volume twice as large as the initial volume, or 0.50 L. (In reality, osmosis across the balloon membrane is not perfect. The solution concentration inside the balloon will be slightly greater than 0.1 M and the volume of the balloon will be slightly less than 0.50 L.)

    13.12  Beaker (b) represents a liquid-liquid emulsion such as milk. Beaker (a) represents a true solution. Beaker (c) has some alignment of molecules and represents a liquid crystal.

    The Solution Process (Section 13.1)

    13.13

    1. False. It is generally true that solute-solvent interactions must balance the sum of solute-solute and solvent-solvent interactions. If solute-solute interactions are significantly stronger than solute-solvent interactions, dissolving will probably not occur. Entropy of mixing can compensate for a slightly endothermic enthalpy of solution, but not a large one.
    2. False. If a solution forms, enthalpy of mixing is exothermic, a negative number.
    3. True.

    13.14

    1. False. NaCl is more soluble in water than it is in benzene because the enthalpy of mixing for NaCl and water is much more exothermic. Ion-dipole interactions between NaCl and water are much stronger than the interactions between ionic NaCl and nonpolar benzene. (And, water is more dense than benzene.)
    2. True
    3. True

    13.15 Analyze/Plan. Decide whether the solute and solvent in question are ionic, polar covalent, or nonpolar covalent. Draw Lewis structures as needed. Then state the appropriate type of solute-solvent interaction. Solve .

    1. CCl 4 , nonpolar; benzene, nonpolar; dispersion forces
    2. methanol, polar with hydrogen bonding; water, polar with hydrogen bonding; hydrogen bonding
    3. KBr, ionic; water, polar; ion-dipole forces
    4. HCl, polar; CH 3 CN, polar; dipole-dipole forces

    13.16  From weakest to strongest solvent-solute interactions:

    (b), dispersion forces < (c), hydrogen bonding < (a), ion-dipole

    13.17

    1. Very soluble.
    2. ΔH mix will be the largest negative number. ΔH solute and ΔH solvent are both positive (endothermic); they represent the energy required to overcome attractive interactions in the solute and solvent, respectively. ΔH mix represents the attractive interactions between solute and solvent particles. In order for ΔH soln to be negative (exothermic), ΔH mix must have a greater magnitude than (ΔH solute + ΔH solvent ).

    13.18

    1. This solution process is endothermic. The enthalpy of the solution is greater than the enthalpy of the unmixed solute plus solvent.
    2. The solution forms because the favorable entropy of mixing outweighs the increase in enthalpy by the solution.

    13.19

    1. Lattice energy is the amount of energy required to completely separate a mole of solid ionic compound into its gaseous ions (Section 8.2 ). For ionic solutes, this corresponds to ΔH solute (solute-solute interactions) in Equation 13.1.
    2. In Equation 13.1, ΔH mix is always exothermic. Formation of attractive interactions, no matter how weak, always lowers the energy of the system, relative to the energy of the isolated particles.

    13.20  ΔH mix is much more negative (exothermic) than ΔH solvent or ΔH solute . Both ΔH solvent and ΔH solute will be endothermic, because separating solvent molecules or solute ions requires energy. Because ΔH soln is exothermic, ΔH mix must be exothermic, and not just more negative than the other two, but more negative than the sum of the other two. This is not surprising, because ΔH mix involves formation of many ion-dipole interactions, strong interparticle forces.


    13.21

    1. ΔH soln is nearly zero. ΔH soln is determined by the relative magnitudes of the “old” solute-solute (ΔH solute ) and solvent-solvent (ΔH solvent ) interactions and the new solute-solvent interactions (ΔH mix ); ΔH soln = ΔH solute + ΔH solvent + ΔH mix . Because the solute and solvent in this case experience very similar London dispersion forces, the energy required to separate them individually and the energy released when they are mixed are approximately equal. ΔH solute + ΔH solvent ≈ –ΔH mix and ΔH soln is nearly zero.
    2. The entropy of the system increases when heptane and hexane form a solution. The change involves taking two pure liquids and forming a (homogeneous) mixture. The result is an increase in randomness or disorder. From part (a), the enthalpy of mixing is nearly zero, so the increase in entropy is the driving force for mixing in all proportions.

    13.22  Statement (c) is the best explanation. Statement (a) is true, but it doesn't explain the enthalpy of solution. Statement (b) is false. With a positive enthalpy of solution, entropy of mixing must be favorable. Statement (d) involves molar mass; molar mass influences dispersion forces that are not important in the solubility of ionic solids.

    Saturated Solutions; Factors Affecting Solubility (Sections 13.2 and 13.3)

    13.23

    1. Supersaturated, because the solution contains more solute than a saturated solution at this temperature.
    2. The bits of glass scraped from the vessel act as a seed crystal, a place where solute molecules can align to form a crystal. The excess chromium nitrate is crystallizing out.
    3. 324 g dissolved at 35 °C – 208 g soluble at 15 °C = 116 g crystals form

    13.24

    1. 1.22 mol MnSO 4 · H 2 O 1 L soln × 169.0 g MnSO 4 · H 2 O 1 mol × 0.100 L = 20.6 g MnSO 4 · H 2 O/100 mL

      The 1.22 M solution is unsaturated.

    2. Add a known mass, say 5.0 g, of MnSO 4 · H 2 O, to the unknown solution. If the solid dissolves, the solution is unsaturated. If there is undissolved MnSO 4 · H 2 O, filter the solution and weigh the solid. If there is less than 5.0 g of solid, some of the added MnSO 4 · H 2 O, dissolved and the unknown solution is unsaturated. If there is exactly 5.0 g, no additional solid dissolved and the unknown is saturated. If there is more than 5.0 g, excess solute has precipitated and the solution is supersaturated.

    13.25 Analyze/Plan .  On Figure 13.15, find the solubility curve for the appropriate solute. Find the intersection of 40 °C and 40 g solute on the graph. If this point is below the solubility curve, more solute can dissolve and the solution is unsaturated. If the intersection is on or above the curve, the solution is saturated. Solve .

    1. unsaturated
    2. saturated
    3. saturate
    4. unsaturated

    13.26

    1. at 30 ° C, 10 g KClO 3 100 g H 2 O × 250 g H 2 O = 25 g KClO 3
    2. 66 g Pb(NO 3 ) 2 100 g H 2 O × 250 g H 2 O = 165 = 1 .7 × 10 2 g Pb(NO 3 ) 2
    3. 3 g Ce 2 ( SO 4 ) 3 100 g H 2 O × 250 g H 2 O = 7 .5 = 8 g Ce 2 ( SO 4 ) 3

    13.27

    1. We expect the liquids water and glycerol to be miscible in all proportions. Glycerol has an –OH group on each C atom in the molecule. This structure facilitates strong hydrogen bonding similar to that in water. When two liquids have very similar intermolecular interactions, entropy of mixing drives the solution process and the two liquids are usually miscible in all proportions (Solution 13.21). Like dissolves like.
    2. Hydrogen bonding, dipole-dipole forces, London dispersion forces

    13.28  The most likely reason is statement (b), although statement (c) also contributes.

    Many substances are called “oil,” but they typically contain molecules composed mostly of carbon and hydrogen with fairly high molecular weights. Both statements contribute to the fact that oil molecules are nonpolar and experience strong dispersion forces. The properties of water are dominated by its strong hydrogen bonding. Oil and water have very dissimilar intermolecular interactions and the liquids are not miscible.

    There are examples of high molecular weight compounds (e.g., sugars and proteins) with many hydrogen-bonding interactions that are soluble in water, so statement (b) is the better answer.

    13.29 Analyze/Plan . Evaluate molecules in the four common laboratory solvents for strength of intermolecular interactions with nonpolar solutes. Solve . Toluene, C 6 H 5 CH 3 , is the best solvent for nonpolar solutes. Without polar groups or nonbonding electron pairs, it forms only dispersion interactions with itself and other molecules. The enthalpy of solution, ΔH soln , is essentially zero (as in Solution 13.21) and solution occurs because of the favorable entropy of mixing.

    13.30  We expect alanine to be more soluble in water than hexane. Alanine has a –COOH and a –NH 2 group available to form hydrogen bonds with water molecules. Although there are some potential dispersion forces between the terminal –CH 3 group of alanine and hexane molecules, we expect the hydrogen bonding between alanine and water to be stronger. Stronger intermolecular attractive forces between alanine and water lead to a more negative ΔH mix and more negative (smaller positive) ΔH soln for water than for hexane.

    13.31

    1. Despite the presence of the –COOH group, stearic acid is more soluble in nonpolar CCl 4 than in polar (hydrogen bonding) water. Dispersion interactions among nonpolar CH 3 (CH 2 ) 16– chains dominate the properties of stearic acid.
    2. Dioxane will be more soluble in water than cyclohexane will, because dioxane can act as a hydrogen bond acceptor.

    13.32  The red part of the molecule, a carboxyl group able to form hydrogen bonds, contributes to its water solubility. The gray and white parts, a phenyl ring and several –CH 3 groups, form a large nonpolar area that contributes to its water insolubility.


    13.33 Analyze/Plan .  Hexane is a nonpolar hydrocarbon that experiences dispersion forces with other nonpolar molecules. Solutes that primarily experience dispersion forces will be more soluble in hexane. Solve .

    1. CCl 4 is more soluble because dispersion forces among nonpolar CCl 4 molecules are similar to dispersion forces in hexane. Ionic bonds in CaCl 2 are unlikely to be broken by weak solute-solvent interactions. For CaCl 2 , ΔH solute is large, relative to ΔH mix .
    2. Benzene, C 6 H 6 , is also a nonpolar hydrocarbon and will be more soluble in hexane. Glycerol experiences hydrogen bonding with itself; these solute-solute interactions are less likely to be overcome by weak solute-solvent interactions.
    3. Octanoic acid, CH 3 (CH 2 ) 6 COOH, will be more soluble than acetic acid CH 3 COOH. Both solutes experience hydrogen bonding by –COOH groups, but octanoic acid has a long, rod-like hydrocarbon chain with dispersion forces similar to those in hexane, facilitating solubility in hexane.

    13.34 Analyze/Plan .  Water, H 2 O, is a polar solvent that forms hydrogen bonds with other H 2 O molecules. The more soluble solute in each case will have intermolecular interactions that are most similar to the hydrogen bonding in H 2 O. Solve .

    1. Glucose, C 6 H 12 O 6 , is more soluble because it is capable of hydrogen bonding (Figure 13.10). Nonpolar C 6 H 12 is capable only of dispersion interactions and does not have strong intermolecular interactions with polar (hydrogen bonding) H 2 O.
    2. Ionic sodium propionate, CH 3 CH 2 COONa, is more soluble. Sodium propionate is a crystalline solid, whereas propionic acid is a liquid. The increase in disorder or entropy when an ionic solid dissolves leads to significant water solubility, despite the strong ion-ion forces (large ΔH solute ) present in the solute (see Solution 13.22).
    3. HCl is more soluble because it is a strong electrolyte and completely ionized in water. Ionization leads to ion-dipole solute-solvent interactions, and an increase in disorder. CH 3 CH 2 Cl is a molecular solute capable of relatively weak dipole-dipole solute-solvent interactions and is much less soluble in water.

    13.35

    1. False. The lower the temperature, the more soluble most gases are in water.
    2. True.
    3. False. In a supersaturated solution all solute remains dissolved.
    4. True. Solubility of liquids and solids usually increases with increasing temperature.

    13.36

    1. True.
    2. False. The solubility of most ionic solids increases as the temperature of the solution increases.
    3. False. The solubility of gases in water decreases as temperature increases because the kinetic energy of the gas particles increases and more solute particles have sufficient energy to escape the solution. Also, interactions between gaseous solutes and water typically do not involve hydrogen bonding.
    4. True. See Figure 13.15.

    13.37 Analyze/Plan .  Follow the logic in Sample Exercise 13.2. Solve .

    S He = 3.7 × 10– 4 M /atm × 1.5 atm = 5.6 × 10– 4 M

    S N 2 = 6.0 × 10 4 M / atm × 1 .5 atm = 9 .0 × 10 4 M

    13.38 6 50 torr × 1 atm 760 torr = 0.855 atm; P O 2 = X O 2 ( P t ) = 0.21 ( 0.855 atm ) = 0.1796 = 0.18 atm

    S O 2 = kP O 2 = 1.38 × 10 3 mol L-atm × 0.1796 atm = 2 .5 × 10 4 M

    Concentrations of Solutions (Section 13.4)

    13.39 Analyze/Plan .  Follow the logic in Sample Exercise 13.3. Solve .

    1. mass % = mass solute total mass solution × 100 = 10.6 g Na 2 SO 4 10.6 g Na 2 SO 4 + 483 g H 2 O × 100 = 2.15 %
    2. ppm = mass solute total mass solution × 10 6 ; 2 .86 g Ag 1 ton ore × 1 ton 2000 lb × 1 lb 453.6 g × 10 6 = 3.15 ppm

    13.40

    1. mass % = mass solute total mass solution × 100

      mass solute = 0 .035 mol I 2 × 253.8 g I 2 1 mol I 2 = 8.883 = 8.9 g I 2

      mass % I 2 = 8.883 g I 2 8.883 g I 2 + 125 g CCl 4 × 100 = 6.635 = 6.6 % I 2

    2. ppm = mass solute total mass solution × 10 6 = 0.0079 g Sr 2 + 1 × 10 3 g H 2 O × 10 6 = 7.9 ppm Sr 2 +

    13.41 Analyze/Plan .  Given masses of CH 3 OH and H 2 O, calculate moles of each component.

    1. Mole fraction CH 3 OH = (mol CH 3 OH)/(total mol)
    2. mass % CH 3 OH = [(g CH 3 OH)/(total mass)] × 100
    3. molality CH 3 OH = (mol CH 3 OH)/(kg H 2 O). Solve .
    1. 14.6 g CH 3 OH × 1 mol CH 3 OH 32.04 g CH 3 OH = 0.4557 = 0.456 mol CH 3 OH

      184 g H 2 O × 1 mol H 2 O 18.02 g H 2 O = 10.211 = 10.2 mol H 2 O

      X CH 3 OH = 0.4557 0.4557 + 10.211 = 0.04272 = 0.0427

    2. mass % CH 3 OH = 14 .6 g CH 3 OH 14 .6 g CH 3 OH + 184 g H 2 O × 100 = 7.35 % CH 3 OH
    3. m = 0 .4557 mol CH 3 OH 0.184 kg H 2 O = 2.477 = 2.48 m CH 3 OH

    13.42

    1. 2 0 .8 g C 6 H 5 OH 94.11 g/mol = 0.2210 = 0.221 mol C 6 H 5 OH

      425 g CH 3 CH 2 OH 46.07 g/mol = 9.2251 = 9.23 mol CH 3 CH 2 OH

      X C 6 H 5 OH = 0.2210 0.2210 + 9.2251 = 0.02340 = 0.0234

    2. mass % = 20 .8 g C 6 H 5 OH 20 .8 g C 6 H 5 OH + 425 g CH 3 CH 2 OH × 100 = 4.67 % C 6 H 5 OH
    3. m = 0 .2210 mol C 6 H 5 OH 0.425 kg CH 3 CH 2 OH = 0.5200 = 0.520 m C 6 H 5 OH

    13.43 Analyze/Plan .  Given mass solute and volume solution, calculate mol solute, then molarity = mol solute/L solution. Or, for dilution, M c × L c = M d × L d . Solve .

    1. M = mol solute L soln ; 0.540 g Mg(NO 3 ) 2 0.2500 L soln × 1 mol Mg(NO 3 ) 2 148.3 g Mg(NO 3 ) 2 = 1.46 × 10 2 M Mg(NO 3 ) 2
    2. 22.4 g LiClO 4 · 3 H 2 O 0.125 L soln × 1 mol LiClO 4 · 3 H 2 O 160.4 g LiClO 4 · 3 H 2 O = 1.12 M LiClO 4 · 3 H 2 O
    3. M c × L c = M d × L d ; 3.50 M HNO 3 × 0.0250 L = × M HNO 3 × 0.250 L 250 mL of 0.350 M HNO 3

    13.44

    1. M = mol solute L soln ; 15 .0 g Al 2 ( SO 4 ) 3 0.250 L soln × 1 mol Al 2 (SO 4 ) 3 342.2 g Al 2 ( SO 4 ) 3 = 0.175 M Al 2 ( SO 4 ) 3
    2. 5.25 g Mn(NO 3 ) 2 · 2 H 2 O 0.175 L soln × 1 mol Mn(NO 3 ) 2 · 2 H 2 O 215.0 g Mn(NO 3 ) 2 · 2 H 2 O = 0.140 M Mn(NO 3 ) 2
    3. M c × L c = M d × L d ; 9.00 M H 2 SO 4 × 0.0350 L = × M H 2 SO 4 × 0.500 L 500 mL of 0.630 M H 2 SO 4

    13.45 Analyze/Plan .  Follow the logic in Sample Exercise 13.4. Solve .

    1. m = mol solute kg solvent ; 8 .66 g C 6 H 6 23.6 g CCl 4 × 1 mol C 6 H 6 78.11 g C 6 H 6 × 1000 g CCl 4 1 kg CCl 4 = 4.70 m C 6 H 6
    2. The density of H 2 O = 0.997 g/mL = 0.997 kg/L.

      4.80 g NaCl 0.350 L H 2 O × 1 mol NaCl 58.44 g NaCl × 1 L H 2 O 0.997 kg H 2 O = 0.235 m NaCl

    13.46

    1. 16.0 mol H 2 O × 18 .02 g H 2 O 1 mol H 2 O = 288.3 g H 2 O = 0 .288 kg H 2 O

      m = 1.12 mol KCl 0.2883 kg H 2 O = 3.8846 = 3.88 m KCl

    2. m = mol solute kg solute ; mol S 8 = m × kg C 10 H 8 = 0.12 m × 0.1000 kg C 10 H 8 = 0.012 mol

      0.012 mol S 8 × 256.5 g S 8 1 mol S 8 = 3.078 = 3.1 g S 8


    13.47 Analyze/Plan .  Assume 1 L of solution. Density gives the total mass of 1 L of solution. The g H 2 SO 4 /L are also given in the problem. Mass % = (mass solute/total mass solution) × 100. Calculate mass solvent from mass solution and mass solute. Calculate moles solute and solvent and use the appropriate definitions to calculate mole fraction, molality, and molarity. Solve .

    1. 571.6 g H 2 SO 4 1 L soln × 1 L soln 1329 g soln = 0.430098 g H 2 SO 4 / g soln

      mass % is thus 0.4301 × 100 = 43.01% H 2 SO 4

    2. In a liter of solution there are 1329 – 571.6 = 757.4 = 757 g H 2 O.

      571.6 g H 2 SO 4 98.09 g/mol = 5.827 mol H 2 SO 4 ; 757 .4 g H 2 O 18.02 g/mol = 42.03 = 42.0 mol H 2 O

      X Η 2 SO 4 = 5.827 42.03 + 5.827 = 0.122

      (The result has 3 sig figs because (g H 2 O) resulting from subtraction is limited to 3 sig figs.)

    3. molality = 5.827 mol H 2 SO 4 0.7574 kg H 2 O = 7.693 = 7.69 m H 2 SO 4
    4. molarity = 5.827 mol H 2 SO 4 1 L soln = 5.827 M H 2 SO 4

    13.48

    1. mass % = mass C 6 H 8 O 6 total mass solution × 100 ;

      80.5 g C 6 H 8 O 6 80.5 g C 6 H 8 O 6 + 210 g H 2 O × 100 = 27.71 = 27.7 % C 6 H 8 O 6

    2. mol C 6 H 8 O 6 = 80.5 g C 6 H 8 O 6 176.1 g/mol = 0.4571 = 0.457 mol C 6 H 8 O 6

      mol H 2 O = 210 g H 2 O 18.02 g/mol = 11.654 = 11.7 mol H 2 O

      X C 6 H 8 O 6 = 0.4571 mol C 6 H 8 O 6 0.4571 mol C 6 H 8 O 6 + 11.654 mol H 2 O = 0.0377

    3. m = 0.4571 mol C 6 H 8 O 6 0.210 kg H 2 O = 2.18 m C 6 H 8 O 6
    4. M = mol C 6 H 8 O 6 L solution ; 290 .5 g soln × 1 mL 1 .22 g × 1 L 1000 mL = 0.2381 = 0.238 L

      M = 0.4571 mol C 6 H 8 O 6 0.2381 L soln = 1.92 M C 6 H 8 O 6

    13.49 Analyze/Plan .  Given: 98.7 mL of CH 3 CN(l), 0.786 g/mL; 22.5 mL CH 3 OH, 0.791 g/mL. Use the density and volume of each component to calculate mass and then moles of each component. Use the definitions to calculate mole fraction, molality, and molarity. Solve .


    1. mol CH 3 CN = 0 .786 g 1 mL × 98.7 mL × 1 mol CH 3 CN 41 .05 g CH 3 CN = 1.8898 = 1.89 mol

      mol CH 3 OH = 0 .791 g 1 mL × 22.5 mL × 1 mol CH 3 OH 32 .04 g CH 3 OH = 0.5555 = 0.556 mol

      X CH 3 OH = 0.5555 mol CH 3 OH 1.8898 mol CH 3 CN + 0 .5555 mol CH 3 OH = 0.227

    2. Assuming CH 3 OH is the solute and CH 3 CN is the solvent,

      98.7 mL CH 3 CN × 0 .786 g 1 mL × 1 kg 1000 g = 0.07758 = 0.0776 kg CH 3 CN

      m CH 3 OH = 0.5555 mol CH 3 OH 0.07758 kg CH 3 CN = 7.1604 = 7.16 m CH 3 OH

    3. The total volume of the solution is 121.2 mL, assuming volumes are additive.

      M = 0.5555 mol CH 3 OH 0.1212 L solution = 4.58 M CH 3 OH

    13.50  Given: 8.10 g C 4 H 4 S, 1.065 g/mL; 250.0 mL C 7 H 8 , 0.867 g/mL

    1. mol C 4 H 4 S = 8.10 g C 4 H 4 S × 1 mol C 4 H 4 S 84.15 g C 4 H 4 S = 0.09626 = 0.0963 mol C 4 H 4 S

      mol C 7 H 8 = 0.867 g 1 mL × 250.0 mL × 1 mol C 7 H 8 92.14 g C 7 H 8 = 2.352 = 2.35 mol

      X C 4 H 4 S = 0.09626 mol C 4 H 4 S 0.09626 mol C 4 H 4 S + 2 .352 mol C 7 H 8 = 0.03932 = 0.0393

    2. m C 4 H 4 S = mol C 4 H 4 S kg C 7 H 8 ; 250 .0 mL × 0 .867 g 1 mL × 1 kg 1000 g = 0.2168 = 0.217 kg C 7 H 8

      m C 4 H 4 S = 0.09626 mol C 4 H 4 S 0.2168 kg C 7 H 8 = 0.444 m C 4 H 4 S

    3. 8. 10 g C 4 H 4 S × 1 mL 1 .065 g = 7.606 = 7.61 mL C 4 H 4 S;

      V soln = 7.61 mL C 4 H 4 S + 250.0 mL C 7 H 8 = 257.6 mL

      M C 4 H 4 S = 0.09626 mol C 4 H 4 S 0.2576 L soln = 0.374 M C 4 H 4 S

    13.51 Analyze/Plan .  Given concentration and volume of solution use definitions of the appropriate concentration units to calculate amount of solute; change amount to moles if needed. Solve .

    1. mol = M × L; 0 .250 mol SrBr 2 1 L soln × 0.600 L = 0 .150 mol SrBr 2
    2. Assume that for dilute aqueous solutions, the mass of the solvent is the mass of solution. Use proportions to get mol KCl.

      0.180 mol KCl 1 kg H 2 O = x mol KCl 0.0864 kg H 2 O ; x = 1 .56 × 10 2 mol KCl


    1. Use proportions to get mass of glucose, then change to mol glucose.

      6.45 g C 6 H 12 O 6 100 g soln = x g C 6 H 12 O 6 124 .0 g soln ; x = 8 .00 g C 6 H 12 O 6

      8.00 g C 6 H 12 O 6 × 1 mol C 6 H 12 O 6 180.2 g C 6 H 12 O 6 = 4.44 × 10 2 mol C 6 H 12 O 6

    13.52

    1. 1.50 mol HNO 3 1 L soln × 0.255 L = 0 .3825 = 0 .383 mol HNO 3
    2. Assume that for dilute aqueous solutions, the mass of the solvent is the mass of solution.

      1.50 mol NaCl 1 kg H 2 O = x mol 50 .0 × 10 6 kg ; x = 7 .50 × 10 5 mol NaCl

    3. 1.50 g C 12 H 22 O 11 100 g soln = x g C 12 H 22 O 11 75.0 g soln ; x = 1 .125 = 1 .13 g C 12 H 22 O 11

      1.125 g C 12 H 22 O 11 × 1 mol C 12 H 22 O 11 342.3 g C 12 H 22 O 11 = 3.287 × 10 3 = 3.29 × 10 3 mol C 12 H 22 O 11

    13.53 Analyze/Plan .  When preparing solution, we must know amount of solute and solvent. Use the appropriate concentration definition to calculate amount of solute. If this amount is in moles, use molar mass to get grams; use mass in grams directly. Amount of solvent can be expressed as total volume or mass of solution. Combine mass solute and solvent to produce the required amount (mass or volume) of solution. Solve .

    1. mol = M × L; 1 .50 × 10 2 mol KBr 1 L soln × 0.75 L × 119 .0 g KBr 1 mol KBr = 1.3 g KBr

      Weigh out 1.3 g KBr, dissolve in water, dilute with stirring to 0.75 L (750 mL).

    2. Mass of solution is required, but density is not specified. Use molality to calculate mass fraction, and then the masses of solute and solvent needed for 125 g of solution.

      0.180 mol KBr 1000 g H 2 O × 119.0 g KBr 1 mol KBr = 21.42 = 21.4 g KBr/kg H 2 O . Thus,

      mass fraction = 21 .42 g KBr 100 0 + 21 .42 = 0.02097 = 0.0210

      In 125 g of the 0.180 m solution, there are

      ( 125 g soln ) × 0.02097 g KBr 1 g soln = 2.621 = 2.62 g KBr

      Weigh out 2.62 g KBr, dissolve it in 125 – 2.62 = 122.38 = 122 g H 2 O to make exactly 125 g of 0.180 m solution.

    3. Using solution density, calculate the total mass of 1.85 L of solution, and from the mass % of KBr, the mass of KBr required.

      1.85 L soln × 1000 mL 1 L × 1.10 g soln 1 mL = 2035 = 2.04 × 10 3 g soln

      0.120 (2035 g soln) = 244.2 = 244 g KBr

      Dissolve 244 g KBr in water, dilute with stirring to 1.85 L.


    1. Calculate moles KBr needed to precipitate 16.0 g AgBr. AgNO 3 is present in excess.

      16.0 g AgBr × 1 mol AgBr 187.8 g AgBr × 1 mol KBr 1 mol AgBr = 0.08520 = 0.0852 mol KBr

      0.0852 mol KBr × 1 L soln 0 .150 mol KBr = 0.568 L soln

      Weigh out 0.0852 mol KBr (10.1 g KBr), dissolve it in a small amount of water, and dilute to 0.568 L.

    13.54

    1. 0.110 mol (NH 4 ) 2 SO 4 1 L soln × 1.50 L × 132 .2 g (NH 4 ) 2 SO 4 1 mol (NH 4 ) 2 SO 4 = 21.81 = 21.8 g (NH 4 ) 2 SO 4

      Weigh 21.8 g (NH 4 ) 2 SO 4 , dissolve in a small amount of water, continue adding water with thorough mixing up to a total solution volume of 1.50 L.

    2. Determine the mass fraction of Na 2 CO 3 in the solution:

      0 .65 mol Na 2 CO 3 1000 g H 2 O × 106.0 g Na 2 CO 3 1 mol Na 2 CO 3 = 68.9 g = 69 g Na 2 CO 3 1000 g H 2 O

      mass fraction = 68 .9 g Na 2 CO 3 1000 g H 2 O + 68 .9 g Na 2 CO 3 = 0.06446 = 0.064

      In 225 g of solution, there are 0.06446(225) = 14.503 = 15 g Na 2 CO 3 .

      Weigh out 15 g Na 2 CO 3 and dissolve it in 225 – 15 = 210 g H 2 O to make exactly 225 g of solution. (210 g H 2 O/0.997 g H 2 O/mL @ 25 °C = 211 mL H 2 O)

      [Carrying 3 sig figs, weigh 14.5 g Na 2 CO 3 and dissolve it in 225 – 14.5 = 210.5 g H 2 O. This produces a solution that is much closer to 0.65 m .]

    3. 1.20 L × 1000 mL 1 L × 1.16 g 1 mL = 1392 g solution; 0.150(1392 g soln) = 209 g Pb(NO 3 ) 2

      Weigh 209 g Pb(NO 3 ) and add (1392 – 209) = 1183 g H 2 O to make exactly (1392 = 1.39 × 10 3 ) g or 1.20 L of solution.

      (1183 g H 2 O/0.997 g/mL @ 25 °C = 1187 mL H 2 O)

    4. Calculate the mol HCl necessary to neutralize 5.5 g Ba(OH) 2 .

      Ba(OH) 2 (s) + 2 HCl(aq) → BaCl 2 (aq) + 2 H 2 O(l)

      5.5 g Ba(OH) 2 + 1 mol Ba(OH) 2 171 g Ba(OH) 2 × 2 mol HCl 1 mol Ba(OH) 2 = 0.0643 = 0.064 mol HCl

      M = mol L ; L = mol M = 0.0643 mol HCl 0.50 M HCl = 0.1287 = 0.13 L = 130 mL

      130 mL of 0.50 M HCl are needed.

      M c × L c = M d × L d ; 6.0 M × L c = 0.50 M × 0.1287 L; L c = 0.01072 L = 11 mL

      Using a pipette, measure exactly 11 mL of 6.0 M HCl and dilute with water to a total volume of 130 mL.


    13.55 Analyze/Plan .  Assume a solution volume of 1.00 L. Calculate the mass of 1.00 L of solution and the mass of HNO 3 in 1.00 L of solution. Mass % = (mass solute/mass solution) × 100. Solve .

    1.00 L × 1000 mL 1 L × 1.42 g soln mL soln = 1.42 × 10 3 g soln

    16 M = 16 mol HNO 3 1 L soln × 63.02 g HNO 3 1 mol HNO 3 = 1008 = 1.0 × 10 3 g HNO 3

    mass % = 1008 g HNO 3 1.42 × 10 3 g soln × 100 = 71 % HNO 3

    13.56 Analyze/Plan .  Assume 1.00 L of solution. Calculate mass of 1 L of solution using density. Calculate mass of NH 3 using mass %, then mol NH 3 in 1.00 L. Solve .

    1.00 L soln × 1000 mL 1 L × 0.90 g soln 1 mL soln = 9.0 × 10 2 g soln/L

    900 g soln 1.00 L soln × 28 g NH 3 100 g soln × 1 mol NH 3 17.03 g NH 3 = 14.80 = 15 mol NH 3 /L soln = 15 M NH 3

    13.57 Analyze .  Given: 80.0% Cu, 20.0% Zn by mass; density = 8750 kg/m 3 . Find: (a) m of Zn (b) M of Zn

    1. Plan .  In the brass alloy, Zn is the solute (lesser component) and Cu is the solvent (greater component). m = mol Zn/kg Cu. 1 m 3 brass alloy weighs 8750 kg. 80.0% is Cu, 20.0% is Zn. Change g Zn → mol Zn and solve for m . Solve .

      8750 kg brass × 80 g Cu 100 g brass = 7.00 × 10 3 kg Cu

      8750 kg brass – 7000 kg Cu = 1750 kg Zn

      1750 kg Zn × 1000 g kg × 1 mol Zn 65 .39 g Zn = 26 , 762.5 = 2.68 × 10 4 mol Zn

      m = 2.676 × 10 4 mol Zn 7000 kg Cu = 3.82 m Zn

    2. Plan . M = mol Zn/L brass. Use mol Zn from part (a). Change 1 m 3 → L brass and calculate M . Solve .

      1 m 3 × ( 10 ) 3 dm 3 m 3 × 1 L 1 dm 3 = 1000 L

      M = 2.676 × 10 4 mol Zn 1000 L brass = 26.76 = 26.8 M Zn

    13.58

    1. 0.0500 mol C 8 H 10 N 4 O 2 1 kg CHCl 3 × 194.2 g C 8 H 10 N 4 O 2 1 mol C 8 H 10 N 4 O 2 = 9.7100 = 9.71 g C 8 H 10 N 4 O 2 /kg CHCl 3

      9. 710 g C 8 H 10 N 4 O 2 9. 710 g C 8 H 10 N 4 O 2 + 1000.00 g CHCl 3 × 100 = 0.9617 = 0.962 % C 8 H 10 N 4 O 2 by mass


    1. 1000 g CHCl 3 × 1 mol CHCl 3 119.4 CHCl 3 = 8.375 = 8.38 mol CHCl 3

      X C 8 H 10 N 4 O 2 = 0.0500 0.0500 + 8.375 = 0.00593

    13.59 Analyze .  Given: 4.6% CO 2 by volume (in air), 1 atm total pressure. Find: partial pressure and molarity of CO 2 in air.

    Plan .  4.6% CO 2 by volume means 4.6 mL of CO 2 could be isolated from 100 mL of air, at the same temperature and pressure. According to Avogadro’s law, equal volumes of gases at the same temperature and pressure contain equal numbers of moles. By inference, the volume ratio of CO 2 to air, 4.6/100 or 0.046, is also the mole ratio. Solve .

    1. P CO 2 = X CO 2 × P t = 0.046 (1 atm) = 0 .046 atm
    2. M = mol CO 2 /L air = n/V.  PV = nRT, M = n/V = P/RT

      M CO 2 = P CO 2 RT = 0.046 atm 310 K × mol-K 0 .08206 L-atm = 1.8 × 10 3 M

    13.60

    1. For gases at the same temperature and pressure, volume % = mol %. The volume and mol % of CO 2 in this breathing air is 4.0%.
    2. P CO 2 = X CO 2 × P t = 0.040 (1 atm) = 0 .040 atm

      M CO 2 = P CO 2 RT = 0.040 atm 310 K × mol-K 0 .08206 L-atm = 1.6 × 10 3 M

    Colligative Properties (Section 13.5)

    13.61

    1. False
    2. True
    3. True
    4. False

    13.62

    1. False. Adding solvent decreases the molality of the solution and elevates the freezing point of the solution.
    2. False. The solid that forms is nearly pure solvent.
    3. False. The more concentrated the solution, the lower the freezing point.
    4. True
    5. True

    13.63 Analyze/Plan .  H 2 O vapor pressure will be determined by the mole fraction of H 2 O in the solution. The vapor pressure of pure H 2 O at 20 °C = 17.5 torr. Solve .

    The density of water at 20 °C is not exactly 1 g/mL. From Appendix B, the density of water at 25 °C is 0.99707 g/mL. For the purpose of this calculation, assume this is the density at 20 °C.

    10.0 g C 6 H 12 O 6 180.15 g/mol = 0.055509 = 0.0555 mol; 997 g H 2 O 18.02 g/mol = 55.327 = 55.3 mol

    P H 2 O = X H 2 O P H 2 O ° = 55.3 mol H 2 O 55.3 + 0.0555 × 17.5 torr = 17.48 = 17 .5 torr


    10.0 g C 12 H 22 O 11 342.3 g/mol = 0.029214 = 0.0292 mol; 997 g H 2 O 18.02 g/mol = 55.327 = 55.3 mol

    P H 2 O = X H 2 O P H 2 O ° = 55.3 mol H 2 O 55.3 + 0.0292 × 17.5 torr = 17.49 = 17 .5 torr

    Because these two solutions are so dilute, they have essentially the same vapor pressure. Generally, the less concentrated solution, the one with fewer moles of solute per kilogram of solvent, will have the higher vapor pressure.

    13.64  Analyze/Plan.  Calculate the vapor pressure predicted by Raoult’s law and compare it to the experimental vapor pressure. Assume ethylene glycol (eg) is the solute. Solve .

    X H 2 O = X eg = 0.500 ; P A = X A P A ° = 0.500 ( 149 ) torr = 74 .5 torr

    The experimental vapor pressure (P A ), 67 torr, is less than the value predicted by Raoult’s law for an ideal solution. The solution is not ideal.

    Check .  An ethylene glycol-water solution has extensive hydrogen bonding, which causes deviation from ideal behavior. We expect the experimental vapor pressure to be less than the ideal value and it is.

    13.65

    1. Analyze/Plan .  H 2 O vapor pressure will be determined by the mole fraction of H 2 O in the solution. The vapor pressure of pure H 2 O at 338 K (65 °C) = 187.5 torr.

      Solve .

      22.5 g C 12 H 22 O 11 342.3 g/mol = 0.06573 = 0.0657 mol; 200 .0 g H 2 O 18.02 g/mol = 11.09878 = 11.10 mol

      P H 2 O = X H 2 O P H 2 O ° = 11.09878 mol H 2 O 11.09878 + 0.06573 × 187.5 torr = 186 .4 torr

    2. Analyze/Plan .  For this problem, it will be convenient to express Raoult’s law in terms of the lowering of the vapor pressure of the solvent, ΔP A .

      ΔP A = P X A P = P (1 – X A ). 1 – X A = X B , the mole fraction of the solute particles

      ΔP A = X B P A ° ; the vapor pressure of the solvent (A) is lowered according to the mole fraction of solute (B) particles present. Solve .

      P H 2 O at 40 o C = 55 .3 torr; 340 g H 2 O 18.02 g/mol = 18.868 = 18.9 mol H 2 O

      X C 3 H 8 O 2 = 2.88 torr 55 .3 torr = y mol C 3 H 8 O 2 y mol C 3 H 8 O 2 + 18.868 mol H 2 O = 0.05208 = 0.0521

      0.05208 = y y + 18 .868 ; 0.05208 y + 0 .98263 = y; 0 .94792 y = 0 .98263,

      y = 1.0366 = 1.04 mol C 3 H 8 O 2

      This result has 3 sig figs because (0.340 kg water) has 3 sig figs.

      1.0366 mol C 3 H 8 O 2 × 76.09 g C 3 H 8 O 2 mol C 3 H 8 O 2 = 78.88 = 78.9 g C 3 H 8 O 2

    1. P C 6 H 6 = 0.2453 ( 75 torr) = 18 .4 torr;   P C 7 H 8 = 0.7547 ( 22 torr) = 16 .6 torr

      In the vapor, X C 6 H 6 = P C 6 H 6 P t = 18.4 torr 18 .4 torr + 16 .6 torr = 0.53 ; X C 7 H 8 = 0.47


    13.66

    1. H 2 O vapor pressure will be determined by the mole fraction of H 2 O in the solution. The vapor pressure of pure H 2 O at 343 K (70 °C) = 233.7 torr.

      28.5 g C 3 H 8 O 3 92.10 g/mol = 0.3094 = 0.309 mol; 125 g H 2 O 18.02 g/mol = 6.937 = 6.94 mol

      P H 2 O = 6.937 mol H 2 O 6.937 + 0.309 × 233.7 torr = 223 .7 = 224 torr

    2. Calculate X B by vapor pressure lowering; X B = ΔP A /P . [See Solution 13.65(b).] Given moles solvent, calculate moles solute from the definition of mole fraction.

      X C 2 H 6 O 2 = 10.0 torr 100 torr = 0.100

      1.00 × 10 3 g C 2 H 5 OH 46.07 g/mol = 21.71 = 21.7 mol C 2 H 5 OH; let y = mol C 2 H 6 O 2

      X C 2 H 6 O 2 = y mol C 2 H 6 O 2 y mol C 2 H 6 O 2 + 21.71 mol C 2 H 5 OH = 0.100 = y y + 21 .71

      0.100 y + 2.171 = y; 0.900 y = 2.171; y = 2.412 = 2.41 mol C 2 H 6 O 2

      2.412 mol C 2 H 6 O 2 × 62.07 g 1 mol = 150 g C 2 H 6 O 2

    13.67 Analyze/Plan .  At 63.5 C, P H 2 O = 175 torr, P Eth = 400 torr . Let G = the mass of H 2 O and/or C 2 H 5 OH. Solve .

    1. X Eth = G 46 .07 g/mol C 2 H 5 OH G 46 .07 g/mol C 2 H 5 OH + G 18 .02 g/mol H 2 O

      Multiplying top and bottom of the right side of the equation by 1/G gives:

      X Eth = 1 / 46.07 1 / 46.07 + 1 / 18.02 = 0.02171 0.02171 + 0.05549 = 0.2812

    2. P t = P Eth + P H 2 O ; P Eth = X Eth × P Eth ; P H 2 O = X H 2 O P H 2 O

      X Eth = 0.2812, P Eth = 0.2812 (400 torr) = 112.48 = 112 torr

      X Η 2 O = 1 0 .2812 = 0 .7188;    P H 2 O = 0.7188 ( 175 torr) = 125 .8 = 126 torr

      P t = 112.5 torr + 125.8 torr = 238.3 = 238 torr

    3. X Eth in vapor = P Eth P total = 112.5 torr 238.3 torr = 0.4721 = 0.472

    13.68

    1. Because C 6 H 6 and C 7 H 8 form an ideal solution, we can use Raoult’s law. Because both components are volatile, both contribute to the total vapor pressure of 35 torr.

      P t = P C 6 H 6 + P C 7 H 8 ; P C 6 H 6 = X C 6 H 6 P C 6 H 6 ; P C 7 H 8 = X C 7 H 8 P C 7 H 8

      X C 7 H 8 = 1 X C 6 H 6 ; P T = X C 6 H 6 P C 6 H 6 + ( 1 X C 6 H 6 ) P C 7 H 8

      35 torr = X C 6 H 6 ( 75 torr) + (1 X C 6 H 6 ) 22 torr

      13 torr = 53 torr ( X C 6 H 6 ) ; X C 6 H 6 = 13 torr 53 torr = 0.2453 = 0.25 ; X C 7 H 8 = 0.7547 = 0.75


    13.69

    1. Because NaCl is a soluble ionic compound and a strong electrolyte, there are 2 mol of dissolved particles for every 1 mol of NaCl solute. C 6 H 12 O 6 is a molecular solute, so there is 1 mol of dissolved particles per mol solute. Boiling point elevation is directly related to total moles of dissolved particles; 0.10 m NaCl has more dissolved particles so its boiling point is higher than 0.10 m C 6 H 12 O 6 .
    2. In solutions of strong electrolytes like NaCl, electrostatic attractions between ions lead to ion pairing. Ion pairing reduces the effective number of particles in solution, decreasing the change in boiling point. The actual boiling point is then lower than the calculated boiling point for a 0.10 m solution.

    13.70 Analyze/Plan .  ΔT b depends on mol dissolved particles. Assume 100 g of each solution, calculate mol solute and mol dissolved particles. Glucose and sucrose are molecular solutes, but NaNO 3 dissociates into 2 mol particles per mol solute. Solve .

    10% by mass means 10 g solute in 100 g solution. If we have 10 g of each solute, the one with the smallest molar mass will have the largest mol solute. The molar masses are: glucose, 180.2 g/mol; sucrose, 342.3 g/mol; NaNO 3 , 85.0 g/mol. NaNO 3 has most mol solute, and twice as many dissolved particles, so it will have the highest boiling point. Sucrose has least mol solute and lowest boiling point. Glucose is intermediate.

    In order of increasing boiling point: 10% sucrose < 10% glucose < 10% NaNO 3 .

    13.71 Analyze/Plan . Rank the solutions in order of increasing boiling point. All solutes are nonvolatile. The more nonvolatile solute particles, the higher the boiling point of the solution. Solve .

    Solve . Because LiBr and Zn(NO 3 ) 2 are electrolytes, the particle concentrations in these solutions are 0.10 m and 0.15 m , respectively (although ion-ion attractive forces may decrease the effective concentrations somewhat). Thus, the order of increasing particle concentration and boiling point is:

    0.050 m LiBr < 0.120 m glucose < 0.050 m Zn(NO 3 ) 2

    13.72  0.030 m phenol > 0.040 m glycerin = 0.020 m KBr. Phenol is very slightly ionized in water, but not enough to match the number of particles in a 0.040 m glycerin solution. Assuming the ideal van’t Hoff factor of 2.00, the KBr solution is 0.040 m in particles, so it has the same freezing point as 0.040 m glycerin, which is a nonelectrolyte. (The measured van’t Hoff factor for 0.040 m KBr will be slightly less than 2. We expect the measured order of freezing points to be 0.030 m phenol > 0.020 m KBr > 0.040 m glycerin.)

    13.73 Analyze/Plan .  ΔT = K ( m ); first, calculate the molality of each solution. Solve .

    1. 0.22 m
    2. 2.45 mol CHCl 3 × 119.4 g CHCl 3 mol CHCl 3 = 292.53 g = 0 .293 kg;

      0.240 mol C 10 H 8 0.29253 kg CHCl 3 = 0.8204 = 0.820 m


    1. 1. 50 g NaCl × 1 mol NaCl 5 8 .44 g NaCl × 2 mol particles 1 mol NaCl = 0.05133 = 0.0513 mol particles

      m = 0.05133 mol NaCl 0.250 kg H 2 O = 0.20534 = 0.205 m

    1. 2.04 g KBr × 1 mol KBr 119.0 g KBr × 2 mol particles 1 mol KBr = 0.03429 = 0.0343 mol particles

      4.82 g C 6 H 12 O 6 × 1 mol C 6 H 12 O 6 180.2 g C 6 H 12 O 6 = 0.02675 = 0.0268 mol particles

      m = ( 0.03429 + 0.02675 ) mol particles 0.188 kg H 2 O = 0.32465 = 0.325 m

    Solve .  Then, fp = T f – K f ( m ); bp = T b + K b ( m ); T in C

    m T f K f (m) fp T b +K b (m) bp
    (a) 0.22 –114.6 –1.99(0.22) = –0.44 –115.0 78.4 1.22(0.22) = 0.27 78.7
    (b) 0.820 −63.5 −4.68(0.820) = −3.84 −67.3 61.2 3.63(0.820) = 2.98 64.2
    (c) 0.205 0.0 −1.86(0.205) = −0.381 −0.4 100.0 0.51(0.205) = 0.10 100.1
    (d) 0.325 0.0 −1.86(0.325) = −0.605 −0.6 100.0 0.51(0.325) = 0.17 100.2

    13.74  ΔT = K( m ); first calculate the molality of the solute particles.

    1. 0.25 m
    2. 20.0 g C 10 H 22 0.0500 kg CHCl 3 × 1 mol C 10 H 22 142.3 g C 10 H 22 = 2.811 = 2.81 m
    3. 3. 50 g NaOH × 1 mol NaOH 4 0 .00 g NaOH × 2 mol particles 1 mol NaOH = 0.1750 = 0.175 mol particles

      m = 0.1750 mol NaCl 0.175 kg H 2 O = 1.000 = 1.00 m

    4. m = 0.45 mol eg + 2(0 .15) mol KBr 0.150 kg H 2 O = 0.75 mol particles 0.150 kg H 2 O = 5.0 m

      Then, fp = T f – K f ( m ); bp = T b + K b ( m ); T in C

    m T f – K f ( m ) fp T b + K b ( m ) bp
    (a) 0.25 −114.6 −1.99(0.25) = −0.50 −115.1 78.4 1.22(0.25) = 0.31 78.7
    (b) 2.81 −63.5 −4.68(2.81) = −13.2 −76.7 61.2 3.63(2.81) = 10.2 71.4
    (c) 1.00 0.0 −1.86(1.00) = −1.86 −1.9 100.0 0.51(1.00) = 0.51 100.5
    (d) 5.0 0.0 −1.86(5.0) = −9.3 −9.3 100.0 0.51(5.0) = 2.6 102.6

    13.75 Analyze . Given freezing point of solution and mass of solvent, calculate mass of solute.

    Plan . Use ΔT f = K f (m) to calculate the required molality, and then apply the definition of molality to calculate moles and grams of C 2 H 6 O 2 .

    Solve . fp of solution = −5.00 °C; fp of solvent (H 2 O) = 0.0 °C


    ΔT f = 5.00 o C = K f ( m ); 5.00 o C = 1.86 o C / m ( m )

    m = 5.00 o C 1 .86 o C/ m = 2.688 = 2.69 m C 2 H 6 O 2

    m = mol C 2 H 6 O 2 kg H 2 O = C 2 H 6 O 2 = m × kg H 2 O

    2.688 m C 2 H 6 O 2 × 1.00 kg H 2 O = 2.688 = 2.69 mol C 2 H 6 O 2

    2.688 m C 2 H 6 O 2 × 62 .07 g C 2 H 6 O 2 1 mol = 166.84 = 167 g C 2 H 6 O 2

    13.76  Use ΔT b = find m of aqueous solution, and then use m to calculate ΔT f and freezing point. K b = 0.51, K f = 1.86.

    bp = 105.0 °C; ΔT b = 105.0 °C – 100.0 °C = 5.0 °C

    ΔT b = K b (m); m = Δ T b K b = 5.0 °C 0.51 = 9.804 = 9.8 m

    ΔT f = 1.86 °C/m × 9.804 m = 18.24 = 18 °C; freezing point = 0.0 °C – 18.24 °C = –18 °C

    13.77 Analyze/Plan .  Π = M RT; T = 25 °C + 273 = 298 K; M = mol C 9 H 8 O 4 /L soln Solve .

    M = 44.2 mg C 9 H 8 O 4 0.358 L × 1 g 1000 mg × 1 mol C 9 H 8 O 4 180.2 g C 9 H 8 O 4 = 6.851 × 10 4 = 6.85 × 10 4 M

    Π = 6.851 × 10 4 mol L × 0.08206 L-atm mol-K × 298 K = 0 .01675 = 0 .0168 atm = 12 .7 torr

    13.78  Π = M RT; T = 20 °C + 273 = 293 K

    M (of ions) = mol NaCl × 2 L soln = 3.4 g NaCl 1 L soln × 1 mol NaCl 58.4 g NaCl × 2 mol ions 1 mol NaCl = 0.116 = 0.12 M

    Π = 0.116 mol L × 0.08206 L-atm mol-K × 293 K = 2 .8 atm

    13.79 Analyze/Plan .  Follow the logic in Sample Exercise 13.10 to calculate the molar mass of adrenaline based on the boiling point data. Use the structure to obtain the molecular formula and molar mass. Compare the two values. Solve .

    Δ T b = K b m ; m = Δ T b K b = + 0.49 5.02 = 0.0976 = 0.098 m adrenaline

    m = mol adrenaline kg CCl 4 = g adrenaline MM adrenaline × kg CCl 4

    M M adrenaline = g adrenaline m × kg CCl 4 = 0.64 g adrenaline 0.0976 m × 0.0360 kg CCl 4 = 1.8 × 10 2 g/mol adrenaline

    Check. The molecular formula is C 9 H 13 NO 3 , MM = 183 g/mol. The values agree to 2 sig figs, the precision of the experimental value.


    13.80 Δ T f = 5.5 4 .1 = 1 .4; m = Δ T f K f = 1.4 5.12 = 0.273 = 0.27 m

    MM lauryl alcohol = g lauryl alcohol m × kg C 6 H 6 = 5.00 g lauryl alcohol 0.273 × 0.100 kg C 6 H 6 = 1 .8 × 10 2 g/mol lauryl alcohol

    13.81 Analyze/Plan .  Follow the logic in Sample Exercise 13.11. Solve .

    Π = M RT; M = Π RT ; T = 25 C o + 273 = 298 K

    M = 0.953 torr × 1 atm 760 torr × mol-K 0 .08206 L-atm × 1 298 K = 5.128 × 10 5 = 5.13 × 10 5 M

    mol = M × L = 5.128 × 10– 5 × 0.210 L = 1.077 × 10– 5 = 1.08 × 10– 5 mol lysozyme

    MM = g mol = 0.150 g 1 .077 × 10 5 mol = 1.39 × 10 4 g/mol lysozyme

    13.82 M = P/RT = 0 .605 atm 298 K × mol-K 0 .08206 L-atm = 0.02474 = 0 .0247 M

    MM = g M × L = 2 .35 g 0.02474 M × 0.250 L = 380 g/mol

    13.83 Analyze/Plan. i = Π (measured) / Π (calculated for a nonelectrolyte);

    Π (calculated) = M RT. Solve .

    Π (calculated) = 0 .010 mol L × 0.08206 L-atm mol-K × 298 K = 0 .2445 = 0 .24 atm

    i = 0.674 atm/0.2445 atm = 2.756 = 2.8

    13.84  If these were ideal solutions, they would have equal ion concentrations and equal ΔT f values. Data in Table 13.4 indicates that the van’t Hoff factors (i) for both salts are less than the ideal values. For 0.030 m NaCl, i is between 1.87 and 1.94, about 1.92. For 0.020 m K 2 SO 4 , i is between 2.32 and 2.70, about 2.62. From Equation 13.15,

    ΔT f (measured) = i × ΔT f (calculated for nonelectrolyte)

    NaCl: ΔT f (measured) = 1.92 × 0.030 m × 1.86 °C/ m = 0.11 °C

    K 2 SO 4 : ΔT f (measured) = 2.62 × 0.020 m × 1.86°C/ m = 0.097 °C

    0.030 m NaCl would have the larger ΔT f .

    (The deviations from ideal behavior are due to ion pairing in the two electrolyte solutions. K 2 SO 4 has more extensive ionpairing and a larger deviation from ideality because of the higher charge on SO 4 2– relative to Cl .)

    Colloids (Section 13.6)

    13.85

    1. No. In the gaseous state, the particles are far apart and intermolecular attractive forces are small. When two gases combine, all terms in Equation 13.1 are essentially zero and the mixture is always homogeneous.

    1. The outline of a light beam passing through a colloid is visible, whereas light passing through a true solution is invisible unless collected on a screen. This is the Tyndall effect. To determine whether Faraday’s (or anyone’s) apparently homogeneous dispersion is a true solution or a colloid, shine a beam of light on it and see if the light is scattered.

    13.86  The best answer is (b) emulsion (Table 13.5).

    13.87  The best emulsifying agent is (d) CH 3 (CH 2 ) 11 COONa. A good emulsifying agent has a polar (or ionic) end to interact with hydrophilic substances, and an nonpolar end to interact with hydrophobic substances. Choices (c) and (d) fit this description, but the ionic end of (d) will help stabilize the colloi387d.

    13.88  The presence of aerosols in the atmosphere decreases the amount of sunlight that arrives at Earth’s surface, compared to an “aerosol-free” atmosphere. All colloids scatter light (the Tyndall effect). Aerosols in the atmosphere scatter the incoming sunlight. Although some of this scattered light will eventually reach Earth, some will not.

    13.89

    1. No. Adsorbed ions stabilize hydrophobic colloids in water. The hydrophobic/ hydrophilic nature of the protein will determine which electrolyte at which concentration will be the most effective precipitating salt.
    2. Stronger. If a protein has been “salted out,” protein-protein interactions are sufficiently strong so that the protein molecules “stick together” and form a solid. Before the electrolyte is added, protein-protein interactions are weaker than protein-water interactions and the protein molecules remained suspended in solution.
    3. The first hypothesis seems plausible, because ion-dipole interactions among electrolytes and water molecules are stronger than dipole-dipole and hydrogen- bonding interactions between water and protein molecules. However, this ignores the strength of ion-dipole interactions between the electrolyte and protein molecules. And, we know from Figure 13.27 that ions are adsorbed on the surface of hydrophobic colloids. With the right protein and electrolyte, the second hypothesis also seems plausible.

      The van’t Hoff effect is a result of ion pairing. We know from Table 13.4 that the effect of ion pairing increases with concentration and the charge on the ions of the electrolyte. If we could measure the charge and adsorbed water content of protein molecules as a function of salt concentration, then we could distinguish between these two hypotheses.

    13.90

    1. Head.
    2. Tail.
    3. The charged –COO head will experience ion-dipole, dipole-dipole, and hydrogen bonding interactions with water. The hydrocarbon tail of sodium stearate will experience dispersion forces with hydrophobic grease.

    Additional Exercises

    13.91

    1. Hydrochloride. The hydrochloride is a salt, an ionic compound, with the possibility of ion-dipole interactions in addition to hydrogen bonding and dipole-dipole interactions. Of the two forms, it will be more soluble in water.
    2. Free base. Both forms have several hydrogen bond receptors (O and N atoms with nonbonded electron pairs), but the free base is less soluble because it does not have the possibility of ion-dipole interactions.
    3. Analyze/Plan . Calculate the molar mass of the free base, then moles and molarity. mol = g/molar mass; M = mol/L. Solve .

      The molar mass of the free base is 303.353 g/mol. 6.70 mL = 0.00670 L ethanol

      1.00 g free base × 1 mol free base 303.353 g free base × 1 0.00670 L ethanol = 0.492 M free base

    4. Use the method from part (c) to calculate molarity of the hydrochloride.

      The molar mass of the hydrochloride is 339.814 g/mol. 0.400 mL = 0.000400 L

      1.00 g hydrochloride × 1 mol hydrochloride 339.814 g free base × 1 0.000400 L water = 7.36 M hydrochloride

    5. Analyze/Plan . According to the chemical reaction given in the exercise, the free base reacts in a 1:1 mole ratio with HCl(aq). Calculate moles of free base in 1.00 kg and then liters of 12.0 M HCl(aq) required. 1.00 kg = 1.00 × 10 3 g. Solve .

      1.00 × 10 3 g free base × 1 mol free base 303.353 g free base × 1 L 12.0 mol HCl = 0.275 L = 275 mL

    13.92

    1. True
    2. False. A saturated solution in contact with undissolved solute exists in a state of dynamic equilibrium. Both dissolving and crystallization occur simultaneously and at the same rates.
    3. True

    13.93  Assume that the density of the solution is 1.00 g/mL.

    1. 4 ppm O 2 = 4 mg O 2 1 kg soln = 4 × 10 3 g O 2 1 L soln × 1 mol O 2 32.0 g O 2 = 1.25 × 10 4 = 1 × 10 4 M
    2. S O 2 = kP O 2 ; P O 2 = S O 2 /k = 1.25 × 10 4 mol L × L-atm 1.71 × 10 3 mol = 0.0731 = 0.07 atm

      0.0731 atm × 760 torr 1 atm = 55.6 = 60 torr

    13.94

    1. S Rn = kP Rn ; k = S Rn / P Rn = 7.27 × 10 3 M /1 atm = 7.27 × 10 3 mol/L- atm
    2. P Rn = χ Rn P total ; P Rn = 3.5 × 10– 6 (32 atm) = 1.12 × 10– 4 = 1.1 × 10– 4 atm

      S Rn = k P Rn ; S Rn = 7.27 × 10 3 mol L-atm × 1.12 × 10 4 atm = 8 .1 × 10 7 M


    13.95  0.10% by mass means 0.10 g glucose/100 g blood.

    1. ppm glucose = g glucose g solution × 10 6 = 0.10 g glucose 100 g blood × 10 6 = 1000 ppm glucose
    2. m = mol glucose/kg solvent. Assume that the mixture of nonglucose components is the “solvent.”

      mass solvent = 100 g blood – 0.10 g glucose = 99.9 g solvent = 0.0999 kg solvent

      mol glucose = 0 .10 g × 1 mol 180 .2 g C 6 H 12 O 6 = 5.55 × 10 4 = 5.6 × 10 4 mol glucose

      m = 5.55 × 10 4 mol glucose 0.0999 kg solvent = 5.6 × 10 3 m glucose

    3. To calculate molarity, solution volume must be known. The density of blood is needed to relate mass and volume.

    13.96 Analyze . Given 13 ppt Au in seawater, find grams of Au in 1.0 × 10 3 gal seawater. The definition of ppt is (mass solute/mass solution) × 10 12 . Plan . Assume seawater is a dilute aqueous solution with a density of 1.00 g/mL. Use the definition of ppt to calculate g Au. Solve .

    13 g Au 1 × 10 12 g soln × 1 .0 g soln mL soln × 1000 mL 1  L × 3 .7854 L gal × 1.0 × 10 3 gal soln = 4.9 × 10 5 g Au

    13.97 Analyze . The definition of ppb is (mass solute/mass solution) × 10 9 . Plan. Use the definition to get g Pb and g solution. Change g Pb to mol Pb, g solution to L solution, calculate molarity. Solve .

    1. 9.0 ppb = 9 .0 g Pb 1 × 10 9 g soln × 10 9

      For dilute aqueous solutions (drinking water) assume that the density of the solution is the density of H 2 O.

      9 .0 g Pb 1 × 10 9 g soln × 1 .0 g soln mL soln × 1000 mL 1  L × 1 mol Pb 207 .2 g Pb = 4.34 × 10 8 M = 4.3 × 10 8 M

    2. Change 60 m 3 H 2 O to cm 3 (mL) H 2 O to g H 2 O (or g soln).

      60 m 3 × 100 3 cm 3 m 3 × 1 g H 2 O cm 3 H 2 O = 6.0 × 10 7 g H 2 O or soln

      9 .0 g Pb 1 × 10 9 g soln × 6.0 × 10 7 g soln = 0 .54 g Pb

    13.98

    1. 1.80 mol LiBr 1 L soln × 86.85 g LiBr 1 mol LiBr = 156.3 = 156 g LiBr

      1 L soln = 826 g soln; g CH 3 CN = 826 – 156.3 = 669.7 = 670 g CH 3 CN

      m LiBr = 1 .80 mol LiBr 0.6697 kg CH 3 CN = 2.69 m

    2. 669.7 g CH 3 CN 41.05 g/mol = 16.31 = 16.3 mol CH 3 CN; X LiBr = 1.80 1.80 + 16.31 = 0.0994
    3. mass % = 669 .7 g CH 3 CN 826 g soln × 100 = 81.1 % CH 3 CN

    13.99 Mole fraction ethyl alcohol, X C 2 H 5 OH = P C 2 H 5 OH P C 2 H 5 OH o = 8 torr 100 torr = 0.08

    620 × 10 3 g C 24 H 50 338.6 g/mol = 1.83 × 10 3 mol C 24 H 50 ; let y = mol C 2 H 5 OH

    X C 2 H 5 OH = 0.08 = y y + 1.83 × 10 3 ; 0 .92 y = 146 .4;  y = 1 .6 × 10 2 mol C 2 H 5 OH

    (Strictly speaking, y should have 1 sig fig because 0.08 has 1 sig fig, but this severely limits the calculation.)

    1.6 × 10 2 mol C 2 H 5 OH × 46 g C 2 H 5 OH 1 mol = 7.4 × 10 3 g or 7 .4 kg C 2 H 5 OH

    13.100 Analyze . Given vapor pressure of both pure water and the aqueous solution and moles H 2 O find moles of solute in the solution.

    Plan . Use vapor pressure lowering, P A = X A P A o , to calculate X A , mole fraction solvent, and then use the definition of mole fraction to calculate moles solute particles. Because NaCl is a strong electrolyte, there is 1 mol NaCl for every 2 mol solute particles.

    Solve .

    X H 2 O = P soln /P H 2 O = 25.7 / 31.8 = 0.80818 = 0.808

    X H 2 O = mol  H 2 O mol ions + mol H 2 O ; 0.80818 = 0.115 ( mol ions + 0.115 )

    0.80818 (0.115 + mol ions) = 0.115; 0.80818 (mol ions) = 0.115 – 0.092940

    mol ions = 0.02206/0.80818 = 0.02730 = 0.0273;

    mol NaCl = mol ions/2 = 0.02730/2 = 0.01365 = 0.0137 mol NaCl

    0.01365 mol NaCl × 58 .443 g C 2 H 5 OH 1 mol = 0.7977 = 0.798 g NaCl

    13.101

    1. The solvent vapor pressure over each solution is determined by the total particle concentrations present in the solutions. When the particle concentrations are equal, the vapor pressures will be equal and equilibrium established. The particle concentration of the nonelectrolyte is just 0.050 M , the ion concentration of the NaCl is 2 × 0.035 M = 0.070 M . Solvent will diffuse from the less concentrated nonelectrolyte solution. The level of the NaCl solution will rise, and the level of the nonelectrolyte solution will fall.
    2. Let x = volume of solvent transferred

      0.050 M × 30.0 mL ( 30.0 x) mL = 0.070 M × 30.0 mL ( 30.0 + x) mL ; 1.5 ( 30.0 + x) = 2 .1(30 .0 x)

      45 + 1.5 x = 63 – 2.1 x; 3.6 x = 18; x = 5.0 = 5 mL transferred

      The volume in the nonelectrolyte beaker is (30.0 – 5.0) = 25.0 mL; in the NaCl beaker (30.0 + 5.0) = 35.0 mL.


    13.102 Analyze/Plan .  A nonelectrolyte is dissolved in ethanol, C 2 H 5 OH, at its normal boiling point, 78.4 °C. At the normal boiling point, the vapor pressure of a liquid equals 1 atm or 760 torr. The resulting solution is not boiling because the solute has raised the boiling point above 78.4 °C. The vapor pressure of the solution at this temperature and pressure is 7.40 × 10 2 or 740 torr. Use Raoult’s law, Equation 13.10, to calculate the mole fraction of ethanol in the solution. Use the definition of mole fraction to calculate the molar mass of the solute. Solve .

    P solution = X solvent P solvent ° ; X ethanol = P solution /P solvent °

    x ethanol = 740 torr/760 torr = 0.97368 = 0.974

    mol ethanol = 100.0 g ethanol/ 46.068 g/mol = 2.1707 = 2.171 mol

    X Ethanol = mol ethanol mol ethanol + mol solute = 2 .1707 mol ethanol 2 .1707 mol ethanol + 9 .15 g solute molar mass solute

    Multiply top and bottom by molar mass solute (MM), then multiply both sides by (2.1707MM + 9.15).

    0.97368(2.1707MM + 9.15) = 2.1707MM

    8.9092 = (2.1707 – 2.1136)MM; MM = 156.03 = 156 g/mol

    13.103

    1. 0.100 m K 2 SO 4 is 0.300 m in particles. H 2 O is the solvent.

      ΔT f = K f m = –1.86(0.300) = –0.558; T f = 0.0 – 0.558 = –0.558 °C = –0.6 °C

    2. ΔT f (nonelectrolyte) = –1.86(0.100) = –0.186; T f = 0.0 – 0.186 = –0.186 °C = –0.2 °C

      T f (measured) = i × T f (nonelectrolyte)

      From Table 13.4, i for 0.100 m K 2 SO 4 = 2.32

      T f (measured) = 2.32(–0.186 °C) = –0.432 °C = –0.4 °C

    13.104

    1. K b = Δ T b m ; Δ T b = 47.46 ° C 46.30 ° C = 1.16 ° C

      m = mol solute kg CS 2 = 0.250 mol 400.0 mL CS 2 × 1 mL CS 2 1.261 g CS 2 × 1000 g 1 kg = 0.4956 = 0.496 m

      K b = 1.16 ° C 0 .4956 m = 2.34 ° C/ m

    2. m = Δ T b K b = ( 47.08 46.30 ) ° C 2.34 ° C/ m = 0.333 = 0.33 m

      m = mol unknown kg CS 2 ; m × kg CS 2 = g unknown MM unknown ; MM = g unknown m × kg CS 2

      50.0 mL CS 2 × 1.261 g CS 2 1 mL × 1 kg 1000 g = 0.06305 = 0.0631 kg CS 2

      MM = 5.39 g unknown 0.333 m × 0.06305 kg CS 2 = 257 = 2.6 × 10 2 g/mol


    13.105 M = Π RT = 57.1 torr 298 K × 1 atm 760 torr × mol-K 0 .08206 L-atm = 3.072 × 10 3 = 3.07 × 10 3 M

    0.036 g solute 100 g H 2 O × 1000 g H 2 O 1 kg H 2 O = 0.36 g solute/kg H 2 O

    Assuming molarity and molality are the same in this dilute solution, we can then say 0.36 g solute = 3.072 × 10– 3 mol; MM = 117 g/mol. Because the salt is completely ionized, the formula weight of the lithium salt is twice this calculated value, or 234 g/mol . The organic portion, C n H 2n+1 O 2– , has a formula weight of 234 – 7 = 227 g. Subtracting 32 for the oxygens, and 1 to make the formula C n H 2n , we have C n H 2n , MM = 194 g/mol. Because each CH 2 unit has a mass of 14, n ≈ 194/14 ≈ 14. The formula for our salt is LiC 14 H 29 O 2 .

    Integrative Exercises

    13.106  Because these are very dilute solutions, assume that the density of the solution ≈ the density of H 2 O ≈ 1.0 g/mL at 25 °C. Then, 100 g solution = 100 g H 2 O = 0.100 kg H 2 O.

    1. CF 4 : 0 .0015 g CF 4 0.100 kg H 2 O × 1 mol CF 4 88.00 g CF 4 = 1.7 × 10 4 m

      CClF 3 : 0 .009 g CClF 3 0.100 kg H 2 O × 1 mol CClF 3 104.46 g CClF 3 = 8.6 × 10 4 m = 9 × 10 4 m

      CCl 2 F 2 : 0 .028 g CCl 2 F 2 0.100 kg H 2 O × 1 mol CCl 2 F 2 120.9 g CCl 2 F 2 = 2.3 × 10 3 m

      CHClF 2 : 0 .30 g CHClF 2 0.100 kg H 2 O × 1 mol CHClF 2 86.47 g CHClF 2 = 3.5 × 10 2 m

    2. Dipole moment. CCl 2 F 2 has the largest molar mass, but it is not the most soluble. None of the molecules are capable of hydrogen bonding with water; F atoms bound to C are not hydrogen bond acceptors and H bound to C is not a hydrogen bond donor.
    3. Air is 21% O 2 by volume. The volume of O 2 (g) in a baby’s lungs is

      0.21(15 mL) = 3.15 = 3.2 mL = 0.0032 L

      Assume air pressure in the lungs is 1 atm and body temperature is 37 °C or 310 K.

      n = PV RT = 1 atm × mol-K 0 .08206 L-atm × 0.00315 L 310 K = 1.238 × 10 4 = 1.2 × 10 4 mol O 2

      A volume of 66 mL of O 2 (g) dissolves in 100 mL of the fluorinated liquid. That is 66% O 2 by volume. [(66/100)100 = 66]

      In a 15 mL volume of the liquid, the volume of O 2 (g) is

      0.66(15 mL) = 9.90 = 9.9 mL = 0.0099 L

      n = PV RT = 1 atm × mol-K 0 .08206 L-atm × 0.00990 L 310 K = 3.892 × 10 4 = 3.9 × 10 4 mol O 2


    13.107

    1. 0.015 g N 2 1 L blood × 1 mol N 2 28.01 g N 2 = 5.355 × 10 4 = 5.4 × 10 4 mol N 2 /L blood
    2. At 100 ft, the partial pressure of N 2 in air is 0.78 (4.0 atm) = 3.12 atm. This is just four times the partial pressure of N 2 at 1.0 atm air pressure. According to Henry’s law, S g = k P g , a fourfold increase in P g results in a fourfold increase in S g , the solubility of the gas. Thus, the solubility of N 2 at 100 ft is 4(5.355 × 10 –4 M ) = 2.142 × 10 –3 = 2.1 × 10 –3 M .
    3. If the diver suddenly surfaces, the amount of N 2 /L blood released is the difference in the solubilities at the two depths:

      (2.142×10– 3 mol/L – 5.355×10– 4 mol/L) = 1.607×10– 3 = 1.6×10– 3 mol N 2 /L blood.

      At surface conditions of 1.0 atm external pressure and 37 °C = 310 K,

      V = nRT P = 1.607 × 10 3 mol × 310 K 1 .0 atm × 0.08206 L-atm mol-K = 0.041 L

      That is, 41 mL of tiny N 2 bubbles are released from each liter of blood.

    13.108  The stronger the intermolecular forces, the higher the heat (enthalpy) of vaporization.

    1. None of the substances are capable of hydrogen bonding in the pure liquid, and they have similar molar masses. All intermolecular forces are van der Waals forces, dipole-dipole, and dispersion forces. In decreasing order of strength of forces: acetone > acetaldehyde > ethylene oxide < cyclopropane

      The first three compounds have dipole-dipole and dispersion forces, the last only dispersion forces.

    2. The order of solubility in hexane should be the reverse of the order above. The least polar substance, cyclopropane, will be most soluble in hexane. Ethanol, CH 3 CH 2 OH, is capable of hydrogen bonding with the three polar compounds. Thus, acetaldehyde, acetone, and ethylene oxide should be more soluble than cyclopropane, but without further information we cannot distinguish among the polar molecules.

    13.109

    1. The central atom and the number of electron-pair domains about it are: (i) Cl, 4; (ii) B, 4; (iii) P, 6; (iv) Al, 4; (v) B, 4
    2. The electron-domain geometry around B in BARF is tetrahedral.
    3. The central P atom in anion (iii) has an expanded octet. As drawn, the central Cl atom in anion (i) also has an expanded octet. Note that multiple resonance structures for ClO 4 can be drawn, including one where Cl obeys the octet rule. The structure shown in this exercise is the one that minimizes formal charge.
    4. BARF is the largest anion; it will have the strongest dispersion forces that promote solubility in nonpolar solvents.

    13.110

    1. Zn(s) + H 2 SO 4 (aq) → ZnSO 4 (aq) + H 2 (g)

      2.050 g Zn × 1 mol Zn 65 .39 g Zn = 0.03135 mol Zn

      1.00 M H 2 SO 4 × 0.0150 L = 0.0150 mol H 2 SO 4

      Because Zn and H 2 SO 4 react in a 1:1 mole ratio, H 2 SO 4 is the limiting reactant; 0.0150 mol of H 2 (g) are produced.

    2. P = nRT V = 0.0150 mol 0.122 L × 0.08206 L-atm mol-K × 298 K = 3.0066 = 3.01 atm
    3. S H 2 = kP H 2 = 7.8 × 10 4 mol L-atm × 3.0066 atm = 0.002345 = 2.3 × 10 3 M

      0.002345 mol H 2 L soln × 0.0150 L = 3.518 × 10 5 = 3.5 × 10 5 mol dissolved H 2

      3.5 × 10 5 mol dissolved H 2 0.0150 mol H 2 produced × 100 = 0.23 % dissolved H 2

      This is approximately 2.3 parts per thousand; for every 10,000 H 2 molecules, 23 are dissolved. It was reasonable to ignore dissolved H 2 (g) in part (b).

    13.111

    1. 1.3 × 10 3 mol CH 4 L soln × 4.0 L = 5.2 × 10 3 mol CH 4

      V = nRT P = 5.2 × 10 3 mol × 298 K 1.0 atm × 0.08206 L-atm mol-K = 0.13 L

    2. No. The three hydrocarbons are all nonpolar. The order of increasing water solubility is the order of increasing polarizability.
    3. Hydrocarbons can experience only dispersion forces with water.
    4. image

      Ethylene is the only one of the hydrocarbons that possesses a π bond. That ethylene is more soluble in water than ethane means that is has stronger dispersion interactions with water and that the π cloud is more polarizable. The π cloud is outside the molecular framework and the σ cloud is inside. The π cloud is more easily deformed by approaching molecules, making it more polarizable than the σ cloud. In common vernacular, the π cloud is mushy and sticks out, exposing it to the attack of surrounding water molecules.

    5. NO is most soluble because it is polar. It is more soluble in water than O 2 or N 2 because it has a dipole moment and has dipole-dipole interactions with water. The three molecules have similar molar masses and their dispersion forces with water are similar.
    6. H 2 S has dipole-dipole and dispersion forces with water. While H–S bonds technically do not qualify for hydrogen bonding, dipole-dipole interactions with water are strong enough so that a small amount of H–S bond dissociation does occur. H 2 S is weakly acidic in water. This encourages the water solubility of H 2 S.

    1. SO 2 has dispersion and dipole-dipole interactions with water. In fact, the dipole-dipole forces are strong enough so that SO 2 reacts with water to form H 2 SO 3 , a weak acid. The large solubility of SO 2 is a sure sign that a chemical process has occurred.

    13.112 The resulting solution is very dilute, so assume ideal behavior. Assume the amount of water consumed in the reaction is negligible. Ignore the solubility of H 2 (g) in the solution (see Solution 3.110).

    1.0 mm 3 × 0. 535 g cm 3 × 1 3 cm 3 10 3 mm 3 = 5.35 × 10 4 = 5.4 × 10 4 g Li

    5. 35 × 10 4 g Li × 1 mol Li 6. 941 g Li = 7.708 × 10 5 = 7.7 × 10 5 mol Li

    mol Li = mol LiOH; 2 mol ions per mol LiOH

    7.708 × 10– 5 mol Li = 7.708 × 10 –5 mol LiOH = 1.542 × 10 –4 mol ions = 1.5 × 10 –4 mol ions

    m = 1. 542 × 10 4 mol ions 0.500 L H 2 O × 1 L 1000 mL × 1 mL H 2 O 0.997 g H 2 O × 1000 g 1 kg = 3.092 × 10 4 = 3.1 × 10 4 m

    ΔT f = K f m = –1.86(3.092 × 10– 4 ) = –5.8 × 10– 4 °C; T f = 0.00000 – 0.00058 = –0.00058 °C

    The freezing point of the LiOH(aq) solution is essentially zero.

    13.113 X CHCl 3 = X C 3 H 6 O = 0.500

    1. For an ideal solution, Raoult’s law is obeyed.

      P t = P CHCl 3 + P C 3 H 6 O ; P CHCl 3 = 0.5 ( 300 torr) = 150 torr

      P C 3 H 6 O = 0.5 ( 360 torr ) = 180 torr ; P t = 150 torr + 180 torr = 330 torr

    2. The mixing of the two liquids is exothermic. According to Coulomb’s law, electrostatic attractive forces lead to an overall lowering of the energy of the system. Thus, when the two liquids mix and hydrogen bonds are formed, the energy of the system is decreased and ΔH soln < 0.

    13.114

    1. True solutions do not scatter light, colloids do. Below the critical micelle concentration, cmc, the mixture of solvent and surfactant is a true solution. Above the cmc, the mixture is a colloid. The micelles are too large to be perfectly mixed in the solvent. They are suspended in the solvent, resulting in a colloid that scatters light.
    2. Surfactant monomers are anions; the “head” carries a negative charge. Below the cmc, each monomer is an independent particle. Above the cmc, many monomers aggregate into one micelle, drastically reducing the effective number of particles “in solution.” (A micelle does have a greater negative charge than a monomer.) This dramatically changes the ionic conductivity.
    3. The interior of a micelle is a hydrophobic environment. If a dye molecule becomes entrapped in a micelle, it will fluoresce. In the absence of micelles, the dye will not fluoresce in an aqueous solution.

      At low sodium stearate concentrations, the fluorescent intensity will be low. As the surfactant concentration increases, fluorescent intensity increases gradually until the cmc is reached (assuming a few micelles form at concentrations below cmc). At the cmc, there is a large increase in fluorescent intensity. At concentrations greater than the cmc, fluorescent intensity remains high and probably increases gradually.


     

    14 Chemical Kinetics

    Visualizing Concepts

    14.1 Analyze/Plan. Consider the chemical reaction that occurs in the cylinders of an automobile engine. How are the droplets related to the reaction, and how does droplet size affect the rate of the reaction?

    Solve. The reaction occurring in the cylinder is the combustion of gasoline. Gasoline is injected into the cylinders in the form of a spray, as shown in the photos. This is a heterogeneous reaction, because gasoline is a liquid and oxygen (from air) is a gas. The rate of a heterogeneous reaction depends on the surface area of the liquid or solid reactant, in this case, the surface area of the droplets in the spray. The smaller the droplets, the greater the surface area exposed to oxygen, the faster the combustion reaction.

    In the case of a clogged injector, larger droplets lead to slower combustion. Uneven combustion in the various cylinders can cause the engine to run roughly and decrease fuel economy.

    14.2 Analyze/Plan. Given the plot of [X] vs time, answer questions about reaction speed and rate. Consider the definitions of average reaction rate and instantaneous rate. Solve.

    1. True. X is a product, because its concentration increases with time.
    2. False. The speed of a reaction is its rate, or how quickly the concentration of a reactant or product changes over time. This graph shows how [X] increases over time. The rate at any particular time, the instantaneous rate, is the slope of the tangent to the curve at that time. Visualizing the tangents at points 0, 1, 2, and 3, we see that the slopes of these lines are decreasing with time. That is, the rate of reaction is decreasing; the reaction is slowing down as time progresses.
    3. True. The average rate of reaction between any two points on the graph is the slope of the line connecting the two points. Points 1 and 2 are earlier in the reaction when more reactants are available, so the average rate of formation of products is greater. As reactants are used up, the rate of X production decreases, and the average rate between points 2 and 3 is smaller.
    4. False. The graph shows the build-up of [X] as the reaction progresses. [X] will not decrease once it reaches its maximum. (In the case of a chemical equilibrium, [X] will increase until reaching its equilibrium concentration, but it will not decrease after equilibrium is established.)

    14.3

    1. Chemical equation (iv), B → 2A, is consistent with the data. The concentration of A increases with time, and the concentration of B decreases with time, so B must be a reactant and A must be a product. The ending concentration of A is approximately twice as large as the starting concentration of B, so mole ratio of A:B is 2:1. The reaction is B → 2A.
    2. Rate = – Δ[B]/Δt = ½Δ[A]/Δt

    14.4 Analyze/Plan. Given a plot of increase in [M] over time, answer questions about reaction rate and progress. Consider the definition of reaction rate. Solve.

    1. Yes. The plot of [M] versus time from t = 0 to t = 15 is a straight line, so [M] increases at a constant rate and the reaction occurs at a constant rate. The rate is zero after t = 15 min.
    2. Yes. [M] does not change after 15 min. This means that no more M is being produced and the reaction is no longer occurring.
    3. Statement (ii) is correct. After t = 15 min, 0.00 mol K and 0.20 mol L remain in solution. When an additional 0.20 mol K is added at 30 min, the second half of the reaction occurs. The plot of [M] versus t looks like the first half of the reaction, from t = 0 to t = 15 min.

    14.5 Analyze. Given three mixtures and the order of reaction in each reactant, determine which mixture will have the fastest initial rate.

    Plan. Write the rate law. Count the number of reactant molecules in each container. The three containers have equal volumes and total numbers of molecules. Use the molecule count as a measure of concentration of NO and O 2 . Calculate the initial rate for each container and compare.

    Solve. Rate = k[NO] 2 [O 2 ]; rate is proportional to [NO] 2 [O 2 ]

    Container [NO] [O 2 ] [NO] 2 [O 2 ] ∞ rate
    (1) 5 4 100
    (2) 7 2 98
    (3) 3 6 54

    The relative rates in containers (1) and (2) are very similar, with (1) having the slightly faster initial rate.

    14.6 Plan. For a first-order reaction, a plot of ln[A] versus time is linear, as shown in the diagram. The slope is –k, and the intercept is [A] 0 . According to the Arrhenius equation, 14.21, k increases with increasing temperature. Solve.

    1. Graphs 1 and 2 have the same slope and thus the same rate constant, k. These experiments are done at the same temperature. The y-intercepts of the two graphs are different; the experiments had different initial concentrations of A.
    2. Graphs 2 and 3 have the same y-intercept and thus the same starting concentration of A. The slopes of the two graphs are different, so their rate constants are different and they occur at different temperatures. Graph 3, with the smaller slope and k value will occur at the lower temperature.

    14.7 Analyze. Given concentrations of reactants and products at two times, as represented in the diagram, find t 1/2 for this first-order reaction.

    Plan. For a first-order reaction, t 1/2 = 0.693/k; t 1/2 depends only on k. Use Equation 14.12 to solve for k. Solve.

    1. Because reactants and products are in the same container, use number of particles as a measure of concentration. The red dots are reactant A, and the blue are product B. [A] 0 = 16, [A] 30 = 4, t = 30 min.

      ln [A] t [A] 0 = –kt. ln(4/16) = –k(30 min); 1 . 3 8 6 3 3 0 min = k;

      k = 0.046210 = 0.0462 min –1

      t 1/2 = 0.693/k = 0.693/0.046210 = 15 min

      By examination, [A] 0 = 16, [A] 30 = 4. After 1 half-life, [A] = 8; after a second half-life, [A] = 4. Thirty minutes represents exactly 2 half-lives, so t 1/2 = 15 min. [This is more straightforward than the calculation, but a less general method.]

    2. After 4 half-lives, [A] t = [A] 0 × 1/2 × 1/2 × 1/2 × 1/2 = [A] 0 /16. In general, after n half-lives, [A] = [A] 0 /2 n .

    14.8

    1. Plot (v) is zero order.
    2. Plot (i) is first order.
    3. Plot (iii) is second order.

    14.9 Analyze/Plan. The reaction profile has a single high point (peak), so the reaction occurs in a single step. This step is necessarily the rate-determining step. Solve.

    1. Total potential energy of the reactants.
    2. E a , activation energy of the reaction. This is the difference in energy between the potential energy of the activated complex (transition state) and the potential energy of the reactants.
    3. ΔE, net energy change for the reaction. This is the difference in energy between the products and reactants. (Under appropriate conditions, this could also be ΔH.) For this reaction, the energy of products is lower than the energy of reactants, and the reaction releases energy to the surroundings.
    4. Total potential energy of the products.

    14.10 Analyze/Plan. On a plot of ln k versus 1/T, the slope is –E a /R and the y-intercept is ln A, where E a is activation energy and A is the frequency factor. Solve.

    1. Blue. The magnitude of the slope of the blue line is greater than that of the red line.
    2. Red. The y-intercept of the red line is greater than that of the blue line.

    14.11

    1. False. The red pathway is slower, because it has the greater activation energy, E a .
    2. True. For both reactions, the difference in potential energy between the products and the activated complex is greater than the difference between the reactants and the activated complex.
    3. True. ΔE is the difference between the energy of the reactants and the energy of the products.

    14.12

    1. NO 2 + F 2 → NO 2 F + F

      NO 2 + F → NO 2 F

    2. 2NO 2 + F 2 → 2NO 2 F
    3. F is the intermediate, because it is produced and then consumed during the reaction.
    4. Rate = k[NO 2 ][F 2 ]

    14.13  This is the profile of a two-step mechanism, A → B and B → C. There is one intermediate, B. Because there are two energy maxima, there are two transition states. The B → C step is faster, because its activation energy is smaller. For the overall reaction A → C, ΔE is negative, because the potential energy of the products is lower than the potential energy of the reactants.

    14.14  The most likely transition state shows the relative geometry of both reactants and products. It is reasonable to assume that multiple bonds, with greater total bond energy, remain intact at the expense of single bonds. In the black-and-white diagram below, open circles represent the red balls and closed circles represent the blue.

    image

    14.15

    1. A 2 + AB + AC → BA 2 + A + AC

      BA 2 + A + AC → A 2 + BA 2 + C

      net: AB + AC → BA 2 + C

    2. A is the intermediate; it is produced and consumed.
    3. A 2 is the catalyst; it is consumed and reproduced.

    14.16

    image

    The reaction is exothermic because the energy of products is lower than the energy of reactants. The two intermediates are formed at different rates because E a (1) ≠ E a (2). To have two intermediates, the mechanism must have at least three steps.


    Reaction Rates (Sections 14.1 and 14.2)

    14.17

    1. Reaction rate is the change in the amount of products or reactants in a given amount of time; it is the speed of a chemical reaction.
    2. Rates depend on concentration of reactants, physical state (or surface area) of reactants, temperature, and reaction activation energy/presence of catalyst.
    3. No, the rate of disappearance of reactants is not necessarily the same as the rate of appearance of products. The stoichiometry of the reaction (mole ratios of reactants and products) must be known to relate rate of disappearance of reactants to rate of appearance of products.

    14.18

    1. M /s
    2. As temperature increases, reaction rate increases.
    3. As a reaction proceeds, the instantaneous reaction rate decreases.

    14.19 Analyze/Plan. Given mol A at a series of times in minutes, calculate mol B produced, molarity of A at each time, change in M of A at each 10 min interval, and Δ M A/s. For this reaction, mol B produced equals mol A consumed. M of A or [A] = mol A/0.100 L. The average rate of disappearance of A for each 10 minute interval is

    Δ [ A ] s = [ A ] 1 [ A ] 0 10 min × 1 min 60 s

    Solve.

    Time (min) Mol A (a) Mol B [A] Δ[A] (b) Rate –(Δ[A]/s)
    0 0.065 0.000 0.65
    10 0.051 0.014 0.51 –0.14 2.3×   10 –4
    20 0.042 0.023 0.42 –0.09 2  ×   10 –4
    30 0.036 0.029 0.36 –0.06 1  ×   10 –4
    40 0.031 0.034 0.31 –0.05 0.8×   10 –4
    1. Δ M B Δ t = ( 0.029 0.014 ) mol/ 0.100 L ( 30 10 ) min × 1 min 60 s = 1.25 × 10 4 = 1.3 × 10 4 M / s

    14.20

    Time (s) Mol A (a) Mol B Δ Mol A (b) Rate –(Δ mol A/s)
    0 0.100 0.000
    40 0.067 0.033 –0.033 8.3 × 10 –4
    80 0.045 0.055 –0.022 5.5 × 10 –4
    120 0.030 0.070 –0.015 3.8 × 10 –4
    160 0.020 0.080 –0.010 2.5 × 10 –4
    1. (ii) The volume of the container must be known to report the rate in units of concentration (mol/L) per time.

    14.21

    1. Analyze/Plan . Follow the logic in Sample Exercises 14.1 and 14.2. Solve.
      Time (s) Time Interval (s) Concentration (M) Δ M Rate ( M /s)
      0 0.0165
      2000 2000 0.0110 –0.0055 28 × 10 –7
      5000 3000 0.00591 –0.0051 17 × 10 –7
      8000 3000 0.00314 –0.00277 9.23 × 10 –7
      12,000 4000 0.00137 –0.00177 4.43 × 10 –7
      15,000 3000 0.00074 –0.00063 2.1 × 10 –7
    2. Δ M Δ t = ( 0 . 0 0 0 7 4 0 . 0 1 6 5 ) M ( 1 5 , 0 0 0 0 ) s = 1 . 0 5 0 7 × 1 0 6 = 1 . 0 5 × 1 0 6 M / s
    3. Δ M Δ t = ( 0.00137 0.0110 ) M ( 12 , 000 2000 ) s = 9.63 × 10 7 M / s

      Δ M Δ t = ( 0.00074 0.00314 ) M ( 15 , 000 8000 ) s = 3.43 × 10 7 M / s

      The average rate between t = 2000 and t = 12,000 s is greater. In general, the rate of a reaction decreases over time.

    4. From the slopes of the lines in the figure at right, the rates are: at 5000 s, 12 × 10 –7 M /s; at 8000 s, 5.8 × 10 –7 M /s.
      image

    14.22

    1. Time (s) Time Interval (min) Concentration ( M ) Δ M Rate ( M /s)
      0.0 1.85
      54.0 54.0 1.58 –0.27 8.3 × 10 –5
      107.0 53.0 1.36 –0.22 6.9 × 10 –5
      215.0 108 1.02 –0.34 5.2 × 10 –5
      430.0 215 0.580 –0.44 3.4 × 10 –5
    2. Δ M Δ t = ( 1.85 0.580 ) M ( 430 0 ) min × 1 min 60 s = 4.9225 × 10 5 = 4.92 × 10 5 M / s

    1. Δ M Δ t = ( 1.02 1.58 ) M ( 215 54 ) min × 1 min 60 s = 5.8 × 10 5 M / s

      Δ M Δ t = ( 0.580 1.36 ) M ( 430 107 ) min × 1 min 60 s = 4.0 × 10 5 M / s

      The average rate between t = 54.0 and t = 215.0 min is greater. In general, the rate of a reaction decreases over time.

    1. From the slopes of the lines in the figure at the right, the rates are: at 75.0 min, 4.2 × 10 –3 M /min, or 7.0 × 10 –5 M /s; at 250 min, 2.1 × 10 –3 M /min or 3.5 × 10 –5 M /s.
      image

    14.23 Analyze/Plan. Follow the logic in Sample Exercise 14.3. Solve.

    1. –Δ[H 2 O 2 ]/Δt = Δ[H 2 ]/Δt = Δ[O 2 ]/Δt
    2. –Δ[N 2 O]/2Δt = Δ[N 2 ]/2Δt = Δ[O 2 ]/Δt

      –Δ[N 2 O]/Δt = Δ[N 2 ]/Δt = 2Δ[O 2 ]/Δt

    3. –Δ[N 2 ]/Δt = Δ[NH 3 ]/2Δt; –Δ[H 2 ]/3Δt = Δ[NH 3 ]/2Δt

      –2Δ[N 2 ]/Δt = Δ[NH 3 ]/Δt; –Δ[H 2 ]/Δt = 3Δ[NH 3 ]/2Δt

    4. –Δ[ C 2 H 5 NH 2 ]/Δt = Δ[ C 2 H 4 ]/Δt = Δ[NH 3 ]/Δt

    14.24

    1. Rate = –Δ[H 2 O]/2Δt = Δ[H 2 ]/2Δt = Δ[O 2 ]/Δt
    2. Rate = –Δ[SO 2 ]/2Δt = –Δ[O 2 ]/Δt = Δ[SO 3 ]/2Δt
    3. Rate = –Δ[NO]/2Δt = –Δ[H 2 ]/2Δt = Δ[N 2 ]/Δt = Δ[H 2 O]/2Δt
    4. Rate = –Δ[N 2 ]/Δt = –Δ[H 2 ]/2Δt = Δ[N 2 H 4 ]/Δt

    14.25 Analyze/Plan. Use Equation 14.4 to relate the rate of disappearance of reactants to the rate of appearance of products. Use this relationship to calculate desired quantities. Solve.

    1. Δ[H 2 O]/2Δt = –Δ[H 2 ]/2Δt = –Δ[O 2 ]/Δt

      H 2 is burning, –Δ[H 2 ]/Δt = 0.48 mol/s

      O 2 is consumed, –Δ[O 2 ]/Δt = –Δ[H 2 ]/2Δt = 0.48 mol/s/2 = 0.24 mol/s

      H 2 O is produced, +Δ[H 2 O]/Δt = –Δ[H 2 ]/Δt = 0.48 mol/s


    1. The change in total pressure is the sum of the changes of each partial pressure. NO and Cl 2 are disappearing and NOCl is appearing.

      –ΔP NO /Δt = –56 torr/min

      Δ P Cl 2 / Δ t = Δ P NO / 2 Δ t = 28 torr/min

      +ΔP NOCl /Δt = –ΔP NO /Δt = +56 torr/min

      ΔP T /Δt = –56 torr/min – 28 torr/min + 56 torr/min = –28 torr/min

    14.26

    1. –Δ[C 2 H 4 ]/Δt = Δ[CO 2 ]/2Δt = Δ[H 2 O]/2Δt

      –2Δ[C 2 H 4 ]/Δt = Δ[CO 2 ]/Δt = Δ[H 2 O]/Δt

      C 2 H 4 is burning, –Δ[C 2 H 4 ]/Δt = 0.036 M /s

      CO 2 and H 2 O are produced, at twice the rate that C 2 H 4 is consumed.

      Δ[CO 2 ]/Δt = Δ[H 2 O]/Δt = 2(0.036) M /s = 0.072 M /s

    2. In this reaction, pressure is a measure of concentration.

      –Δ[N 2 H 4 ]/Δt = –Δ[H 2 ]/Δt = Δ[NH 3 ]/2Δt

      N 2 H 4 is consumed, –Δ[N 2 H 4 ]/Δt = 74 torr/h

      H 2 is consumed, –Δ[H 2 ]/Δt = 74 torr/h

      NH 3 is produced at twice the rate that N 2 H 4 and H 2 are consumed,

      Δ[NH 3 ]/Δt = –2Δ[N 2 H 4 ]/Δt = 2(74) torr/h = 148 torr/h

      ΔP T /Δt = (+148 torr/h – 74 torr/h – 74 torr/h) = 0 torr/h

    Rate Laws (Section 14.3)

    14.27 Analyze/Plan. Follow the logic in Sample Exercise 14.5. Solve.

    1. If [A] is doubled, there will be no change in the rate or the rate constant. The overall rate is unchanged because [A] does not appear in the rate law; the rate constant changes only with a change in temperature.
    2. The reaction is zero order in A, second order in B, and second order overall.
    3. Units of k = M / s M 2 = M 1 s 1

    14.28

    1. Rate = k[A][C] 2
    2. Rate is proportional to [A], rate doubles
    3. Rate is not affected by [B], no change
    4. Rate changes as [C] 2 , rate increases by a factor of 3 2 or 9
    5. Rate increases by a factor of (3)(3) 2 = 27
    6. Rate decreases by a factor of (1/2)(1/2) 2 = 1/8

    14.29 Analyze/Plan. Follow the logic in Sample Exercise 14.5. Solve.

    1. Rate = k[N 2 O 5 ] = 4.82 × 10 –3 s –1 [N 2 O 5 ]
    2. Rate = 4.82 × 10 –3 s –1 (0.0240 M ) = 1.16 × 10 –4 M /s
    3. Rate = 4.82 × 10 –3 s –1 (0.0480 M ) = 2.31 × 10 –4 M /s

      When the concentration of N 2 O 5 doubles, the rate of the reaction doubles.

    4. Rate = 4.82 × 10 –3 s –1 (0.0120 M ) = 5.78 × 10 –5 M /s

      When the concentration of N 2 O 5 is halved, the rate of the reaction is halved.

    14.30

    1. Rate = k[H 2 ][NO] 2
    2. Rate = (6.0 × 10 4 M –2 s –1 ) (0.035 M ) 2 (0.015 M ) = 1.1 M /s
    3. Rate = (6.0 × 10 4 M –2 s –1 ) (0.10 M ) 2 (0.010 M ) = 6.0 M /s
    4. Rate = (6.0 × 10 4 M –2 s –1 ) (0.010 M ) 2 (0.030 M ) = 0.18 M /s

    14.31 Analyze/Plan. Write the rate law and rearrange to solve for k. Use the given data to calculate k, including units. Solve.

    (a, b) Rate = k[CH 3 Br][OH ]; k = rate [ CH 3 Br ] [ OH ]
    at 2 9 8 K , k = 0 . 0 4 3 2 M / s ( 5 . 0 × 1 0 3 M ) ( 0 . 0 5 0 M ) = 1 . 7 × 1 0 2 M 1 s 1
    1. Because the rate law is first order in [OH ], if [OH ] is tripled, the rate triples.
    1. If [OH ] and [CH 3 Br] both triple, the rate increases by a factor of (3)(3) = 9.

    14.32

    (a, b) Rate = k [ C 2 H 5 Br ] [ OH ] ; k = rate [ C 2 H 5 Br ] [ OH ]
    at 298 K , k = 1.7 × 10 7 M / s [ 0.0477 M ] [ 0.100 M ] = 3.6 × 10 5 M 1 s 1
    1. Adding an equal volume of ethyl alcohol reduces both [C 2 H 5 Br] and [OH ] by a factor of two. New rate = (1/2)(1/2) = 1/4 of old rate.

    14.33 Analyze/Plan. Follow the logic in Sample Exercise 14.6. Solve.

    1. From the data given, when [OCl ] doubles, rate doubles. When [I ] doubles, rate doubles. The reaction is first order in both [OCl ] and [I ]. Rate = k[OCl ][I ].
    2. Using the first set of data:

      k = rate [ OCl ] [ I ] = 1.36 × 10 4 M / s ( 1.5 × 10 3 M ) ( 1.5 × 10 3 M ) = 60.444 = 60 M 1 s 1

    3. Rate = 60.444 M -s ( 2.0 × 10 3 M ) ( 5.0 × 10 4 M ) = 6.0444 × 10 5 = 6.0 × 10 5 M /s

    14.34

    1. From the data given, when [ClO 2 ] increases by a factor of 3 (experiment 2 to experiment 1), the rate increases by a factor of 9. When [OH ] increases by a factor of 3 (experiment 2 to experiment 3), the rate increases by a factor of 3. The reaction is second order in [ClO 2 ] and first order in [OH ]. Rate = k[ClO 2 ] 2 [OH ].

    1. Using data from experiment 2:

      k = rate [ ClO 2 ] 2 [ OH ] = 0.00276 M / s ( 0.020 M ) 2 ( 0.030 M ) = 2.3 × 10 2 M 2 s 1

    1. Rate = 2.3 × 10 2 M –2 s –1 (0.100 M ) 2 (0.050 M ) = 0.115 = 0.12 M /s

    14.35 Analyze/Plan. Follow the logic in Sample Exercise 14.6 to deduce the rate law. Rearrange the rate law to solve for k and deduce units. Calculate a k value for each set of concentrations and then average the three values. Solve.

    1. Doubling [NH 3 ] while holding [BF 3 ] constant doubles the rate (experiments 1 and 2). Doubling [BF 3 ] while holding [NH 3 ] constant doubles the rate (experiments 4 and 5).

      Thus, the reaction is first order in both BF 3 and NH 3 ; rate = k[BF 3 ][NH 3 ].

    2. The reaction is second order overall.
    3. From experiment 1: k = 0.2130 M / s ( 0.250 M ) ( 0.250 M ) = 3.41 M 1 s 1

      (Any of the five sets of initial concentrations and rates could be used to calculate the rate constant k. The average of these 5 values is k avg = 3.408 = 3.41 M –1 s –1 .)

    4. Rate = 3.408 M –1 s –1 (0.100 M )(0.500 M ) = 0.1704 = 0.170 M /s.

    14.36 Analyze/Plan. Follow the logic in Sample Exercise 14.6 to deduce the rate law. Rearrange the rate law to solve for k and deduce units. Calculate a k value for each set of concentrations and then average the three values. Solve.

    1. Doubling [NO] while holding [O 2 ] constant increases the rate by a factor of 4 (experiments 1 and 2). Doubling [O 2 ] while holding [NO] constant doubles the rate (experiments 2 and 3). The reaction is second order in [NO] and first order in [O 2 ]. Rate = k[NO] 2 [O 2 ].
    (b, c) From experiment 1: k 1 = 1.41 × 10 2 M / s ( 0.0126 M ) 2 ( 0.0125 M ) = 7105 = 7.11 × 10 3 M 2 s 1
    k 2 = 0.113/(0.0252) 2 (0.0250) = 7118 = 7.12 × 10 3 M –2 s –1
    k 3 = 5.64 × 10 –2 /(0.0252) 2 (0.125) = 7105 = 7.11 × 10 3 M –2 s –1
    k avg = (7105 + 7118 + 7105)/3 = 7109 = 7.11 × 10 3 M –2 s –1
    1. Rate = 7.109 × 10 3 M –2 s –1 (0.0750 M ) 2 (0.0100 M ) = 0.3999 = 0.400 M /s.
    1. The data are given in terms of the disappearance of NO. Use Equation 14.4 to relate the disappearance of NO to the disappearance of O 2 .

      –Δ[NO]/2Δt = – [O 2 ]/Δt

      For the concentrations given in part (d), Δ[NO]/Δt = 0.400 M /s.

      Δ [O 2 ]/Δt = Δ[NO]/2Δt = 0.400 M /s/2 = 0.200 M /s

    14.37 Analyze/Plan. Follow the logic in Sample Exercise 4.6 to deduce the rate law. Rearrange the rate law to solve for k and deduce units. Calculate a k value for each set of concentrations and then average the three values. Solve.


    1. Increasing [NO] by a factor of 2.5 while holding [Br 2 ] constant (experiments 1 and 2) increases the rate by a factor 6.25 or (2.5) 2 . Increasing [Br 2 ] by a factor of 2.5 while holding [NO] constant increases the rate by a factor of 2.5. The rate law for the appearance of NOBr is: Rate = Δ[NOBr]/Δt = k[NO] 2 [Br 2 ].
    2. From experiment 1: k 1 = 24 M / s ( 0.10 M ) 2 ( 0.20 M ) = 1.20 × 10 4 = 1.2 × 10 4 M 2 s 1

      k 2 = 150/(0.25) 2 (0.20) = 1.20 × 10 4 = 1.2 × 10 4 M –2 s –1

      k 3 = 60/(0.10) 2 (0.50) = 1.20 × 10 4 = 1.2 × 10 4 M –2 s –1

      k 4 = 735/(0.35) 2 (0.50) = 1.2 × 10 4 = 1.2 × 10 4 M –2 s –1

      k avg = (1.2 × 10 4 + 1.2 × 10 4 + 1.2 × 10 4 + 1.2 × 10 4 )/4 = 1.2 × 10 4 M –2 s –1

    3. Use the reaction stoichiometry and Equation 14.4 to relate the designated rates. Δ[NOBr]/2Δt = –Δ[Br 2 ]/Δt; the rate of disappearance of Br 2 is half the rate of appearance of NOBr.
    4. Note that the data are given in terms of appearance of NOBr.

      Δ [ Br 2 ] Δ t = k [ NO ] 2 [ Br 2 ] 2 = 1.2 × 10 4 2 M 2 s × ( 0.075 M ) 2 × ( 0.250 M ) = 8.4 M /s

    14.38

    1. Increasing [S 2 O 8 2– ] by a factor of 1.5 while holding [I ] constant increases the rate by a factor of 1.5 (Experiments 1 and 2). Doubling [S 2 O 8 2– ] and increasing [I ] by a factor of 1.5 triples the rate (2 × 1.5 = 3, experiments 1 and 3). Thus the reaction is first order in both [S 2 O 8 2– ] and [I ]; rate = k [S 2 O 8 2– ] [I ].
    2. k = rate/[S 2 O 8 2– ] [I ]

      k 1 = 2.6 × 10 –6 M /s/(0.018 M )(0.036 M ) = 4.01 × 10 –3 = 4.0 × 10 –3 M –1 s –1

      k 2 = 3.9 × 10 –6 /(0.027)(0.036) = 4.01 × 10 –3 = 4.01 × 10 –3 = 4.0 × 10 –3 M –1 s –1

      k 3 = 7.8 × 10 –6 /(0.036)(0.054) = 4.01 × 10 –3 = 4.01 × 10 –3 = 4.0 × 10 –3 M –1 s –1

      k 4 = 1.4 × 10 –5 /(0.050)(0.072) = 3.89 × 10 –3 = 3.9 × 10 –3 M –1 s –1

      k avg = 3.98 × 10 –3 = 4.0 × 10 –3 M –1 s –1

    3. –Δ[S 2 O 8 2– ]/Δt = –Δ[I ]/3Δt; the rate of disappearance of S 2 O 8 2– is one-third the rate of disappearance of I .
    4. Note that the data are given in terms of disappearance of S 2 O 8 2– .

      Δ [ I ] Δ t = 3 Δ [ S 2 O 8 2 ] Δ t = 3 ( 3.98 × 10 3 M 1 s 1 ) ( 0.025 M ) ( 0.050 M ) = 1.5 × 10 5 M / s

    Change of Concentration with Time (Section 14.4)

    14.39

    1. A graph of ln[A] versus time yields a straight line for a first-order reaction.
    2. On graph of ln[A] versus time, the rate constant, k, is the (–slope) of the straight line.

    14.40

    1. A graph of 1/[A] versus time yields a straight line for a second-order reaction.
    2. On a graph of 1/[A] versus time, the slope of the straight line is the rate constant, k.
    3. The half-life of a second-order reaction remains the same as the reaction proceeds. According to Equation 14.19, the half-life of a second-order reaction does depend on [A] 0 , t 1/2 = 1/k[A] 0 , but this quantity is constant over the course of the reaction.

    14.41 Analyze/Plan. The half-life of a first-order reaction depends only on the rate constant, t 1/2 = 0.693/k. Use this relationship to calculate k for a given t 1/2 , and, at a different temperature, t 1/2 given k. Solve.

    1. t 1/2 = 2.3 × 10 5 s; t 1/2 = 0.693/k, k = 0.693/t 1/2

      k = 0.693/2.3 × 10 5 s = 3.0 × 10 –6 s –1

    2. k = 2.2 × 10 –5 s –1 . t 1/2 = 0.693/2.2 × 10 –5 s –1 = 3.15 × 10 4 = 3.2 × 10 4 s

    14.42

    1. For a first-order reaction, t 1/2 = 0.693/k.

      t 1/2 = 0.693/0.271 s –1 = 2.5572 = 2.56 s

    2. For a first-order reaction, ln[A] t – ln[A] 0 = –kt. ln[A] t = –kt + ln[A] 0

      [A] 0 = 0.050 M I 2 , t = 5.12 s, k = 0.271 s –1

      ln[I 2 ] = –0.271 s –1 (5.12 s) + ln(0.050)

      ln[I 2 ] = –4.3833, [I 2 ] = 0.0125 M

      Check. 5.12 s is 2 half-lives. [I 2 ] should be reduced by a factor of 4, and it is.

    14.43 Analyze/Plan. Follow the logic in Sample Exercise 14.7. In this reaction, pressure is a measure of concentration. In (a) we are given k, [A] 0 , t and asked to find [A] t , using Equation 14.13, the integrated form of the first-order rate law. In (b), [A t ] = 0.1[A 0 ], find t. Solve.

    1. ln P t = –kt + ln P 0 ; P 0 = 450 torr; t = 60 s

      ln P 60 = –4.5 × 10 –2 s –1 (60) + ln (450) = –2.70 + 6.109 = 3.409

      P 60 = 30.24 = 30 torr

    2. P t = 0.10 P 0 ; ln (P t /P 0 ) = –kt

      ln (0.10 P 0 /P 0 ) = –kt, ln (0.10) = –kt; –ln (0.10)/k = t

      t = –(–2.303)/4.5 × 10 –2 s –1 = 51.2 = 51 s

    Check. From part (a), the pressure at 60 s is 30 torr, P t ~ 0.07 P 0 . In part (b) we calculate the time where P t = 0.10 P 0 to be 51 s. This time should be smaller than 60 s, and it is. Data and results in the two parts are consistent.

    14.44

    1. Using Equation 14.13 for a first-order reaction: ln [A] t = –kt + ln [A] 0

      5.0 min = 300 s; [N 2 O 5 ] 0 = (0.0250 mol/2.0 L) = 0.0125 = 0.013 M

      ln [N 2 O 5 ] 300 = –(6.82 × 10 –3 s –1 )(300 s) + ln (0.0125)

      ln [N 2 O 5 ] 300 = –2.0460 + (–4.3820) = –6.4280 = –6.43

      [N 2 O 5 ] 300 = 1.616 × 10 –3 = 1.6 × 10 –3 M

      mol N 2 O 5 = 1.616 × 10 –3 M × 2.0 L = 3.2 × 10 –3 mol

    2. [N 2 O 5 ] t = 0.010 mol/2.0 L = 0.0050 M ; [N 2 O 5 ] 0 = 0.0125 M

      ln (0.0050) = –(6.82 × 10 –3 s –1 ) (t) + ln (0.0125)

      t = [ ln ( 0.0050 ) ln ( 0.0125 ) ] ( 6.82 × 10 3 s 1 ) = 134.35 = 1.3 × 10 2 s × 1 min 60 s = 2.24 = 2.2 min

    3. t 1/2 = 0.693/k = 0.693/6.82 × 10 –3 s –1 = 101.6 = 102 s or 1.69 min

    14.45 Analyze/Plan . Given reaction order, various values for t and P t , find the rate constant for the reaction at this temperature. For a first-order reaction, a graph of ln P versus t is linear with as slope of –k. Solve.

    Time (s) P S O 2 C l 2 ln P S O 2 C l 2
    0 1.000 0
    2500 0.947 –0.0545
    5000 0.895 –0.111
    7500 0.848 –0.165
    10,000 0.803 –0.219
    image

    Graph ln P SO 2 Cl 2 versus time. (Pressure is a satisfactory unit for a gas, because the concentration in moles/liter is proportional to P.) The graph is linear with slope –2.19 × 10 –5 s –1 as shown on the figure. The rate constant k = –slope = 2.19 × 10 –5 s –1 .

    14.46

    Time (s) P C H 3 N C l n P C H 3 N C
    0 502 6.219
    2,000 335 5.814
    5,000 180 5.193
    8,000 95.5 4.559
    12,000 41.7 3.731
    15,000 22.4 3.109
    image

    A graph of ln P vs t is linear with a slope of –2.08 × 10 –4 s –1 . The rate constant, k, = –slope = 2.08 × 10 –4 s –1 . Half-life = t 1/2 = 0.693/k = 3.33 × 10 3 s.

    14.47 Analyze/Plan. Given: mol A, t. Change mol to M at various times. Make both first- and second-order plots to see which is linear. Solve.

    1. Time (min) mol A [A] ( M ) ln[A] 1/mol A
      0 0.065 0.65 –0.43 1.5
      10 0.051 0.51 –0.67 2.0
      20 0.042 0.42 –0.87 2.4
      30 0.036 0.36 –1.02 2.8
      40 0.031 0.31 –1.17 3.2
      image

      The plot of 1/[A] vs time is linear, so the reaction is second order in [A].


    1. For a second-order reaction, a plot of 1/[A] versus t is linear with slope k.

      k = slope = (3.2 – 2.0) M –1 / 30 min = 0.040 M –1 min –1

      (The best fit to the line yields slope = 0.042 M –1 min –1 .)

    1. t 1/2 = 1/k[A] 0 = 1/(0.040 M –1 min –1 )(0.65 M ) = 38.46 = 38 min

      (Using the “best-fit” slope, t 1/2 = 37 min.)

    14.48

    1. Make both first- and second-order plots to see which is linear. Moles is a satisfactory concentration unit, because volume is constant.
      Time (s) mol A ln (mol A) 1/mol A
      0 0.1000 –2.303 10.00
      40 0.067 –2.70 14.9
      80 0.045 –3.10 22.2
      120 0.030 –3.51 33.3
      160 0.020 –3.91 50.0
      image

      The plot of ln (mol A) vs time is linear, so the reaction is first order in A.

    2. k = –slope = – [–3.91 – (–2.70)]/120 = 0.010083 = 0.0101 s –1

      (The best fit to this line yields the same value for the slope, 0.01006 = 0.0101 s –1 )

    3. t 1/2 = 0.693/k = 0.693/0.010083 s –1 = 68.7 s

    14.49 Analyze/Plan. Make both first- and second-order plots to see which is linear. Solve.

    1. Time (s) [NO 2 ]( M ) ln [NO 2 ] 1/[NO 2 ]
      0.0 0.100 –2.303 10.0
      5.0 0.017 –4.08 59
      10.0 0.0090 –4.71 110
      15.0 0.0062 –5.08 160
      20.0 0.0047 –5.36 210
      image

      The plot of 1/[NO 2 ] versus time is linear, so the reaction is second order in NO 2 .


    1. The slope of the line is (210 – 59) M –1 / 15.0 s = 10.07 = 10 M –1 s –1 = k. (The slope of the best-fit line is 10.02 = 10 M –1 s –1 .)
    1. From the results above, the rate law is: rate = k[NO 2 ] 2 = 10 M –1 s –1 [NO 2 ] 2 Using the rate law, calculate the rate at each of the given initial concentrations.

      Rate @ 0.200 M = 10 M –1 s –1 [NO 2 ] 2 = 10 M –1 s –1 [0.200 M ] 2 = 0.400 M /s

      Rate @ 0.100 M = 10 M –1 s –1 [NO 2 ] 2 = 10 M –1 s –1 [0.100 M ] 2 = 0.100 M /s

      Rate @ 0.050 M = 10 M –1 s –1 [NO 2 ] 2 = 10 M –1 s –1 [0.050 M ] 2 = 0.025 M /s

    14.50

    1. Make both first- and second-order plots to see which is linear.
      Time (min) [C 12 H 22 O 11 ]( M ) ln [C 12 H 22 O 11 ] 1/[C 12 H 22 O 11 ]
      0 0.316 –1.152 3.16
      39 0.274 –1.295 3.65
      80 0.238 –1.435 4.20
      140 0.190 –1.661 5.26
      210 0.146 –1.924 6.85
      image

      The plot of ln [C 12 H 22 O 11 ] is linear, so the reaction is first order in C 12 H 22 O 11 .

    2. k = –slope = –[–1.924 – (–1.295)]/171 min = 3.68 × 10 –3 min –1

      (The slope of the best-fit line is –3.67 × 10 –3 min –1 .)

    3. For a reaction zero order in sucrose, the rate does not change as [sucrose] changes. A plot of [sucrose] versus time is linear with negative slope, until all reactant is consumed. [sucrose] t = –kt + [sucrose] 0

      @39 min, [sucrose] = –3.68 × 10 –3 min –1 (39 min) + 0.316 M = 0.17 M

      @80 min, [sucrose] = –3.68 × 10 –3 min –1 (80 min) + 0.316 M = 0.022 M

      @140 min, [sucrose] = –3.68 × 10 –3 min –1 (140 min) + 0.316 M = 0 M

      @210 min, [sucrose] = 0 M . All sucrose is consumed at (0.316/3.68 × 10 –3 min –1 ) = 85.9 min.

    Temperature and Rate (Section 14.5)

    14.51

    1. The energy of the collision and the orientation of the molecules when they collide determine whether a reaction will occur.
    2. Assuming other conditions remain the same, the rate and therefore the rate constant usually increase with an increase in reaction temperature.

    1. The fraction of molecules with energy greater than the activation energy changes most dramatically with temperature. Frequency of collision and the orientation factor are lumped into the frequency factor, A, which is considered to be constant with temperature.

    14.52

    1. The orientation factor is less important in H + Cl → HCl, because the reactants are monatomic and spherical (nondirectional); all collision orientations are equally effective.
    2. The orientation factor does not depend on temperature. It is combined with frequency of collision into the frequency factor, A, which is essentially constant with temperature.

    14.53 Analyze/Plan. Given the temperature and energy, use Equation 14.20 to calculate the fraction of Ar atoms that have at least this energy. Solve.

    f = e E a /RT E a = 10.0 kJ/mol = 1.00 × 10 4 J/mol ; T = 400 K ( 127 ° C )

    E a / RT = 1.00 × 10 4 J/mol 400 K × mol-K 8.314 J = 3.0070 = 3.01

    f = e– 3.0070 = 4.9 × 10 –2

    At 400 K, approximately 1 out of 20 molecules has this kinetic energy.

    14.54

    1. f = e E a /RT E a = 160 kJ/mol = 1.60 × 10 5 J/mol, T = 500 K

      E a / RT = 1.60 × 10 5 J/mol 500 K × mol-K 8.314 J = 38.489 = 38.5

      f = e– 38.489 = 1.924 × 10 –17 = 2 × 10 –17

    2. E a / RT = 1.60 × 10 5 J/mol 520 K × mol-K 8.314 J = 37.009 = 37.0

      f = e –37.009 = 8.45712 × 10 –17 = 8.46 × 10 –17

      f at 520 K f at 500 K = 8.46 × 10 17 1.92 × 10 17 = 4.41

      An increase of 20 K means that 4.41 times more molecules have this energy.

    14.55 Analyze/Plan. Use the definitions of activation energy (E max – E react ) and ΔE (E prod – E react ) to sketch the graph and calculate E a for the reverse reaction. Solve.

    1. image
    2. E a (reverse) = 73 kJ

    14.56 Analyze/Plan. Use the definitions of activation energy (E max – E react ) and ΔE (E prod – E react ) to sketch the graph and calculate E a for the reverse reaction. Solve.

    1. image
    2. E a (reverse) = 18 kJ

    14.57

    1. False. If you compare two reactions with similar collision factors, the one with the larger activation energy will be slower .
    2. False. A reaction that has a small rate constant will have either a small frequency factor (A), a large activation energy (E a ), or both.
    3. True.

    14.58

    1. False. If you measure the rate constant for a reaction at different temperatures, you can calculate the overall activation energy, E a , for the reaction.
    2. False. Exothermic reactions are not necessarily faster than endothermic reactions. (The rate of a reaction is not determined by the overall enthalpy change going from reactants to products.)
    3. False. If you double the temperature for a reaction, there is no change to the activation energy, E a .

    14.59  The order of slowest reaction to fastest reaction is: rate (c) < rate (a) < rate (b). Assuming all collision factors (A) to be the same, reaction rate depends only on E a ; it is independent of ΔE.

    14.60  E a for the reverse reaction is:

    1. 45 – (–25) = 70 kJ
    2. 35 – (–10) = 45 kJ
    3. 55 – 10 = 45 kJ

    Based on the magnitude of E a , the reverse of reactions (b) and (c) occur at the same rate, which is faster than the reverse of reaction (a).

    14.61 Analyze/Plan. Given k 1 , at T 1 , calculate k 2 at T 2 . Change T to Kelvins, then use the Equation 14.23 to calculate k 2 . Solve.

    T 1 = 20 °C + 273 = 293 K;   T 2 = 60 °C + 273 = 333 K; k 1 = 2.75 × 10 –2 s –1

    1. ln ( k 1 k 2 ) = E a R ( 1 333 1 293 ) = 75.5 × 10 3 J/mol 8.314 J/mol ( 4.1 × 10 4 )

      ln ( k 1 / k 2 ) = 3 . 7 2 2 9 = 3 . 7 ; k 1 / k 2 = 0 . 0 2 4 2 = 0 . 0 2 ; k 2 = 0 . 0 2 7 5 s 1 0 . 0 2 4 2 = 1 . 1 4 = 1 s 1

    2. ln ( k 1 k 2 ) = 1 2 5 × 1 0 3 J / mol 8 . 3 1 4 J / mol ( 1 3 3 3 1 2 9 3 ) = 6 . 1 6 3 8 = 6 . 2

      k 1 / k 2 = 2 . 1 0 4 × 1 0 3 = 2 × 1 0 3 ; k 2 = 0 . 0 2 7 5 s 1 2 . 1 0 4 × 1 0 3 = 1 3 . 0 7 = 1 × 1 0 s 1


    1. The method in parts (a) and (b) assumes that the collision model and thus the Arrhenious equation describe the kinetics of the reactions. That is, activation energy is constant over the temperature range under consideration. There is no assumption about temperature dependence of the frequency factor, because it drops out of the difference equation by subtraction.

    14.62  T 1 = 737 °C + 273 = 1010 K, k 1 = 0.0796 M –1 s –1 ;

    T 2 = 947 °C + 273 = 1220 K, k 2 = 0.0815 M –1 s –1

    ln ( k 1 k 2 ) = E a R ( 1 T 2 1 T 1 )

    ln ( 0.0796 0.0815 ) = E a 8.314 J/mol ( 1 1220 1 1010 )

    –0.023589 = E a ( 1.704 × 10 4 ) 8.314 J/mol

    E a = 8.314 ( 0.023589 ) J/mol ( 1.704 × 10 4 ) = 1.151 × 10 3 J/mol = 1.15 kJ/mol

    14.63 Analyze/Plan . Follow the logic in Sample Exercise 14.11. Solve.

    k ln k T(K) 1/T(× 10 3 )
    0.0521 –2.955 288 3.47
    0.101 –2.293 298 3.36
    0.184 –1.693 308 3.25
    0.332 –1.103 318 3.14
    image

    The slope, –5.64 × 10 3 , equals –E a /R. Thus, E a = 5.64 × 10 3 × 8.314 J/mol = 46.9 kJ/mol.

    14.64

    k ln k T(K) 1/T(× 10 3 )
    0.028 –3.58 600 1.67
    0.22 –1.51 650 1.54
    1.3 0.26 700 1.43
    6.0 1.79 750 1.33
    23 3.14 800 1.25
    image

    Using the relationship ln k = ln A – E a /RT, the slope, –15.94 × 10 3 = –16 × 10 3 , is –E a /R. E a = 15.94 × 10 3 × 8.314 J/mol = 1.3 × 10 2 kJ/mol. To calculate A, we will use the rate data at 700 K. From the equation given above, 0.262 = ln A – 15.94 × 10 3 /700;

    ln A = 0.262 + 22.771. A = 1.0 × 10 10 .


    Reaction Mechanisms (Section 14.6)

    14.65

    1. An elementary reaction is a process that occurs in a single event; the order is given by the coefficients in the balanced equation for the reaction.
    2. A unimolecular elementary reaction involves only one reactant molecule; the activated complex is derived from a single molecule. A bimolecular elementary reaction involves two reactant molecules in the activated complex and the overall process.
    3. A reaction mechanism is a series of elementary reactions that describe how an overall reaction occurs and explain the experimentally determined rate law.
    4. A rate-determining step is the slowest step in a reaction mechanism. It limits the overall reaction rate.

    14.66

    1. No. An intermediate is a substance that is produced and then consumed during a chemical reaction. It could be a product in the first step of a reaction mechanism, but not a reactant.
    2. On a reaction profile, an intermediate is a valley. It is a product of one step in the mechanism and a reactant in the next.
    3. Unimolecular. An elementary reaction occurs in a single step. The order of an elementary reaction is given by the coefficients in the chemical reaction. For the decomposition of Cl 2 , the coefficient of Cl 2 is one.
    4. A transition state is a high energy complex formed when one or more reactants collide and distort in a way that can lead to formation of product(s). An intermediate is the product of an early elementary reaction in a multistep reaction mechanism. A transition state occurs at an energy maximum or peak of a reaction profile as in Figure 14.19. An intermediate exists at an energy minimum or trough of a reaction profile. Every reaction, single- or multi-step, has a transition state. Only multistep reactions have intermediates.

    14.67 Analyze/Plan. Elementary reactions occur as a single step, so the molecularity is determined by the number of reactant molecules; the rate law reflects reactant stoichiometry. Solve.

    1. unimolecular, rate = k[Cl 2 ]
    2. bimolecular, rate = k[OCl ][H 2 O]
    3. bimolecular, rate = k[NO][Cl 2 ]

    14.68

    1. bimolecular, rate = k[NO] 2
    2. unimolecular, rate = k[C 3 H 6 ]
    3. unimolecular, rate = k[SO 3 ]

    14.69 Analyze/Plan. Use the definitions of the term intermediate , along with the characteristics of reaction profiles, to answer the questions. Solve.

    This is a three-step mechanism, A → B, B → C, and C → D.

    1. There are 2 intermediates, B and C.
    2. There are 3 energy maxima in the reaction profile, so there are 3 transition states.

    1. Step C → D has the lowest activation energy, so it is fastest.
    1. The energy of D is slightly greater than the energy of A, so ΔE for the overall reaction is positive.

    14.70

    1. Two elementary reactions; two energy maxima
    2. One intermediate; one energy minimum between reactants and products
    3. The second step is rate-limiting; second energy maximum and E a is larger.
    4. For the overall reaction, ΔE is negative, because the potential energy of the products is lower than the potential energy of the reactants.

    14.71 Analyze/Plan. Follow the logic in Sample Exercise 14.14. Solve.

    1. H 2 ( g ) + ICl ( g ) HI ( g ) + HCl ( g )

      HI ( g ) + ICl ( g ) I 2 ( g ) + HCl ( g ) _ H 2 ( g ) + 2 ICl ( g ) I 2 ( g ) + 2 HCl ( g )

    2. Intermediates are produced and consumed during reaction. HI is the intermediate.
    3. The slow step determines the rate law for the overall reaction. If the first step is slow, the observed rate law is: Rate = k[H 2 ][ICl].

    14.72

    1. 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g)
    2. IO (aq) is the intermediate.
    3. Rate = k[H 2 O 2 ] [I ]

    14.73

    1. Analyze . Given data on concentration of a reactant versus time, determine whether the proposed reaction mechanism is consistent with the data.

      Plan. Based on the graph, decide the order of reaction with respect to [NO]. Write the two possible rate laws, depending on which step is rate-determining. Decide if one of the rate laws, and thus the mechanism, is consistent with the rate data.

      Solve. The graph of 1/[NO] versus time is linear with positive slope, indicating that the reaction is second order in [NO]. The rate law will include [NO] 2 .

      If the first step is slow, the observed rate law is the rate law for this step:

      rate = k[NO][Cl 2 ]. Because the observed rate law is second order in [NO], the second step must be slow relative to the first step. Follow the logic in Sample Exercise 14.15 for determining the rate law of a mechanism with a fast initial step.

      From the rate-determining second step, rate = k[NOCl 2 ][NO].

      Assuming the first step is a fast equilibrium, k 1 [NO][Cl 2 ] = k – 1 [NOCl 2 ].

      Solving for [NOCl 2 ] in terms of [NO][Cl 2 ], [ NOCl 2 ] = k 1 k 1 [ NO ] [ Cl 2 ]

      Rate = k 2 k 1 k 1 [ NO ] [ Cl 2 ] [ NO ] = [ NO ] 2 [ Cl 2 ]

      This rate law is second order in [NO]. It is consistent with the observed data.

    2. The linear plot guarantees that the overall rate law will include [NO] 2 . Because the data were obtained at constant [Cl 2 ], we have no information about reaction order with respect to [Cl 2 ].

    14.74

    1. i. HBr + O 2 HOOBr

      ii. HOOBr + HBr 2 HOBr

      iii . 2 HOBr + 2 HBr 2 H 2 O + 2 Br 2 _ 4 HBr + O 2 2 H 2 O + 2 Br 2

    2. The observed rate law is: rate = k[HBr][O 2 ], the rate law for the first elementary step. The first step must be rate-determining.
    3. HOOBr and HOBr are both intermediates; HOOBr is produced in i and consumed in ii and HOBr is produced in ii and consumed in iii.
    4. Because the first step is rate-determining, it is possible that neither of the intermediates accumulates enough to be detected. This does not disprove the mechanism, but indicates that steps ii and iii are very fast, relative to step i.

    Catalysis (Section 14.7)

    14.75

    1. A catalyst is a substance that changes (usually increases) the speed of a chemical reaction without undergoing a permanent chemical change itself.
    2. A homogeneous catalyst is in the same phase as the reactants; a heterogeneous catalyst is in a different phase and is usually a solid.
    3. A catalyst has no effect on the overall enthalpy change for a reaction. A catalyst does affect activation energy, E a , which is one way that it changes reaction rate. It can also affect the frequency factor, A.

    14.76

    1. The smaller the particle size of a solid catalyst, the greater the surface area. The greater the surface area, the more active sites and the greater the increase in reaction rate.
    2. Adsorption is the binding of reactants onto the surface of the heterogeneous catalyst. It is usually the first step in the catalyzed reaction.

    14.77  KBr(s) is added to H 2 O 2 (aq) at t = 0. Assume the KBr(s) dissolves instantly. As the reaction proceeds, the Br catalyst is consumed and then regenerated.

    image

    14.78  For an acid-catalyzed reaction in solution, H + is a homogeneous catalyst. It is consumed and then regenerated during the reaction. (This assumes that H + is present in excess and that H + is not a reactant, that the reactants are neither acids nor bases.) The [H + ] is a maximum at t = 0 and when the reaction is complete.

    image

    14.79

    1. 2 [ NO 2 ( g ) + SO 2 ( g ) NO ( g ) + SO 3 ( g ) ]

      2 NO ( g ) + O 2 ( g ) 2 NO 2 ( g ) 2 SO 2 ( g ) + O 2 ( g ) 2 SO 3 ( g ) ¯

    2. NO 2 (g) is a catalyst because it is consumed and then reproduced in the reaction sequence.
    3. NO(g) is an intermediate, because it is produced and then consumed during the reaction.
    4. Because NO 2 is in the same state as the other reactants, this is homogeneous catalysis.

    14.80

    1. 2 [ NO ( g ) + N 2 O ( g ) N 2 ( g ) + NO 2 ( g ) ]

      2 NO 2 ( g ) 2 NO ( g ) + O 2 ( g ) _ 2 N 2 O ( g ) 2 N 2 ( g ) + O 2 ( g )

    2. NO serves as a catalyst in this reaction. It is present when the reaction sequence begins and after the last step is completed.
    3. No. The proposed mechanism cannot be ruled out if there is no build-up of NO 2 . In this reaction, NO 2 functions as an intermediate; it is produced and then consumed during the reaction. If there is no measurable build-up of NO 2 , the first step is slow relative to the second. As soon as NO 2 is produced by the slow first step, it is consumed by the faster second step.

    14.81

    1. When using a powdered metal catalyst, only a small percentage of the metal atoms are at the surface of the bulk material and catalytically active. Use of chemically stable supports such as alumina and silica makes it possible to obtain very large surface areas per unit mass of the precious metal catalyst. This is so because the metal can be deposited in a very thin, even monomolecular, layer on the surface of the support.
    2. The greater the surface area of the catalyst, the more reaction sites, and the greater the rate of the catalyzed reaction.

    14.82

    1. Catalytic converters are heterogeneous catalysts that adsorb gaseous CO and hydrocarbons and speed up their oxidation to CO 2 (g) and H 2 O(g). They also adsorb nitrogen oxides, NO x , and speed up their reduction to N 2 (g) and O 2 (g). If a catalytic converter is working effectively, the exhaust gas should have very small amounts of the undesirable gases CO, (NO) x , and hydrocarbons.
    2. The high temperatures could increase the rate of the desired catalytic reactions given in part (a). It could also increase the rate of undesirable reactions such as corrosion, which decrease the lifetime of the catalytic converter.
    3. The rate of flow of exhaust gases over the converter will determine the rate of adsorption of CO, (NO) x , and hydrocarbons onto the catalyst and thus the rate of conversion to desired products. Too fast an exhaust flow leads to less than maximum adsorption. A very slow flow leads to back pressure and potential damage to the exhaust system. Clearly the flow rate must be adjusted to balance chemical and mechanical efficiency of the catalytic converter.

    14.83  As illustrated in Figure 14.23, the two C–H bonds that exist on each carbon of the ethylene molecule before adsorption are retained in the process in which a D atom is added to each C (assuming we use D 2 rather than H 2 ). To put two deuteriums on a single carbon, it is necessary that one of the already existing C–H bonds in ethylene be broken while the molecule is adsorbed, so the H atom moves off as an adsorbed atom, and is replaced by a D. This requires a larger activation energy than simply adsorbing C 2 H 4 and adding one D atom to each carbon.

    14.84  Just as the π electrons in C 2 H 4 are attracted to the surface of a hydrogenation catalyst, the nonbonding electron density on S causes compounds of S to be attracted to these same surfaces. Strong interactions could cause the sulfur compounds to be permanently attached to the surface, blocking active sites and reducing adsorption of alkenes for hydrogenation.

    14.85  Let k and E a equal the rate constant and activation energy without the enzyme (uncatalyzed). Let k c and E ac equal the rate constant and activation energy with the enzyme (catalyzed). A is the same for the uncatalyzed and catalyzed reactions. The difference in activation energies is E ac – E a , k c = 1.0 × 10 6 s –1 , k = 0.039 s –1 , T = 25 °C = 298 K.

    According to Equation 14.22, ln k = E a /RT + ln A. Subtracting ln k from ln k c

    ln k c ln k = [ E ac RT ] + ln A [ E a RT ] ln A

    ln ( k c /k ) = E a E ac RT ; E a E ac = RT ln ( k c / k ) E a E ac = 8.314 J mol-K × 298 K × ln 1.0 × 10 6 0.039 = 42 , 267 J = 42.267 kJ = 42 kJ

    Carbonic anyhdrase lowers the activation energy of the reaction by 42 kJ.

    14.86

    1. (NH 2 ) 2 C=O(aq) + H 2 O(l) → CO 2 (g) + 2 NH 3 (aq)
    2. From Solution 14.85, E a E a c = R T ln ( k c / k ) . k c = 3.4 × 10 4 s –1 , k = 4.15 × 10 –5 s –1 , T = 100 °C = 373 K.

      E a E ac = 8.314 J mol-K × 373 K × ln 3.4 × 10 4 4.15 × 10 5 = 63 , 647 J = 63.647 kJ = 64 kJ

    3. Because reaction rate always increases with increasing temperature, we expect the rate of the catalyzed reaction to be significantly greater at 100 °C than at 21 °C.
    4. The 64 kJ difference between activation energies for the catalyzed and uncatalyzed reaction calculated in part (b) is a minimum difference. Because we expect the value of k c to be significantly greater than 3.4 × 10 4 s –1 , the difference in activation energies for the catalyzed and uncatalyzed reactions will be greater than 64 kJ.

    14.87 Analyze/Plan. Let k = the rate constant for the uncatalyzed reaction,

    k c = the rate constant for the catalyzed reaction

    According to Equation 14.22, ln k = –E a /RT + ln A

    Subtracting ln k from ln k c ,

    ln k c ln k = [ 55 kJ/mol RT + ln A ] [ 95 kJ/mol RT + ln A ] . S o l v e .


    1. RT = 8.314 J/mol-K × 298 K × 1 kJ/1000 J = 2.478 kJ/mol; ln A is the same for both reactions.

      ln ( k c / k ) = 95 kJ/mol 55 kJ/mol 2.478 kJ/mol ; k c / k = 1.024 × 10 7 = 1 × 10 7

      The catalyzed reaction is approximately 10,000,000 (ten million) times faster at 25 °C.

    2. RT = 8.314 J/mol-K × 398 K × 1 kJ/1000 J = 3.309 kJ/mol

      ln (k c / k) = 40 kJ/mol 3 .309 kJ/mol ; k c / k = 1.778 × 10 5 = 2 × 10 5

      The catalyzed reaction is 200,000 times faster at 125 °C.

    14.88  Let k and E a equal the rate constant and activation energy for the uncatalyzed reaction. Let k c and E ac equal the rate constant and activation energy of the catalyzed reaction. A is the same for the uncatalyzed and catalyzed reactions. k c /k = 1 × 10 5 , T = 37 °C = 310 K.

    According to Equation 14.22, ln k = –E a /RT + ln A. Subtracting ln k from ln k c

    ln k c ln k = [ E ac RT ] + ln A [ E a RT ] ln A

    ln ( k c / k ) = E a E a c R T ; E a E a c = R T ln ( k c / k ) E a E a c = 8 . 3 1 4 J mol - k × 3 1 0 k × ln ( 1 × 1 0 5 ) = 2 . 9 6 6 × 1 0 4 J = 2 9 . 6 6 k J = 3 × 1 0 1 k J

    The enzyme must lower the activation energy by 30 kJ to achieve a 1 × 10 5 -fold increase in reaction rate.

    Additional Exercises

    14.89

    1. False. If this is not an elementary reaction, we need more information to write the correct rate law.
    2. True. If the reaction is elementary, the rate law in part (a) is correct, and the reaction is second order overall.
    3. False. If the reaction is elementary, the reverse reaction is second order.
    4. False. The relationship between the forward and reverse activation energies depends on whether ΔE for the reaction is positive or negative.

    14.90 Rate = Δ [ H 2 S ] Δ t = Δ [ Cl ] 2 Δ t = k [ H 2 S ] [ Cl 2 ]

    Δ [ H 2 s ] Δ t = ( 3. 5 × 1 0 2 M 1 s 1 ) ( 2 . 0 × 1 0 4 M ) ( 0 . 0 2 5 M ) = 1 . 7 5 × 1 0 7 = 1 . 8 × 1 0 7 M / s

    Δ [ Cl ] Δ t = 2 Δ [ H 2 S ] Δ t = 2 ( 1.75 × 10 7 M / s ) = 3.5 × 10 7 M / s


    14.91

    1. R a T E = Δ [ N O ] 2 Δ t = Δ [ O 2 ] Δ t = 9 . 3 × 1 0 5 M / s 2 = 4 . 7 × 1 0 5 M / s
    (b, c) Rate = k[NO] 2 [O 2 ]; k = rate/[NO] 2 [O 2 ]
    k = 4.7 × 10 5 M /s ( 0.040 M ) 2 ( 0.035 M ) = 0.8393 = 0.84 M 2 s 1
    1. Because the reaction is second order in NO, if the [NO] is increased by a factor of 1.8, the rate would increase by a factor of 1.8 2 , or (3.24) = 3.2.

    14.92 Analyze/Plan. Using the relationship rate = k[A] x , determine the value of x that produces a rate law to match the described situation. Solve.

    1. x = 0. The rate of reaction does not depend on [A] 0 , so the reaction is zero-order in A.
    2. x = 2. When [A] 0 increases by a factor of 3, rate increases by a factor of (3) 2 = 9.
    3. x = 3. When [A] 0 increases by a factor of 2, rate increases by a factor of (2) 3 = 8.

    14.93

    1. The rate increases by a factor of nine when [C 2 O 4 2– ] triples (compare experiments 1 and 2). The rate doubles when [HgCl 2 ] doubles (compare experiments 2 and 3). The apparent rate law is: Rate = k[HgCl 2 ][C 2 O 4 2– ] 2
    2. k = rate [ HgCl 2 ] [ C 2 O 4 2 ] 2 Using the data for Experiment 1,

      k = ( 3.2 × 10 5 M /s ) [ 0.164 M ] [ 0.15 M ] 2 = 8.672 × 10 3 = 8.7 × 10 3 M 2 s 1

    3. Rate = (8.672 × 10 –3 M –2 s –1 )(0.100 M )(0.25 M ) 2 = 5.4 × 10 –5 M /s

    14.94

    1. Compare experiments 2 and 3, where the [X] 0 is the same and [Z] 0 increases by a factor of 1.5. The rate increases by a factor or 2.25, or (1.5) 2 . The reaction is second order in [Z]. Next compare experiments 1 and 2, where both [X] 0 and [Z] 0 increase by a factor of two. The rate increases by a factor of eight. We know that doubling [Z] 0 increases rate by a factor of four; doubling [X] 0 then increases rate by a factor of two. The reaction is first order in [X] 0 . The apparent rate law is: Rate = k[X][Z] 2

      Check. Compare experiments 1 and 3. Here [X] 0 doubles and [Z] 0 triples. If our rate law is correct, the rate should increase by a factor of (2)(3) 2 = 18. This is indeed the relationship between the rates of Experiments 1 and 3.

    2. k = rate [ X ] [ Z ] 2 . Using the data for Experiment 2,

      k = ( 3.2 × 10 2 M / s ) [ 0.50 M ] [ 0.50 M ] 2 = 2.560 × 10 3 = 2.6 × 10 3 M 2 s 1

      The same value is obtained using data from the other two experiments.

    3. Rate = (2.56 × 10 3 M –2 s –1 )(0.75 M )(1.25 M ) 2 = 3.0 × 10 3 M /s

    14.95

    1. Because the units of the rate constant are M –1 s –1 , the reaction is second order overall and second order in NO 2 .
    2. If [NO 2 ] 0 = 0.100 M and [NO 2 ] t = 0.025 M , use the integrated form of the second order rate equation, 1 [ A ] t = kt + 1 [ A ] 0 , Equation 14.14, to solve for t.

      1 0.025 M = 0.63 M 1 s 1 ( t ) + 1 0.100 M ; ( 40 10 ) M 1 0.63 M 1 s 1 = t = 47.62 = 48 s .

    14.96  For a first-order reaction, t 1/2 = 0.633/k. For a second-order reaction, t 1/2 = 1/k[A] 0 . Half-life is constant over the course of the reaction for first-order reactions. Although second-order half-life does not appear to depend on t, the value of “[A] 0” does change over the course of the reaction, and t 1/2 increases with time. (For a zero reaction, t 1/2 decreases with time.)

    The rate law for reaction (1) must be first order, because that is the only reaction type that has a constant half-life. Reaction (2), where t 1/2 increases with time, is second order.

    14.97 Analyze/Plan. Given k and [A] 0 , use the integrated form of the first-order rate law to calculate [A] at t = 660 s. Then, Rate = k[A].

    For a first-order reaction, ln[A] t – ln[A] 0 = –kt. ln[A] t = –kt + ln[A] 0

    [A] 0 = 2.5 × 10 –2 M , t = 660 s, k = 3.2 × 10 –3 s –1

    ln [A] = –3.2 × 10 –3 s –1 (660 s) + ln(0.025)

    ln [A] = –5.8009, [A] = 3.025 × 10 –3 M = 3.0 × 10 –3 M

    Rate = k[A] = (3.2 × 10 –3 s –1 )( 3.025 × 10 –3 M ) = 9.7 × 10 –6 M /s

    14.98

    1. t 1/2 = 0.693/k = 0.693/7.0 × 10 –4 s –1 = 990 = 9.9 × 10 2 s
    2. Use Equation 14.23 to calculate the desired temperature, T 2 . Assuming the same initial concentration of H 2 O 2 , if the rate of reaction doubles, the rate constant, k, doubles. Then,

      k 2 = 2 k 1 , k 1 /k 2 = 1/2; E a = 75 kJ/mol = 7.5 × 10 4 J/mol; T 1 = 300 K, T 2 = ?

      ln ( k 1 k 2 ) = E a R [ 1 T 2 1 T 1 ] ; ln ( 1 / 2 ) = 7.5 × 10 4 J/mol 8.314 J/mol [ 1 T 2 1 300 ]

      0.69315 = 9.0209 × 10 3 [ 1 T 2 3.3333 × 10 3 ] ; 1 T 2 = 0.69315 9.0209 × 10 3 + 3.3333 × 10 3

      T 2 = 307 K

      Note that a relatively small change in temperature is required to double the reaction rate.

    14.99 Analyze. Given rate constants for the decay of two radioisotopes, determine half-lives, decay rates, and amount remaining after three half-lives. Plan . Determine reaction order. Based on reaction-order, select the appropriate relationships for (a) rate constant and half-life and (c) rate-constant, time and concentration. In this example, mass is a measure of concentration.


    Solve. Decay of radioisotopes is a first-order process, because only one species is involved and the decay is not initiated by collision.

    1. For a first-order process, t 1/2 = 0.693/k.

      241 Am: t 1/2 = 0.693/1.6 × 10 –3 yr –1 = 433.1 = 4.3 × 10 2 yr

      125 I: t 1/2 = 0.693/0.011 day –1 = 63.00 = 63 days

    2. For a given sample size, half of the 241 Am sample decays in 433 years, whereas half of the 125 I sample decays in 63 days. 125 I decays at a much faster rate.
    3. For a first-order process, ln[A] t – ln[A] 0 = –kt. ln[A] t = –kt + ln[A] 0 .

      [A] 0 = 1.0 mg; t = 3 t 1/2 .

      241 Am: t = 3 t 1/2 = 3(433.1 yr) = 1.299 × 10 3 = 1.3 × 10 3 yr

      ln [Am] t = –1.6 × 10 –3 yr –1 (1.299 × 10 3 yr) – ln (1.0) = –2.079 – 0 = –2.08

      [Am] t = 0.125 = 0.13 mg

      or, mass 241 Am remaining = 1.0 mg/2 3 = 0.125 = 0.13 mg

      125 I: For the same size starting sample and number of elapsed half-lives, the same mass, 0.13 mg 125 I, will remain. (The difference is that the elapsed time of 3 half-lives (for 125 I is 3(63) = 189 days = 0.52 yr, versus 433 yr for 241 Am.)

    4. Again, for a first-order process, ln[A] t – ln[A] 0 = –kt. ln[A] t = –kt + ln[A] 0 .

      [A] 0 = 1.0 mg; t = 4 days.

      k Am = 1.6 × 10 –3 yr –1 (1 yr/365 days) = 4.3836 × 10 –6 = 4.4 × 10 –6 day –1

      241 Am: ln [Am] t = –4.4 × 10 –6 day –1 (4 days) – ln (1.0) = –1.7 × 10– 5– 0 = –1.7 × 10 –5

      The amount of 241 Am remaining after 4 days is 0.99998 mg, to three significant figures, 1.00 mg.

      125 I: ln [I] t = –0.011 day –1 (4 days) – ln (1.0) = –0.044 – 0 = –0.044

      The amount of 125 I remaining after 4 days is 0.956 mg.

    14.100

    1. k = (8.56 × 10 –5 M /s)/(0.200 M ) = 4.28 × 10 –4 s –1
    2. ln [urea] = –(4.28 × 10 –4 s –1 × 4.00 × 10 3 s) + ln (0.500)

      ln [urea] = –1.712 – 0.693 = –2.405 = –2.41; [urea] = 0.0903 = 0.090 M

    3. t 1/2 = 0.693/k = 0.693/4.28 × 10 –4 s –1 = 1.62 × 10 3 s

    14.101

    1. A = εbc, Equation 14.5. A = 0.605, ε = 5.60 × 10 3 cm– 1 M –1 , b = 1.00 cm

      c = A ε b = 0.605 ( 5.60 × 10 3 cm 1 M 1 ) ( 1.00 cm ) = 1.080 × 10 4 = 1.08 × 10 4 M

    2. Calculate [c] t using Beer’s law. We calculated [c] 0 in part (a). Use Equation 14.13 to calculate k.

      A 30 = ε bc 30 ; c 30 = A 30 ε b = 0.250 ( 5.60 × 10 3 cm 1 M 1 ) ( 1.00 cm ) = 4.464 × 10 5 M

      ln [ c ] t = kt + ln [ c ] 0 ; ln [ c ] 0 ln [ c ] t t = k ; t = 30 min × 60 s min = 1800 s

      k = (ln (1.080 × 10 –4 ) – ln (4.464 × 10 –5 ) )/1800 s = 4.910 × 10 –4 = 4.91 × 10 –4 s –1


    1. For a first-order reaction, t 1/2 = 0.693/k.

      t 1/2 = 0.693/4.910 × 10 –4 s –1 = 1.411 × 10 3 = 1.41 × 10 3 s = 23.5 min

    1. A t = 0.100; calculate c t using Beer’s law, then t from the first-order integrated rate equation.

      c t = A ε b = 0.100 ( 5.60 × 10 3 cm 1 M 1 ) ( 1.00 cm ) = 1.786 × 10 5 = 1.79 × 10 5 M t = ln [ c ] 0 ln [ c ] t k = ln ( 1.080 × 10 4 ) ln ( 1.786 × 10 5 ) 4.910 × 10 4 s 1

      t = 3.666 × 10 3 = 3.67 × 10 3 s = 61.1 min

    14.102  Calculate [dye] at each time, using Beer’s law, A = εbc; calculate ln[dye] and 1/[dye] and plot these quantities vs time in two separate graphs. The straight-line plot indicates the order of reaction with respect to [dye].

    A 0 = ε bc 0 ; c 0 = A 0 ε b = 1.254 ( 4.7 × 10 4 cm 1 M 1 ) ( 1.00 cm ) = 2.668 × 10 5 = 2.7 × 10 5 M

    Time (min) A at 608 nm [dye] ln [dye] 1/[dye]
    0 1.254 2.7 × 10 –5 –10.53 3.7 × 10 4
    30 0.941 2.0 × 10 –5 –10.82 5.0 × 10 4
    60 0.752 1.6 × 10 –5 –11.04 6.3 × 10 4
    90 0.672 1.4 × 10 –5 –11.16 7.0 × 10 4
    120 0.545 1.2 × 10 –5 –11.36 8.6 × 10 4
    image

    Although the graphs are not absolutely definitive, the plot of 1/[dye] vs time appears to be more linear. (The data point at t = 90 min us “out of line” in both plots and is suspect. More precision and accuracy in the experimental data would be helpful.) Assuming the reaction is second order with respect to the dye, the rate law is: Rate = k[dye] 2 .

    k = slope = (8.6 × 10 4 – 3.7 × 10 4 ) M –1 /(120–0)min = 4.1 × 10 2 M –1 min –1

    (The best-fit slope and k value is 3.9 × 10 2 M –1 min –1 .)


    14.103

    Time (s) [C 5 H 6 ] ( M ) ln[C 5 H 6 ] 1/[C 5 H 6 ]
    0 0.0400 –3.219 25.0
    50 0.0300 –3.507 33.3
    100 0.0240 –3.730 41.7
    150 0.0200 –3.912 50.0
    200 0.0174 –4.051 57.5
    image
    image
    image
    1. The plot of 1/[C 5 H 6 ] vs time is linear and the reaction is second order.
    2. The slope of the line in the plot of 1/[C 5 H 6 ] vs time is the value of k. k = slope = (50.0 – 25.0) M –1 /(150–0)s = 0.167 M –1 s –1

      (The best-fit slope and k value is 0.163 M –1 s –1 .)

    14.104

    ln k 1/T
    –24.17 3.33 × 10 –3
    –20.72 3.13 × 10 –3
    –17.32 2.94 × 10 –3
    –15.24 2.82 × 10 –3
    image

    The calculated slope is –1.751 × 10 4 . The activation energy E a , equals – (slope) × (8.314 J/mol). Thus, E a = 1.8 × 10 4 (8.314) = 1.5 × 10 5 J/mol = 1.5 × 10 2 kJ/mol. (The best-fit slope is –1.76 × 10 4 = –1.8 × 10 4 and the value of E a is 1.5 × 10 2 kJ/mol.)

    14.105 Analyze . Given the time required to reach the “sour” point, at two different temperatures, estimate the activation energy for the reaction.

    Plan/Solve . The warmer reaction at 28 °C is faster than the cooler reaction at 5 °C. The ratio of the rates of reaction is the inverse ratio of times.

    rate 28 rate 5 = k 28 [ sour milk ] 28 k 5 [ sour milk ] 5 = t 5 t 28 = 48 hr 4 hr = 12

    Assume the concentration of sour milk at 28 °C is the same as the concentration of sour milk at 5 °C. That is, [sour milk] 28 = [sour milk] 5 . Then, the ratio of rate constants, k 28 /k 5 , equals 12.


    T 1 = 28 °C = 301 K; T 2 = 5 °C = 278 K; k 1 /k 2 = 12

    ln ( k 1 k 2 ) = E a R [ 1 T 2 1 T 1 ] ; ln ( 12 ) = E a 8.314 J/mol [ 1 278 1 301 ]

    E a = ln ( 12 ) ( 8.314 J/mol ) 2.749 × 10 4 = 7.516 × 10 4 J = 75 kJ/mol

    14.106

    1. T 1 = 77 °F; °C = 5/9 (°F – 32) = 5/9 (77 – 32) = 25 °C = 298 K

      T 2 = 59 °F; °C = 5/9 (59–32) = 15 °C = 288 K; k 1 /k 2 = 6

      ln ( k 1 k 2 ) = E a R [ 1 T 2 1 T 1 ] ; ln ( 6 ) = E a 8.314 J/mol [ 1 288 1 298 ]

      E a = ln ( 6 ) ( 8.314 J/mol ) 1.165 × 10 4 = 1.28 × 10 5 J = 1.3 × 10 2 kJ/mol

      T 1 = 77 °F = 25 °C = 298 K; T 2 = 41 °F = 5 °C = 278 K, k 1 /k 2 = 40

      ln ( 40 ) = E a 8.314 J/mol [ 1 278 1 298 ] ; E a = ln ( 40 ) ( 8.314 J/mol ) 2.414 × 10 4

      E a = 1.27 × 10 5 J = 1.3 × 10 2 kJ/mol

      The values are amazingly consistent, considering the precision of the data.

    2. For a first-order reaction, t 1/2 = 0.693/k, k = 0.693/t 1/2

      k 1 at 298 K = 0.693/2.7 yr = 0.257 = 0.26 yr –1

      T 1 = 298 K, T 2 = 273 – 15 °C = 258 K

      ln ( 0.257 k 2 ) = 1.27 × 10 5 J 8.314 J/mol [ 1 258 1 298 ] = 7.94727 = 7.95

      0.257/k 2 = e 7.94727 = 2.828 × 10 3 ; k 2 = 0.257/2.828 × 10 3 = 9.088 × 10 –5 = 9.1 × 10 –5 yr –1

      t 1/2 = 0.693/k = 0.693/9.088 × 10 –5 = 7.625 × 10 3 yr = 7.6 × 10 3 yr

    14.107

    1. NO ( g ) + NO ( g ) N 2 O 2 ( g )

      N 2 O 2 ( g ) + H 2 ( g ) N 2 O ( g ) + H 2 O ( g ) 2 NO ( g ) + N 2 O 2 ( g ) + H 2 ( g ) N 2 O 2 ( g ) + N 2 O ( g ) + H 2 O ( g ) 2 NO ( g ) + H 2 ( g ) N 2 O ( g ) + H 2 O ( g )

    2. First reaction: –Δ [NO]/Δt = k[NO] [NO] = k[NO] 2

      Second reaction: –Δ [H 2 ]/Δt = k[H 2 ][N 2 O 2 ]

    3. N 2 O 2 is the intermediate: it is produced in the first step and consumed in the second.
    4. Because [H 2 ] appears in the rate law, the second step must be slow relative to the first.

    14.108

    1. Cl ( g ) + O 3 (g ) ClO ( g ) + O 2 ( g ) Cl ( g ) + O(g ) Cl ( g ) + O 2 ( g ) Cl ( g ) + O 3 (g ) + Cl ( g ) + O(g ) Cl ( g ) + O 2 ( g ) + Cl ( g ) + O 2 ( g ) O 3 ( g ) + O ( g ) 2 O 2 ( g )
    2. Cl(g) is the catalyst. It is consumed in the first step and reproduced in the second.
    3. ClO(g) is the intermediate. It is produced in the first step and consumed in the second.

    14.109

    1. O 3 ( g ) O 2 ( g ) + O ( g ) O ( g ) +O 3 ( g ) 2 O 2 ( g ) 2 O 3 ( g ) 3 O 2 ( g )
    2. Follow the logic in Sample Exercise 14.15 for determining the rate law of a mechanism with a fast initial step.

      From the rate determining second step, rate = k 2 [O][O 3 ]

      If the first step is a fast equilibrium, k 1 [O 3 ] = k – 1 [O 2 ][O]

      Solving for [O] in terms of [O 3 ] and [O 2 ], [ O ] = k 1 [ O 3 ] k 1 [ O 2 ]

      Rate = k 2 k 1 [ O 3 ] [ O 3 ] k 1 [ O 2 ] = k [ O 3 ] 2 [ O 2 ]

    3. O(g) is an intermediate; it is produced and then consumed during the reaction.
    4. If the reaction occurred in a single step, the rate law would change. It would be rate = k[O 3 ] 2 .

    14.110

    1. Cl 2 ( g ) 2 Cl ( g ) Cl ( g ) + CHCl 3 ( g ) HCl ( g ) + CCl 3 ( g ) Cl ( g ) + CCl 3 ( g ) CCl 4 ( g ) Cl 2 ( g ) + 2 Cl ( g ) + CHCl 3 ( g ) + CCl 3 ( g ) 2 Cl ( g ) + HCl ( g ) + CCl 3 ( g ) + CCl 4 ( g ) Cl 2 ( g ) + CHCl 3 ( g ) HCl ( g ) + CCl 4 ( g )
    2. Cl(g), CCl 3 (g)
    3. Reaction 1 - unimolecular, Reaction 2 - bimolecular, Reaction 3 - bimolecular
    4. Reaction 2, the slow step, is rate determining.
    5. If Reaction 2 is rate determining, rate = k 2 [CHCl 3 ][Cl]. Cl is an intermediate formed in reaction 1, an equilibrium. By definition, the rates of the forward and reverse processes are equal; k 1 [Cl 2 ] = k – 1 [Cl] 2 . Solving for [Cl] in terms of [Cl 2 ],

      [ Cl ] 2 = k 1 k 1 [ Cl 2 ] ; [ Cl ] = ( k 1 k 1 [ Cl 2 ] ) 1 / 2

      Substituting into the overall rate law

      rate = k 2 ( k 1 k 1 ) 1 / 2 [ CHCl 3 ] [ Cl 2 ] 1 / 2 = k [ CHCl 3 ] [ Cl 2 ] 1 / 2 (The overall order is 3/2.)

    14.111

    1. Rate = k 1 [A][B]
    2. Follow the logic in Sample Exercise 14.15.

      Rate = k 2 [A][X]; k 1 [A][B] = k– 1 [C][X]; [ X ] = k 1 k 1 [ A ] [ B ] [ C ]

      Substituting for [X] in the rate expression,

      Rate = k 2 [ A ] k 1 k 1 [ A ] [ B ] [ C ] = k [ A ] 2 [ B ] [ C ]

    3. (iii) The result of part (b) might be surprising because [C] appears in the rate law, and [C] has a negative reaction order.

    14.112

    1. (CH 3 ) 3 AuPH 3 → C 2 H 6 + (CH 3 )AuPH 3
    2. (CH 3 ) 3 Au, (CH 3 )Au, and PH 3 are intermediates.
    3. Reaction 1 is unimolecular, Reaction 2 is unimolecular, Reaction 3 is bimolecular.
    4. Reaction 2, the slow one, is rate determining.
    5. If Reaction 2 is rate determining, Rate = k 2 [(CH 3 ) 3 Au].

      (CH 3 ) 3 Au is an intermediate formed in Reaction 1, an equilibrium. By definition, the rates of the forward and reverse processes in Reaction 1 are equal:

      k 1 [(CH 3 ) 3 AuPH 3 ] = k – 1 [(CH 3 ) 3 Au][PH 3 ]; solving for [(CH 3 ) 3 Au],

      [ ( CH 3 ) 3 Au ] = k 1 [ ( CH 3 ) 3 AuPH 3 ] k 1 [ PH 3 ]

      Substituting into the rate law

      Rate = ( k 2 k 1 k 1 ) [ ( CH 3 ) 3 AuPH 3 ] [ PH 3 ] = k [ ( CH 3 ) 3 AuPH 3 ] [ PH 3 ]

    6. The rate is inversely proportional to [PH 3 ], so adding PH 3 to the (CH 3 ) 3 AuPH 3 solution would decrease the rate of the reaction.

    14.113 Analyze/Plan. Use the structure and unit cell edge of Pt, along with the formulas for volume and surface area of a sphere, to calculate the number of Pt atoms in a 2-nm sphere and on the surface of a 2-nm sphere.

    1. For a Pt sphere with a 2.0 nm diameter, radius = 1.0 nm.

      V = 4 / 3 π r 3 = 4 π ( 1.0 nm ) 3 3 × 10 3 Å 3 1 3 nm 3 = 4.188879 × 10 3 = 4.2 × 10 3 Å 3

      In a face-centered cubic metal structure, there are 4 metal atoms per unit cell. The volume of the unit cell is (3.924 Å) 3 = 60.42 Å 3

      4 Pt atoms 60.42 Å 3 × 4.1889 × 10 3 Å 3 = 277.3 = 2.8 × 10 2 Pt atoms in a 2.0 -nm sphere

    2. Assume that the “footprint” of an atom is its cross-sectional area, the area of a circle with the radius of the atom. The area of this circle is πr 2 . The diameter, d, of a Pt atom is 2.8 Å, so r = d/2 = 1.4 Å. The footprint of the Pt atom is then π(1.4 Å) 2 = 6.1575 = 6.2 Å 2

      The surface area of the 2.0-nm sphere is

      4 π r 2 = 4 π ( 1.0 nm ) 2 × 10 2 Å 2 1 2 nm 2 = 12.56637 × 10 2 = 1.3 × 10 3 Å 2

      1 Pt atoms 6.1575 Å 2 × 1.2566 × 10 3 Å 2 = 204.1 = 2.0 × 10 2 surface Pt atoms on a 2.0 -nm sphere .

    3. 204 surface Pt atoms 277 total Pt atoms × 100 = 74 % Pt atoms on the surface

    1. For a 5.0-nm Pt sphere, radius = 2.5 nm

      V = 4 / 3 π r 3 = 4 π ( 2.50 nm ) 3 3 × 10 3 Å 3 1 3 nm 3 = 65.4498 × 10 3 = 6.5 × 10 4 Å 3

      4 Pt atoms 60.42 Å 3 × 65.4498 × 10 3 Å 3 = 4333 = 4.3 × 10 3 Pt atoms in a 5.0-nm sphere

      The surface area of the 5.0-nm sphere is

      4 π r 2 = 4 π ( 2.5 nm ) 2 × 10 2 Å 2 1 2 nm 2 = 7853.98 = 7.9 × 10 3 Å 2

      1 Pt atoms 6.1575 Å 2 × 7.854 × 10 3 Å 2 = 1275.5 = 1.3 × 10 3 surface Pt atoms on a 5.0 -nm sphere

      1276 surface Pt atoms 4333 total Pt atoms × 100 = 29 % Pt atoms on the surface

      The calculations in parts (b) and (d) overestimate the number of Pt atoms on the surface of the sphere, because they do not account for empty space between atoms. For the purpose of comparison, it is most important that we use the same method for both spheres.

      [Alternatively, use one face of a face-entered cubic unit cell as a model for the surface area that Pt atoms will occupy. On a face, there is the cross-section of 1 Pt atoms in the center and ¼ Pt atom at each corner. This amounts to the cross-sections of two Pt atoms in (3.924 Å) 2 = 15.398 = 15.4 Å 2 .]

      2 Pt atoms 15.40 Å 3 × 1.2566 × 10 3 Å 2 = 163.2 = 1.6 × 10 2 surface Pt atoms on a 2.0 -nm sphere .

      163 surface Pt atoms 277 total Pt atoms × 100 = 59 % Pt atoms on the surface

      [Similarly, in a 5.0-nm sphere, there are 1.0 × 10 3 Pt atoms on the surface, 24% of the total Pt atoms.]

    2. Both surface models predict that the 2.0-nm sphere will be more catalytically active, because it has a much greater percentage of its atoms on the surface, where they can participate in the chemical reaction.

    14.114 Enzyme : carbonic anhydrase; substrate: carbonic acid (H 2 CO 3 );

    turnover number : 1 × 10 7 molecules/s.

    14.115  Let k and E a equal the rate constant and activation energy for the uncatalyzed reaction. Let k c and E ac equal the rate constant and activation energy of the enzyme-catalyzed reaction. Assume A is the same for the uncatalyzed and catalyzed reactions.

    k c /k = 5000, T = 37 °C = 310 K.

    According to Equation 14.22, ln k = –E a /RT + ln A. Subtracting ln k from ln k c

    ln k c ln k = [ E ac RT ] + ln A [ E a RT ] ln A

    ln ( k c / k ) = E a E a c R T ; E a E a c = R T ln ( k c / k ) E a E a c = 8 . 3 1 4 J mol - k × 3 1 0 k × ln ( 5 0 0 0 ) = 2 . 1 9 5 × 1 0 4 J / mol = 2 1 . 9 5 k J / mol = 2 2 k J / mol

    The enzyme must lower the activation energy by 22 kJ/mol to increase the reaction rate by a factor of 5000.


    14.116

    1. The rate law for the slow step is Rate = k 2 [ES], where ES is an intermediate. Use relationships from the fast equilibrium step to substitute for [ES].

      rate of the forward reaction = k 1 [E][S]

      rate of the reverse reaction = k– 1 [ES]

      For an equilibrium, rate forward = rate reverse, k 1 [E][S] = k– 1 [ES];

      [ES] = k 1 k 1 [ E ] [ S ] ; Rate = k 2 [ES] = k 2 k 1 k 1 [ E ] [ S ] = k[E][S]

    2. E + I ⇌ EI

    Integrative Exercises

    14.117 Analyze/Plan. 2 N 2 O 5 → 4 NO 2 + O 2 Rate = k[N 2 O 5 ] = 1.0 × 10 –5 s –1 [N 2 O 5 ]

    Use the integrated rate law for a first-order reaction, Equation 14.13, to calculate k[N 2 O 5 ] at 20.0 h. Build a stoichiometry table to determine mol O 2 produced in 20.0 h. Assuming that O 2 (g) is insoluble in chloroform, calculate the pressure of O 2 in the 10.0 L container. Solve.

    2 0 . 0 h × 6 0 m i n 1 h × 6 0 s 1 m i n = 7 . 2 0 × 1 0 4 s ; [ N 2 O 5 ] 0 = 0 . 6 0 0 M

    ln [A] t – ln [A] 0 = –kt; ln [N 2 O 5 ] t = –kt + ln [N 2 O 5 ] 0

    ln [N 2 O 5 ] t = –1.0 × 10 –5 s –1 (7.20 × 10 4 s) + ln(0.600) = –0.720 – 0.511 = –1.231

    [N 2 O 5 ] t = e –1.231 = 0.292 M

    N 2 O 5 was present initially as 1.00 L of 0.600 M solution.

    mol N 2 O 5 = M × L = 0.600 mol N 2 O 5 initial, 0.292 mol N 2 O 5 at 20.0 h

    2 N 2 O 5 4 NO 2 + O 2
    t = 0 0.600 mol 0 0
    change –0.308 mol 0.616 mol 0.154 mol
    t = 20 h 0.292 mol 0.616 mol 0.154 mol

    [Note that the reaction stoichiometry is applied to the “change” line.]

    PV = nRT; P = nRT/V; V = 10.0 L, T = 45 °C = 318 K, n = 0.154 mol

    P = 0.154 mol × 318 K 10.0 L × 0.08206 L-atm mol-K = 0.402 atm

    14.118

    1. ln k = –E a /RT + ln A; E a = 86.8 kJ/mol = 8.68 × 10 4 J/mol; T = 35 °C + 273 = 308 K; A = 2.10 × 10 11 M –1 s –1

      ln k = 8.68 × 10 4 J/mol 308 K × mol-K 8.314 J + ln ( 2.10 × 10 11 M 1 s 1 )

      ln k = –33.8968 + 26.0704 = –7.8264; k = 3.99 × 10 –4 M –1 s –1


    1. 0 . 3 3 5 g k OH 0 . 2 5 0 L s o ln × 1 mol k OH 5 6 . 1 g kOH = 0 . 0 2 3 8 9 = 0 . 0 2 3 9 M kOH

      1 . 4 5 3 g c 2 H 5 I 0 . 2 5 0 L so ln × 1 mol c 2 H 5 I 1 5 6 . 0 g c 2 H 5 I = 0 . 0 3 7 2 6 = 0 . 0 3 7 3 M c 2 H 5 I

      If equal volumes of the two solutions are mixed, the initial concentrations in the reaction mixture are 0.01194 M KOH and 0.01863 M C 2 H 5 I. Assuming the reaction is first order in each reactant:

      Rate = k[C 2 H 5 I][OH ] = 3.99 × 10 –4 M –1 s –1 (0.01194 M )(0.01863 M ) = 8.88 × 10 –8 M /s

    1. Because C 2 H 5 I and OH react in a 1 : 1 mole ratio and equal volumes of the solutions are mixed, the reactant with the smaller concentration, KOH, is the limiting reactant.
    1. T = 50 °C + 273 = 323 K

      ln k = 8.68 × 10 4 J/mol 323 K × mol-K 8.314 J + ln ( 2.10 × 10 11 M 1 s 1 )

      ln k = –32.3227 + 26.0704 = –6.2523; k = 1.93 × 10 –3 M –1 s –1

    14.119  Obtaining data for an “Arrhenius” plot like this requires running the reaction several times, each at a different temperature. Different rates and rate constants (k) are obtained at each temperature. We expect a straight line, according to the relationship:

    ln k = –E a /RT + ln A. The slope of the graph is –E a and the y-intercept is the orientation factor, A. The graph in the exercise demonstrates these characteristics, times two! The graph indicates that reaction requires two different activation energies, depending on temperature.

    Assuming that reactants and products are the same at all temperatures, the reaction proceeds through different pathways, depending on temperature. This could mean two totally different reaction mechanisms, or a multi-step mechanism where different steps are rate-determining at different temperatures.

    14.120

    1. ln k = –E a /RT + ln A, Equation 14.22. E a = 6.3 kJ/mol = 6.3 × 10 3 J/mol

      T = 100 °C + 273 = 373 K

      ln k = 6.3 × 10 3 J/mol 8.314 J/K-mol × 373 K + ln ( 6.0 × 10 8 M 1 s 1 )

      ln k = –2.032 + 20.212 = 18.181 = 18.2; k = 7.87 × 10 7 = 8 × 10 7 M –1 s –1

    2. NO, 11 valence e–, 5.5 e– pair (Assume the less electronegative N atom will be electron deficient.)
      image

      ONF, 18 valence e , 9 e pr

      image

      The resonance form on the right is a very minor contributor to the true bonding picture, because of high formal charges and the unlikely double bond involving F.


    1. ONF has trigonal planar electron domain geometry, which leads to a “bent” structure with a bond angle of approximately 120°.
    1. image
    1. The electron deficient NO molecule is attracted to electron-rich F 2 , so the driving force for formation of the transition state is greater than simple random collisions.

    14.121

    1. Δ H rxn o = 2 Δ H f o H 2 O ( g ) + 2 Δ H f o Br 2 ( g ) 4 Δ H f o HBr ( g ) Δ H f o O 2 ( g )

      Δ H rxn ° = 2 ( 241.82 ) + 2 ( 30.71 ) 4 ( 36.23 ) ( 0 ) = 277.30 kJ

    2. Because the rate of the uncatalyzed reaction is very slow at room temperature, the magnitude of the activation energy for the rate-determining first step must be quite large. At room temperature, the reactant molecules have a distribution of kinetic energies (Chapter 10), but very few molecules even at the high end of the distribution have sufficient energy to form an activated complex. E a for this step must be much greater than 3/2 RT, the average kinetic energy of the sample431.
    3. 20 e , 10 e pr
      image

      The intermediate resembles hydrogen peroxide, H 2 O 2 .

    14.122

    1. D(Cl–Cl) = 242 kJ/mol Cl 2

      242 kJ mol Cl 2 × 1000 J kJ × 1 mol 6.022 × 10 23 molecules = 4.019 × 10 19 = 4.02 × 10 19 J

      λ = hc/E = 6.626 × 10 34 J-s × 2.998 × 10 8 m/s 4.019 × 10 19 J = 4.94 × 10 7 m

      This wavelength, 494 nm, is in the visible portion of the spectrum.

    2. image

    1. Because D(Cl–Cl) is 242 kJ/mol, CH 4 (g) + Cl 2 (g) should be about 242 kJ below the starting point on the diagram. For the reaction

      CH 4 (g) + Cl 2 (g) → CH 3 (g) + HCl(g) + Cl(g), E a is 242 + 17 = 259 kJ.

      (From bond dissociation enthalpies, ΔH for the overall reaction

      CH 4 (g) + Cl 2 (g) → CH 3 Cl(g) + Cl(g) is –104 kJ, so the graph above is simply a sketch of the relative energies of some of the steps in the process.)

    1. CH 3 , 7 valence e , odd electron species
      image
    1. This sequence is called a chain reaction because Cl radicals are regenerated in Reaction 4, perpetuating the reaction. Absence of Cl terminates the reaction, so Cl + Cl → Cl 2 is a termination step.

    14.123

    1. A generic Lewis structure for a primary amine is shown below. There are four electron domains about nitrogen, so the hybridization is sp 3 . The hybrid orbital picture is shown on the right.
      image
    2. A reactant that is attracted to the lone pair of electrons on nitrogen will produce a tetrahedral intermediate. This can be a moiety with a full, partial, or even transient positive charge. Steric hindrance will not be large, because two of the atoms bound to nitrogen are small hydrogens.

    14.124

    1. Molecule NO NO 2 N 2
      Valence e 11 17 10
      e pairs 5.5 8.5 5
      Lewis structure
      image
      image
      image
      Bond order 2 1.5 3
      Bond energy 607 kJ/mol 404 kJ/mol 941 kJ/mol
    2. The bond energies in the table above are from Table 8.3 of the text. The NO 2 molecule has two resonance forms and the bond order is 1.5. To obtain an approximate bond energy, average the energies for N=O and N–O:

      (607 kJ + 201 kJ)/2 = –404 kJ/bond in NO 2

      We know that resonance stabilizes a molecule, so the actual bond energy is probably somewhat greater than this value.


      Use Avogadro’s number to calculate energy in J/bond, then λ = hc/E to calculate wavelength and region of the electromagnetic spectrum. These are the maximum wavelengths that would cause complete bond dissociation.

      941 kJ mol N 2 × 1000 J kJ × 1 mol 6.022 × 10 23 molecules = 1.5626 × 10 18 = 1.56 × 10 18 J

      λ = hc/E = 6.626 × 10 34 J-s × 2.998 × 10 8 m/s 1.5626 × 10 18 J = 1.27 × 10 7 m

      For NO, the energy per bond is 1.01 × 10 –18 J and the wavelength is 1.97 × 10 –7 m.

      For NO 2 , the energy per bond is 6.71×10– 19 J and the wavelength is 2.96 × 10 –7 m.

      The three “bond dissociation” wavelengths, 127 nm, 197 nm, and 296 nm, are all in the ultraviolet region, near but not in the visible range of 400-700 nm. We expect longer wavelength electronic excitations for the three gases to be in the visible and near UV, in the same relative order as the bond dissociation wavelengths.

    1. The experiment requires a UV-VIS spectrometer and a gas flow cell that can be attached to the exhaust stream. According to Beer’s law, as the concentration of absorbing species decreases, so does the absorbance. By monitoring a different wavelength of maximum absorption (longer than 297 nm) for each gas, we can measure the concentration of each gas at some point in time. We would monitor the stream before the catalytic converter to establish starting concentrations, then after the converter to observe changes. If the catalyst is working, we expect the two longer wavelength peaks for NO and NO 2 to decrease in size, and the shorter wavelength peak for N 2 to increase.

    14.125

    1. No. Ethane, C 2 H 6 , has only sigma bonds. The electron density is all localized along the C–H and C–C bonds. The molecule has no pi electrons to interact with the metal surface.
    2. Yes. Ammonia, NH 3 , has a nonbonding electron pair on nitrogen. This electron domain extends away from the sigma framework of the N–H bonds and is available to interact with the metal surface of the catalyst.

     

    15 Chemical Equilibrium

    Visualizing Concepts

    15.1

    1. k f > k r . According to the Arrhenius equation [14.21], k = Ae E a / RT . As the magnitude of E a increases, k decreases. On the energy profile, E a is the difference in energy between the starting point and the energy at the top of the barrier. Clearly this difference is smaller for the forward reaction, so k f > k r .
    2. From the Equation [15.5], the equilibrium constant = k f /k r . Because k f > k r , the equilibrium constant for the process shown in the energy profile is greater than 1.

    15.2   Yes, the system is in equilibrium in boxes 4 and 5. The first box is pure reactant A. As the reaction proceeds, some A changes to B. In the fourth and fifth boxes, the relative amounts (concentrations) of A and B are constant. Although the reaction is ongoing the rates of A → B and B → A are equal, and the relative amounts of A and B are constant.

    15.3 Analyze. Given box diagram and reaction type, determine whether K greater or smaller than one for the equilibrium mixture depicted in the box.

    Plan. Assign species in the box to reactants and products. Write an equilibrium expression in terms of concentrations. Find the relationship between numbers of moles (molecules in the diagram) and concentration. Calculate K.

    Solve. Let red = A, blue = X, red and blue pairs = AX. (The colors of A and X are arbitrary.) There are 3A, 2B, and 8AX in the box.

    M = mol/L. Because each particle represents one mole, we can use numbers of particles in place of moles in the molarity formula. (The mole is a counting unit for particles, so mol ratios and particle ratios are equivalent.) V = 1 L, so in this case, [A] = number of A particles.

    K = [ AX ] [ A ] [ X ] ; [ AX ] = 8/V = 8 ; [ A ] = 3 /V = 3 ; [ X ] = 2 / V = 2 . K = 8 [3][2] = 8 6 = 1.33

    K is greater than one.

    15.4 Analyze/Plan. Given that element A = red and element B = blue, evaluate the species in the reactant and product boxes, and write the reaction. Answer the remaining questions based on the balanced equation. Solve .

    1. reactants: 4A 2 + 4B; products: 4A 2 B

      balanced equation: A 2 + B → A 2 B

    2. K c = [ A 2 B ] [A 2 ][B]

    1. Evaluate concentrations and the value of K for the box in equilibrium, the one on the right of the diagram. [A 2 B] = 0.4 M ; [A 2 ] = 0.1 M ; [B] = 0.1 M

      K c = [ A 2 B ] [A 2 ][B] = [ 0 .4 ] [0 .1][0 .1] = 40

    1. Δn = ∑n(prod) – ∑n(react) = 1 – 2 = –1.
    1. K p = K c (RT) Δ n , Equation 15.15. Assume a temperature of 25 o C = 298 K.

      K p = 40[(0 .0821)(298)] 1 = 40 (0 .0821)(298) = 1.635 = 2

    15.5   The reaction A(g) + B(g) ⇌ AB(g) has the larger equilibrium constant. At 40s and 50s, there are more diatomic products and fewer monoatomic reactants than in the other reaction. An equilibrium constant is the ratio of concentrations of products to concentrations of reactants at equilibrium. Another way to say this is that the A(g) + B(g) reaction favors products more than the X(g) + Y(g) reaction. (The X(g) + Y(g) reaction reaches equilibrium more quickly than the A(g) + B(g) reaction, but this is a matter of kinetics, not equilibrium constants.)

    15.6 Analyze/Plan. The reaction with the largest equilibrium constant has the largest ratio of products to reactants. Count product and reactant molecules. Calculate ratios and compare. Solve.

    K = [ C 2 H 4 X 2 ] [ C 2 H 4 ] [ X 2 ] . Use numbers of molecules as an adequate measure of concentration.

    (Although the volume terms don’t cancel, they are the same for all parts. For the purpose of comparison, we can ignore volume.) Solve.

    1. 8 C 2 H 4 Cl 2 ,  2 Cl 2 ,  2 C 2 H 4 .  K = 8 ( 2 ) ( 2 ) = 2
    2. 6 C 2 H 4 Br 2 ,  4 Br 2 ,  4 C 2 H 4 .  K = 6 ( 4 ) ( 4 ) = 0.375 = 0.4
    3. 3 C 2 H 4 I 2 ,  7 I 2 ,  7 C 2 H 4 .  K = 3 ( 7 ) ( 7 ) = 0.0612 = 0.06

    From the smallest to the largest equilibrium constant, (c) < (b) < (a).

    Check. By inspection, there are the fewest product molecules and the most reactant molecules in (c); most product and least reactant in (a).

    15.7   Statement (b) is definitely true. The reaction is a heterogeneous equilibrium for which K p = P O 2 . The larger volume of vessel B requires that more PbO 2 (s) must decompose to produce the equilibrium pressure of O 2 (g). (After heating, both vessels will have less than the 5.0 g PbO 2 (s) that was present initially. If the “less” refers to the amount of solid present initially, statement (a) is also true.)

    15.8 Analyze. Given box diagrams, reaction type, and value of K c , determine whether each reaction mixture is at equilibrium.

    Plan. Analyze the contents of each box, express them as concentrations (see Solution 15.3). Write the equilibrium expression, calculate Q for each mixture, and compare it to K c . If Q = K, the mixture is at equilibrium. If Q < K, the reaction shifts right (more product). If Q > K, the reaction shifts left (more reactant).


    Solve. K c = [ AB ] 2 [A 2 ] [B 2 ] .

    For this reaction, Δn = 0, so the volume terms cancel in the equilibrium expression. In this case, the number of each kind of particle can be used as a representation of moles (see Solution 5.3) and molarity.

    1. Mixture (i): 1A 2 , 1B 2 , 6AB; Q = 6 2 ( 1 ) ( 1 ) = 36

      Q > K c , the mixture is not at equilibrium.

      Mixture (ii): 3A 2 , 2B 2 , 3AB; Q = 3 2 ( 3 ) ( 2 ) = 1.5

      Q = K c , the mixture is at equilibrium.

      Mixture (iii): 3A 2 , 3B 2 , 2AB; Q = 2 2 ( 3 ) ( 3 ) = 0.44

      Q < K c , the mixture is not at equilibrium.

    2. Mixture (i) proceeds toward reactants.

      Mixture (iii) proceeds toward products.

    15.9   For the reaction A 2 (g) + B(g) ⇌A(g) + AB(g), Δ n = 0 and K p = K c . We can evaluate the equilibrium expression in terms of concentration. Also, because Δn = 0, the volume terms in the expression cancel and we can use number of particles as a measure of moles and molarity. The mixture contains 2A, 4AB, and 2A 2 .

    K c = [ A ] [ AB ] [A 2 ] [B] = ( 2 ) ( 4 ) ( 2 ) ( B ) = 2 ; B = 2

    2 B atoms should be added to the diagram.

    15.10 Analyze. Given the diagram and reaction type, calculate the equilibrium constant K c .

    Plan. Analyze the contents of the cylinder. Express them as concentrations, using number of particles as a measure of moles, and V = 2 L. Write the equilibrium expression in terms of concentration and calculate K c . Solve.

    1. The mixture contains 2A 2 , 2B, 4AB. [A 2 ] = 2/2 = 1, [B] = 2/2 = 1, [AB] = 4/2 = 2.

      K c = [AB] 2 [ A 2 ] [ B ] 2 = ( 2 ) 2 ( 1 ) ( 1 ) 2 = 4

    2. A decrease in volume favors the reaction with fewer moles of gas. This reaction has two moles of gas in products and three in reactants, so a decrease in volume favors products. The number of AB (product) molecules will increase.

      Note that a change in volume does not change the value of K c . If V decreases, the number of AB molecules must increase to maintain the equilibrium value of K c .


    15.11  If temperature increases, K of an endothermic reaction increases and K of an exothermic reaction decreases. Calculate the value of K for the two temperatures and compare. For this reaction, Δn = 0 and K p = K c . We can ignore volume and use number of particles as a measure of moles and molarity. K c = [A][AB]/[A 2 ][B]

    1. 300 K, 3A, 5AB, 1A 2 , 1B; K c = (3)(5)/(1)(1) = 15
    2. 500 K, 1A, 3AB, 3A 2 , 3B; K c = (1)(3)/(3)(3) = 0.33

    K c decreases as T increases, so the reaction is exothermic.

    15.12

    1. Exothermic. In both reaction mixtures (orange and blue), [AB] decreases as T increases.
    2. In the reaction, there are fewer moles of gas in products than reactants, so greater pressure favors production of products. At any single temperature, [AB] is greater at P = y than at P = x. Because the concentration of the product, AB, is greater at P = y, P = y is the greater pressure.

    Equilibrium; The Equilibrium Constant (Sections 15.1–15.4)

    15.13 Analyze/Plan. Given the forward and reverse rate constants, calculate the equilibrium constant using Equation 15.5. At equilibrium, the rates of the forward and reverse reactions are equal. Write the rate laws for the forward and reverse reactions and use their equality to answer part (b). Solve.

    1. K c = k f k r , Equation 15.5; K c = 4.7 × 10 3 s 1 5.8 × 10 1 s 1 = 8.1 × 10 3

      [For this reaction, K p = K c = 8.1 × 10 −3 ]

    2. At equilibrium, the partial pressure of A is greater than the partial pressure of B.

      rate f = rate r ; k f [A] = k r [B]

      At equilibrium, the two rates are equal. Because k f < k r , [A] must be greater than [B] and the partial pressure of A is greater than the partial pressure of B.

    15.14

    1. The reactant I 2 (g) predominates at equilibrium. The value of K c is much less than one, which means that the denominator of the K expression is much larger than the numerator.
    2. The reverse reaction has the greater rate constant. K c = k f /k r ; if K c is small, k r is larger than k f and the reverse reaction has the greater rate constant.

    15.15 Analyze/Plan. Follow the logic in Sample Exercises 15.1 and 15.5. Solve.

    1. K c = [ N 2 O ] [ NO 2 ] [ NO ] 3
    2. K c = [ CS 2 ] [ H 2 ] 4 [ CH 4 ] [ H 2 S ] 2
    3. K c = [ CO ] 4 [ Ni(CO) 4 ]
    4. K c = [ H + ] [ F ] [ HF ]
    5. K c = [ Ag + ] 2 [ Zn 2 + ]
    6. K c = [ H + ] [ OH ]
    7. K c = [ H + ] 2 [ OH ] 2

    homogeneous: (a), (b), (d), (f), (g); heterogeneous: (c), (e)


    15.16

    1. K c = [ O 2 ] 3 [ O 3 ] 2
    2. K c = 1 [ Cl 2 ] 2
    3. K c = [ C 2 H 6 ] 2 [ O 2 ] [ C 2 H 4 ] 2 [ H 2 O ] 2
    4. K c = [ CH 4 ] [H 2 ] 2
    5. K c = [ Cl 2 ] 2 [ HCl ] 4 [ O 2 ]
    6. K c = [ CO 2 ] 16 [ H 2 O ] 18 [ O 2 ] 25
    7. K c = [ CO 2 ] 16 [ O 2 ] 25

    homogeneous: (a), (c); heterogeneous: (b), (d), (e), (f), (g)

    15.17 Analyze. Given the value of K c or K p , predict the contents of the equilibrium mixture.

    Plan. If K c or K p ≫ 1, products dominate; if K c or K p ≪ 1, reactants dominate. Solve.

    1. mostly reactants (K c ≪ 1)
    2. mostly products (K p ≫ 1)

    15.18

    1. equilibrium lies to right, favoring products (K p ≫ 1)
    2. equilibrium lies to left, favoring reactants (K c ≪ 1)

    15.19

    1. True.
    2. False. A single-headed arrow indicates that the reaction “goes to completion,” that the equilibrium constant is extremely large.
    3. False. The value of the equilibrium constant gives no information about the speed of a reaction.

    15.20

    1. False. The value of Δn is not zero.
    2. False. Equilibrium constants are not expressed with units.
    3. True. For a gas phase equilibrium, increasing pressure (by decreasing volume) favors the reaction that produces fewer moles of gas. In this equilibrium, the forward reaction has fewer moles of gas. When the forward reaction is favored, the value of K increases.

    15.21 Analyze/Plan. Follow the logic in Sample Exercise 15.2. Solve.

    PCl 3 (g) + Cl 2 (g) ⇌ PCl 5 (g), K c = 0.042. Δ n = 1 2 = 1

    K p = K c (RT) Δ n = 0.042(RT) −1 = 0.042/RT

    K p = 0.042 ( 0.08206 ) ( 500 ) = 0.001024 = 1.0 × 10 3

    15.22  SO 2 (g) + Cl 2 (g) ⇌ SO 2 Cl 2 (g), K p = 34.5. Δ n = 1 2 = 1

    K p = K c (RT) Δ n ; 34.5 = K c (RT) −1 = K c /RT;

    K c = 34.5 RT = 34.5(0.08206)(303) = 857.81 = 858


    15.23 Analyze. Given K c for a chemical reaction, calculate K c for the reverse reaction.

    Plan. Evaluate which species are favored by examining the magnitude of K c . The equilibrium expressions for the reaction and its reverse are the reciprocals of each other, and the values of K c are also reciprocal. Solve.

    1. For the reaction as written, K c < 1, which means that reactants are favored. At this temperature, the equilibrium favors NO and Br 2 .
    2. K c ( forward) = [ NOBr ] 2 [ NO ] 2 [ Br 2 ] = 1.3 × 10 2

      K c ( reverse) = [ NO] 2 [Br 2 ] [ NOBr] 2 = 1 1.3 × 10 2 = 76.92 = 77

    3. K c2 ( reverse) = [ NO] [Br 2 ] 1 / 2 [ NOBr] = ( K c ( reverse)) 1/2 = ( 76.92 ) 1 / 2 = 8.8

    15.24  2 H 2 (g) + S 2 (g) ⇌ 2 H 2 S(g), K c = 1.08 × 10 7 at 700 °C. Δn = 2 – 3 = −1.

    1. K p = K c (RT) Δ n ; T = 700 °C + 273 = 973 K.

      K p = 1.08 × 10 7 ( RT ) 1 = 1.08 × 10 7 ( 0.08206 ) ( 973 ) = 1.35 × 10 5

    2. Mostly H 2 S. Both K p and K c are much greater than one, so the product, H 2 S, is favored at equilibrium.
    3. H 2 (g) + ½ S 2 (g) ⇌ H 2 S(g); K c2 = [H 2 S ] [H 2 ] [S 2 ] 1 / 2

      K c2 = (K c ) 1/2 = (1.08 × 10 7 ) 1/2 = 3.29 × 10 3

      K p2 = (K p ) 1/2 = (1.35 × 10 5 ) 1/2 = 368

    15.25 Analyze. Given K p for a reaction, calculate K p for a related reaction.

    Plan. The algebraic relationship between the K p values is the same as the algebraic relationship between equilibrium expressions.

    Solve. K p = P SO 3 P SO 2 × P O 2 1 / 2 = 1.85

    1. K p = P SO 2 × P O 2 1 / 2 P SO 3 = 1 1.85 = 0.541
    2. K p = P SO 3 2 P SO 2 2 × P O 2 = ( 1.85 ) 2 = 3.4225 = 3.42
    3. K p = K c (RT) Δ n ; Δn = 2 – 3 = –1; T = 1000 K

      K p = K c (RT) −1 = K c /RT; K c = K p (RT)

      K c = 3.4225(0.08206)(1000) = 280.85 = 281


    15.26 K p = P HCl 4 × P O 2 P Cl 2 2 × P H 2 O 2 = 0 .0752

    1. K p = P Cl 2 2 × P H 2 O 2 P HCl 4 × P O 2 = 1 0 .0752 = 13.298 = 13.3
    2. K p = P HCl 2 × P O 2 1/2 P Cl 2 × P H 2 O = ( 0.0752 ) 1 / 2 = 0.2742 = 0.274
    3. K p = K c (RT) Δ n ; Δn = 2.5 – 2 = 0.5; T = 480 °C + 273 = 753 K

      K p = K c (RT) 1/2 , K c = K p /(RT) 1/2 = 0.2742/[0.08206 × 753] 1/2 = 0.03488 = 0.0349

    15.27 Analyze/Plan. Follow the logic in Sample Exercise 15.4. Solve.

    CoO(s) + H 2 (g) ⇌ Co(s) + H 2 O(g)         K 1 = 67

    Co(s) + CO 2 (g) ⇌ CoO(s) + CO(g)            K 2 = 1/490

    CoO(s) + H 2 (g) + Co(s) + CO 2 (g) ⇌ Co(s) + H 2 O (g) + CoO(s) + CO(g)

    H 2 (g) + CO 2 (g) ⇌ H 2 O(g)+ CO(g)

    K c = K 1 × K 2 = 67 × 1 490 = 0.1367 = 0.14

    15.28        2NO(g) + Br 2 (g) ⇌ 2NOBr(g)           K 1 = 2.0

    N 2 (g) + O 2 (g) ⇌ 2NO(g)         K 2 = 1 2.1 × 10 30

    2NO(g) + Br 2 (g) + N 2 (g) + O 2 (g) ⇌ 2NOBr(g) + 2NO(g)

    N 2 (g) + O 2 (g) + Br 2 (g) ⇌ 2NOBr(g)

    K c = K 1 × K 2 = 2.0 × 1 2.1 × 10 30 = 9.524 × 10 31 = 9.5 × 10 31

    15.29 Analyze/Plan. Follow the logic in Sample Exercise 15.5. Solve.

    1. K p = P O 2
    2. K c = [Hg(solv)] 4 [O 2 (solv)]

    15.30

    1. K p = 1 / P SO 2
    2. K c = [Na 2 SO 3 ] [Na 2 O ] [SO 2 ]

    Calculating Equilibrium Constants (Section 15.5)

    15.31 Analyze/Plan. Calculate molarity of reactants and products. Follow the logic in Sample Exercise 15.7 using concentrations rather than pressures. Solve.

    [ CH 3 OH ] = 0.0406 mol 2.00 L = 0.0203 M

    [ CO ] = 0.170 mol CO 2.00 L = 0.0850 M ; [ H 2 ] = 0.302 mol H 2 2.00 L = 0.151 M

    K c = [ CH 3 OH ] [ CO ] [ H 2 ] 2 = 0.0203 ( 0.0850 ) ( 0.151 ) 2 = 10.4743 = 10.5


    15.32 K c = [ H 2 ] [I 2 ] [HI ] 2 = ( 4.79 × 10 4 ) ( 4.79 × 10 4 ) ( 3.53 × 10 3 ) 2 = 0.018413 = 0.0184

    15.33 Analyze/Plan. Follow the logic in Sample Exercise 15.7. Solve.

    1. 2NO(g) + Cl 2 (g) ⇌ 2NOCl(g)

      K p = P NOCl 2 P NO 2 × P Cl 2 = ( 0.28 ) 2 ( 0.095 ) 2 ( 0.171 ) = 50.80 = 51

    2. K p = K c (RT) Δ n ; Δn = 2 – 3 = –1; K p = K c (RT) −1 = K c /(RT)

      K c = K p (RT) = 50.80(0.08206 × 500) = 2.1 × 10 3

    15.34

    1. K p = P PCl 5 P PCl 3 × P Cl 2 = 1.30 atm 0 .124 atm × 0.157 atm = 66 .8
    2. Because K p > 1, products (the numerator of the K p expression) are favored over reactants (the denominator of the K p expression).
    3. K p = K c (RT) Δ n ; Δn = 1 – 2 = –1; K p = K c (RT) −1 = K c /(RT)

      K c = K p (RT) = 66.8(0.08206 × 450) = 2.5 × 10 3

    15.35 Analyze/Plan. Follow the logic in Sample Exercise 15.8. Because the container volume is 1.0 L, mol = M . Solve.

    1. First calculate the change in [NO], 0.062 – 0.10 = –0.038 = –0.04 M . From the stoichiometry of the reaction, calculate the changes in the other pressures. Finally, calculate the equilibrium pressures.
      2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g)
      initial 0.10 M 0.050 M 0 M 0.10 M
      change –0.038 M –0.038 M +0.019 M +0.038 M
      equil. 0.062 M 0.012 M 0.019 M 0.138 M

      Strictly speaking, the change in [NO] has two decimal places and thus one sig fig. This limits equilibrium pressures to one sig fig for all but H 2 O, and K c to one sig fig. We compute the extra figures and then round.

    2. K c = [ N 2 ] [ H 2 O ] 2 [NO] 2 [ H 2 ] 2 = ( 0.019 ) ( 0.138 ) 2 ( 0.062 ) 2 ( 0.012 ) 2 = ( 0.02 ) ( 0.14 ) 2 ( 0.06 ) 2 ( 0.01 ) 2 = 653.7 = 7 × 10 2

    15.36

    1. Calculate the initial concentrations of H 2 (g) and Br 2 (g) and the equilibrium concentration of H 2 (g). M = mol/L.

      [ H 2 ] init = 1.374 g H 2 × 1 mol H 2 2.0159 g H 2 × 1 2.00 L = 0.34079 = 0.341 M

      [ Br 2 ] = 70.31 g Br 2 × 1 mol Br 2 159.81 g Br 2 × 1 2.00 L = 0.21998 = 0.220 M

      [ H 2 ] equil = 0.566 g H 2 × 1 mol H 2 2.0159 g H 2 × 1 2.00 L = 0.14038 = 0.140 M

      H 2 (g) + Br 2 (g) 2HBr(g)
      initial 0.34079 M 0.21998 M 0
      change –0.20041 M –0.20041 M +2(0.20041) M
      equil. 0.14038 M 0.01957 M 0.40082 M

      The change in H 2 is (0.34079 – 0.14038 = 0.20041 = 0.200). The changes in [Br 2 ] and [HBr] are set by stoichiometry, resulting in the equilibrium concentrations shown in the table.


    1. K c = [ HBr ] 2 [ H 2 ] [ Br 2 ] = ( 0.40082 ) 2 ( 0.14038 ) ( 0.01957 ) = ( 0.401 ) 2 ( 0.140 ) ( 0.020 ) = 58.48 = 58

      The equilibrium concentration of Br 2 has 3 decimal places and 2 sig figs, so the value of K c has 2 sig figs.

    15.37 Analyze/Plan. Follow the logic in Sample Exercise 15.8, using partial pressures, rather than concentrations. Solve.

    1. P = nRT/V; P CO 2 = 0.2000 mol × 500 K 2 .000L × 0.08206 L-atm mol-K = 4.1030 = 4.10 atm

      P H 2 = 0.1000 mol × 500 K 2 .000L × 0.08206 L-atm mol-K = 2.0515 = 2.05 atm

      P H 2 O = 0.1600 × 500 K 2 .000L × 0.08206 L-atm mol-K = 3.2824 = 3.28 atm

    2. The change in P H 2 O is 3.51 – 3.28 = 0.2276 = 0.23 atm. From the reaction stoichiometry, calculate the change in the other pressures and the equilibrium pressures.
      CO 2 (g) + H 2 (g) CO(g) + H 2 O(g)
      initial 4.10 atm 2.05 atm 0 atm 3.28 atm
      change –0.23 atm –0.23 atm +0.23 +0.23 atm
      equil. 3.87 atm 1.82 atm 0.23 atm 3.51 atm
    3. K p = P CO × P H 2 O P CO 2 × P H 2 = ( 0.23 ) ( 3.51 ) ( 3.87 ) ( 1.82 ) = 0.1146 = 0.11

      Without intermediate rounding, equilibrium pressures are P H 2 O = 3.51 , P CO = 0.2276 , P H 2 = 1.8239 , P CO 2 = 3.8754 and K p = 0.1130 = 0.11 , in good agree-ment with the value above.

    4. K p = K c (RT) Δ n ; Δn = 2 – 2 = 0; K p = K c (RT) 0 ; K c = K p = 0.11

    15.38

    1. N 2 O 4 (g) 2NO 2 (g)
      initial 1.500 atm 1.000 atm
      change +0.244 atm –0.488 atm
      equil. 1.744 atm 0.512 atm

      The change in P NO 2 is (1.000 – 0.512) = –0.488 atm, so the change in P N 2 O 4 is +(0.488/2) = +0.244 atm.

    2. K p = P NO 2 2 P N 2 O 4 = ( 0.512 ) 2 ( 1.744 ) = 0.1503 = 0.150
    3. K p = K c (RT) Δ n ; Δn = 2 – 1 = 1; K p = K c (RT) 1 = K c (RT)

      K c = K p /(RT) = 0.1503/(0.08206 × 298) = 6.15 × 10 −3

    15.39 Analyze/Plan. Follow the logic in Sample Exercise 15.8. m M = 10 −3 M

    X(aq) + Y(aq) XY(aq)
    initial 1.0 m M 1.0 m M 0
    change –0.80 m M –0.80 m M +0.80 m M
    equil. 0.20 m M 0.20 m M 0.80 m M

    K c = [ XY ] [ X ] [ Y ] = ( 0.80 × 10 3 ) ( 0.20 × 10 3 ) ( 0.20 × 10 3 ) = 2.0 × 10 4

    15.40  The initial concentrations of drug candidate and protein are the same in the two experiments, and the two reactions have the same stoichiometry. At equilibrium, the concentration of B-protein complex is greater than the concentration of A-protein complex, so drug B is the better choice for further research. Calculation of equilibrium constants for the two reactions confirms this conclusion.

    A(aq) + protein(aq) A-protein(aq)
    initial 2.00 × 10 −6 m M 1.50 × 10 −6 m M 0
    change –1.00 × 10 −6 m M –1.00 × 10 −6 m M +1.00 × 10 −6 m M
    equil. 1.00 × 10 −6 m M 0.50 × 10 −6 m M 1.00 × 10 −6 m M

    K c = [ A-protein ] [ A ] [ protein ] = ( 1.00 × 10 6 ) ( 1.00 × 10 6 ) ( 0.50 × 10 6 ) = 2.0 × 10 6

    B(aq) + protein(aq) B-protein(aq)
    initial 2.00 × 10 −6 m M 1.50 × 10 −6 m M 0
    change –1.40 × 10 −6 m M –1.40 × 10 −6 m M +1.40 × 10 −6 m M
    equil. 0.60 × 10 −6 m M 0.10 × 10 −6 m M 1.40 × 10 −6 m M

    K c = [ B protein ] [ B ] [ protein ] = ( 1.40 × 10 6 ) ( 0.60 × 10 6 ) ( 0.10 × 10 6 ) = 2.3 × 10 7


    Applications of Equilibrium Constants (Section 15.6)

    15.41

    1. If Q c < K c , the reaction will proceed in the direction of more products, to the right.
    2. If Q c = K c , the system is in equilibrium; the concentrations used to calculate Q must be equilibrium concentrations.

    15.42

    1. If Q c > K c , the reaction will proceed in the direction of more reactants, to the left.
    2. Q c = 0 if the concentration of any product is zero.

    15.43 Analyze/Plan. Follow the logic in Sample Exercise 15.9. We are given molarities, so we calculate Q directly and decide on the direction to equilibrium. Solve.

    K c = [ CO ] [ Cl 2 ] [ COCl 2 ] = 2.19 × 10 10 at 100 ° C

    1. Q = ( 3.3 × 10 6 ) ( 6.62 × 10 6 ) ( 2.00 × 10 3 ) = 1.1 × 10 8 ; Q > K

      The reaction will proceed left to attain equilibrium.

    2. Q = ( 1.1 × 10 7 ) ( 2.25 × 10 6 ) ( 4.50 × 10 2 ) = 5.5 × 10 12 ; Q < K

      The reaction will proceed right to attain equilibrium.

    3. Q = ( 1.48 × 10 6 ) 2 ( 0.0100 ) = 2.19 × 10 10 ; Q = K

      The reaction is at equilibrium.

    15.44  Calculate the reaction quotient in each case, compare with

    K p = P NH 3 2 P N 2 × P H 2 3 = 4.51 × 10 5

    1. Q = ( 98 ) 2 ( 45 ) ( 55 ) 3 = 1.3 × 10 3

      Because Q > K p , the reaction will shift toward reactants to achieve equilibrium.

    2. Q = ( 57 ) 2 ( 143 ) ( 0 ) 3 =

      Because Q > K p , reaction must shift toward reactants to achieve equilibrium. There must be some H 2 present to achieve equilibrium. In this example, the only source of H 2 is the decomposition of NH 3 .

    3. Q = ( 13 ) 2 ( 27 ) ( 82 ) 3 = 1.1 × 10 5

      Q is only slightly less than K p , so the reaction will shift slightly toward products to achieve equilibrium.


    15.45 Analyze/Plan. We are given concentrations, so write the K c expression and solve for [Cl 2 ]. Change molarity to partial pressure using the ideal gas equation and the definition of molarity. Solve.

    K c = [ SO 2 ] [ Cl 2 ] [ SO 2 Cl 2 ] ; [ Cl 2 ] = K c [ SO 2 Cl 2 ] [SO 2 ] = ( 0.078 ) ( 0.108 ) 0.052 = 0.16200 = 0.16 M

    PV = nRT, P = n V RT; n V = M ; P = M RT; T = 100 ° C + 273 = 373 K

    P Cl 2 = 0.16200 mol L × 0.08206 L-atm mol-K × 373 K = 4.959 = 5.0 atm

    Check. K c = ( 0.052 ) ( 0.162 ) ( 0.108 ) = 0.078. Our values are self-consistent.

    15.46 K p = P SO 3 2 P SO 2 2 × P O 2 ; P SO 3 = ( K p × P SO 2 2 × P O 2 ) 1 / 2 = [ ( 0.345 ) ( 0.135 ) 2 ( 0.455 ) ] 1 / 2 = 0.0535 atm

    15.47 Analyze/Plan. Write the equilibrium constant expression. In each case, change masses to molarities, solve for the equilibrium molarity of the desired component, and calculate mass of that substance present at equilibrium. Solve.

    K c = [ Br ] 2 [ Br 2 ] = 1.04 × 10 3

    [ Br 2 ] = 0.245 g Br 2 0.200 L × 1 mol Br 2 159.8 g Br 2 = 0.007666 = 0.00767 M

    [Br] = (K c [Br 2 ]) 1/2 = [(1.04 × 10 −3 )(0.007666)] 1/2 = 0.002824 = 0.00282 M

    0.002824 mol Br L × 0.200 L × 79.90 g Br mol = 0.0451 g Br(g)

    Check. K c = (0.002824) 2 /(0.007666) = 1.04 × 10 −3

    15.48 K c = [ HI ] 2 [H 2 ] [I 2 ] = 55.3 ; [ HI ] = ( K c [ H 2 ] [ I 2 ] ) 1 / 2

    [ H 2 ] = 0.056 g H 2 2.00 L × 1 mol H 2 2.016 g H 2 = 0.01389 = 0.014 M

    [ I 2 ] = 4.36 g I 2 2.00 L × 1 mol I 2 253 .8 gI 2 = 0.008589 = 0.00859 M

    [HI] = [(55.3) (0.01389) (0.008589)] 1/2 = 0.08122 = 0.081 M

    0.08122 M HI × 2.00 L × 127.9 g HI mol HI = 20.78 = 21 g HI

    Check. K c = ( 0.08122 ) 2 ( 0.01389 ) ( 0.008589 ) = 55.3


    15.49 Analyze/Plan. Write the equilibrium constant expression. In each case, change masses to molarities, solve for the equilibrium molarity of the desired component, and calculate mass of that substance present at equilibrium. Solve.

    K c = [ I ] 2 [I 2 ] = 3.1 × 10 5

    [ I ] = 2.67 × 10 2 g I 10.0 L × 1 mol I 126 .9 g I = 2.1040 × 10 5 = 2.10 × 10 5 M

    [I 2 ] = [ I ] 2 K c = ( 2.104 × 10 5 ) 2 3.1 × 10 5 = 1.428 × 10 5 = 1.43 × 10 5 M

    1.428 × 10 5 mol I 2 L × 10.0 L × 253.8 g I 2 mol I 2 = 0.0362 g I 2

    Check. K c = ( 2.104 × 10 5 ) 2 1.428 × 10 5 = 3.1 × 10 5

    15.50 PV = nRT; P = gRT MM V

    P SO 3 = 1.17 g SO 3 80.06 g/mol × 0.08206 L-atm mol-K × 700 K 2.00 L = 0.4197 = 0.420 atm

    P O 2 = 0.105 g O 2 32.00 g/mol × 0.08206 L-atm mol-K × 700 K 2.00 L = 0.09424 = 0.0942 atm

    K p = 3.0 × 10 4 = P SO 3 2 P SO 2 2 × P O 2 ; P SO 2 = [ P SO 3 2 / ( K p ) ( P O 2 ) ] 1 / 2

    P SO 2 = [ ( 0.4197 ) 2 / ( 3.0 × 10 4 ) ( 0.09424 ) ] 1 / 2 = 7.894 × 10 3 = 7.9 × 10 3 atm

    g SO 2 = MM PV RT = 64.06 g SO 2 mol SO 2 × mol-K 0 .08206 L-atm × 7.894 × 10 3 atm × 2.00 L 700 K = 0.01761 = 0.018 g SO 2

    Check. K p = [(0.4197) 2 /(7.894 × 10− 3 ) 2 (0.09424] = 3.0 × 10 4

    15.51 Analyze/Plan. Follow the logic in Sample Exercise 15.11. Because molarity of NO is given directly, we can construct the equilibrium table straight away. Solve.

    2NO(g) N 2 (g) + O 2 (g) K c = [ N 2 ] [ O 2 ] [ NO ] 2 = 2.4 × 10 3
    initial 0.175 M 0 0
    change –2x +x +x
    equil. 0.175 – 2x +x +x

    2.4 × 10 3 = x 2 ( 0.175 2 x) 2 ; ( 2.4 × 10 3 ) 1 / 2 = x 0 .175 2 x


    x = (2.4 × 10 3 ) 1/2 (0.175 – 2x); x = 8.573 – 97.98x; 98.98x = 8.573, x = 0.08662 = 0.087 M

    [N 2 ] = [O 2 ] = 0.087 M ; [NO] = 0.175 – 2(0.08662) = 0.00177 = 0.002 M

    Check. K c = (0.08662) 2 /(0.00177) 2 = 2.4 × 10 3

    15.52  [Br 2 ] = 0.25 mol/3.0 L = 0.08333 = 0.083 M ; [Cl 2 ] = 0.55 mol/3.0 L = 0.1833 = 0.18 M

    Br 2 (g) + Cl 2 (g) 2BrCl(g) K c = [ BrCl ] 2 [ Br 2 ] [ Cl 2 ] = 7.0
    initial 0.083 M 0.18 M 0
    change –x –x +2x
    equil. (0.083 – x) (0.18 – x) +2x

    7.0 = ( 2 x) 2 ( 0.08333 x)(0 .1833 x) ; 4x 2 = 7.0(0.0153 – 0.2666x + x 2 ); 0 = 0.1069 – 1.8662x + 3x 2

    x = 1.8662 ± ( 1.8662 ) 2 4 ( 3 ) ( 0.1069 ) 2 ( 3 ) = 0.06387 = 0.064 M

    (The 0.56 M quadratic solution is not chemically meaningful.)

    [BrCl] = 2x = 0.1277 = 0.13 M ; [Br 2 ] = 0.08333 – 0.06387 = 0.01946 = 0.019 M

    [Cl 2 ] = 0.1833 – 0.06387 = 0.1195 = 0.12 M

    Check. K c = (0.1277) 2 /(0.01946)(0.1195) = 7.0125 = 7.0

    15.53 Analyze/Plan. Write the K p expression, substitute the stated pressure relationship, and solve for P Br 2 . Solve .

    K p = P NO 2 × P Br 2 P NOBr 2

    When P NOBr = P NO , these terms cancel and P Br 2 = K p = 0.416 atm . This is true for all cases where P NOBr = P NO .

    15.54  K c = [NH 3 ][H 2 S] = 1.2 × 10 −4 . Because of the stoichiometry, equilibrium concentrations of H 2 S and NH 3 will be equal; call this quantity y. Then, y 2 = 1.2 × 10 −4 , y = 0.010954 = 0.011 M .

    15.55

    1. CaSO 4 (s) ⇌ Ca 2+ (aq) + SO 4 2– (aq) K c = [ Ca 2 + ] [ SO 4 2 ] = 2.4 × 10 5

      At equilibrium, [Ca 2+ ] = [SO 4 2– ] = x

      K c = 2.4 × 10 −5 = x 2 ; x = 4.9 × 10 −3 M Ca 2+ and SO 4 2–

    2. A saturated solution of CaSO 4 (aq) is 4.9 × 10 −3 M .

      1.4 L of this solution contain:

      4 .9 × 10 3 mol L × 1.4 L × 136.14 g CaSO 4 mol = 0.9337 = 0.94 g CaSO 4

      A bit more than 1.0 g CaSO 4 is needed to have some undissolved CaSO 4 (s) in equilibrium with 1.4 L of saturated solution.


    15.56

    1. Analyze/Plan. If only PH 3 BCl 3 (s) is present initially, the equation requires that the equilibrium concentrations of PH 3 (g) and BCl 3 (g) are equal. Write the K c expression and solve for x = [PH 3 ] = [BCl 3 ]. Solve.

      K c = [PH 3 ][BCl 3 ]; 1.87 × 10 −3 = x 2 ; x = 0.043243 = 0.0432 M PH 3 and BCl 3

    2. Because the mole ratios are 1:1:1, mol PH 3 BCl 3 (s) required = mol PH 3 or BCl 3 produced.

      0 .043243 mol PH 3 L × 0.250 L = 0.01081 = 0.0108 mol PH 3 = 0.0108 mol PH 3 BCl 3

      0 .01081 mol PH 3 BCl 3 × 151 .2 g PH 3 BCl 3 1 mol PH 3 BCl 3 = 1.6346 = 1.63 g PH 3 BCl 3

      In fact, some PH 3 BCl 3 (s) must remain for the system to be in equilibrium, so a bit more than 1.63 g PH 3 BCl 3 is needed.

    15.57 Analyze/Plan. Follow the approach in Solution 15.51. Calculate [IBr] from mol IBr and construct the equilibrium table.

    Solve. [IBr] = 0.500 mol/2.00 L = 0.250 M

    Because no I 2 or Br 2 was present initially, the amounts present at equilibrium are produced by the reverse reaction and stoichiometrically equal. Let these amounts equal x. The amount of HBr that reacts is then 2x. Substitute the equilibrium molarities (in terms of x) into the equilibrium expression and solve for x.

    I 2 + Br 2 2IBr K c = [ IBr ] 2 [ I 2 ][Br 2 ] = 280
    initial 0 M 0 M 0.250 M
    change +x M +x M –2x M
    equil. x M x M (0.250 – 2x) M

    K c = 280 = ( 0.250 2 x) 2 x 2 ; taking the square root of both sides

    16.733 = 0.250 2 x x ; 16.733 x + 2x = 0.250 ; 18.733 x = 0.250

    x = 0.013345 = 0.0133 M ; [I 2 ] = [Br 2 ] = 0.0133 M

    [IBr] = 0.250 – 2(0.013345) = 0.2233 = 0.223 M

    C h e c k . ( 0.2233 ) 2 ( 0.013345 ) 2 = 280. Our values are self - consistent .

    15.58  CaCrO 4 (s) ⇌ Ca 2+ (aq) + CrO 4 2– (aq) K c = [ Ca 2 + ] [ CrO 4 2 ] = 7.1 × 10 4

    At equilibrium, [Ca 2+ ] = [CrO 4 2– ] = x

    K c = 7.1 × 10 −4 = x 2 , x = 0.0266 = 0.027 M Ca 2+ and CrO 4 2–


    15.59 Analyze/Plan. Follow the logic in sample Exercise 15.11, using torr in place of M . We are given K p , so we use pressure in torr directly in the equilibrium expression.

    CH 4 (g) + I 2 (g) CH 3 I(g) + HI(g)
    initial 105.1 torr 7.96 torr 0 torr 0 torr
    change –x torr –x torr +x torr +x torr
    equil. 105.1–x torr 7.96–x torr +x torr +x torr

    K p = 2.26 × 10 4 = x 2 ( 105.1 x) ( 7.96 x) ; x 2 = 2.26 × 10 4 ( 836.6 113.1 x + x 2 )

    0.999774 x 2 + 0.02555 x 0.18907 = 0 ; x = 0.02555 ± ( 0.02555 ) 2 4 ( 0.999774 ) ( 0.18907 ) 2 ( 0.999774 )

    x = 0.422 torr (The negative solution is not chemically meaningful.)

    at equilibirum:   P CH 3 I = P HI = 0 .422 torr;  P CH 4 = 104.7 torr ; P I 2 = 7.54 torr

    15.60

    CH 3 COOH(solv) + CH 3 CH 2 OH(solv) ⇌ CH 3 COOCH 2 CH 3 (g) + H 2 O(solv)
    initial 0.275 M 3.85 M 0 M 0 M
    change –x M –x M +x M +x M
    equil. 0.275–x M 3.85–x M +x M +x M

    K c = 6.68 = x 2 ( 0.275 x) ( 3.85 x) ; x 2 = 6.68 ( 1.059 4.125 x + x 2 )

    0 = 5. 68 x 2 27.56 x + 7.072 ; x = 27.56 ± ( 27.56 ) 2 4 ( 5.68 ) ( 7.072 ) 2 ( 5.68 ) = 0.27185 = 0.272 M

    (The 4.58 M quadratic solution is not chemically meaningful.)

    0 .27185 mol ethyl acetate L × 15.0 L × 88 .10 g ethyl acetate mol = 359.25 = 359 g ethyl acetate

    LeChâtelier’s Principle (Section 15.7)

    15.61 Analyze/Plan. Follow the logic in Sample Exercise 15.12. Solve.

    1. Shift equilibrium to the right; more SO 3 (g) is formed, the amount of SO 2 (g) decreases.
    2. Heating an exothermic reaction decreases the value of K. More SO 2 and O 2 will form, the amount of SO 3 will decrease. This is fundamentally different than shifting the relative amounts of reactants and products to maintain K; here, the equilibrium position itself changes.
    3. Because Δn = –1, a change in volume will affect the equilibrium position and favor the side with more moles of gas. The amounts of SO 2 and O 2 increase and the amount of SO 3 decreases; equilibrium shifts to the left.
    4. No effect. Speeds up the forward and reverse reactions equally.

    1. No effect. The noble gas does not appear in the equilibrium expression; the partial pressures of reactants and products do not change upon addition of a noble gas.
    1. Shift equilibrium to the right; amounts of SO 2 and O 2 decrease.

    15.62  4 NH 3 (g) + 5 O 2 (g) ⇌ 4 NO(g) + 6 H 2 O(g)

    1. increase [NH 3 ], increase yield NO
    2. increase [H 2 O], decrease yield NO
    3. decrease [O 2 ], decrease yield NO
    4. decrease container volume, decrease yield NO (fewer moles gas in reactants)
    5. add catalyst, no change
    6. increase temperature, decrease yield NO (reaction is exothermic)

    15.63 Analyze/Plan. Given certain changes to a reaction system, determine the effect on K p , if any. Only changes in temperature cause changes to the value of K p . Solve.

    1. no effect
    2. no effect
    3. no effect
    4. increase equilibrium constant
    5. no effect

    15.64

    1. The reaction must be endothermic (+ΔH) if heating increases the fraction of products.
    2. There must be more moles of gas in the products if increasing the volume of the vessel increases the fraction of products.

    15.65 Analyze/Plan. Use Hess’s law, Δ H ° = Σ Δ H f o products Σ Δ H f o reactants, to calculate ΔH o . According to the sign of ΔH°, describe the effect of temperature on the value of K. According to the value of Δn, describe the effect of changes to container volume. Solve.

    1. Δ H ° = Δ H f o NO 2 (g) + Δ H f o N 2 O(g) 3 Δ H f o NO(g)

      ΔH° = 33.84 kJ + 81.6 kJ – 3(90.37 kJ) = –155.7 kJ

    2. Because the reaction is exothermic, the equilibrium constant will decrease with increasing temperature.
    3. A change in volume at constant temperature will affect the fraction of products in the equilibrium mixture because Δn does not equal zero. An increase in container volume would favor reactants, whereas a decrease in volume would favor products.

    15.66

    1. Δ H ° = Δ H f o CH 3 OH(g) Δ H f o CO(g) 2 Δ H f o H 2 (g)

      = –201.2 kJ – (–110.5 kJ) – 0 kJ

      = –90.7 kJ

    2. The reaction is exothermic; an increase in temperature would decrease the value of K and decrease the yield. A low temperature is needed to maximize yield.
    3. Increasing total pressure would increase the partial pressure of each gas, shifting the equilibrium toward products. The extent of conversion to CH 3 OH increases as the total pressure increases.

    15.67  For this reaction, there are more moles of product gas than moles of reactant gas. An increase in total pressure increases the partial pressure of each gas, shifting the equilibrium toward reactants. An increase in pressure favors formation of ozone.

    15.68

    1. Low temperature. For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium.
    2. No. Because there are equal numbers of moles of gas in the products and reactants, the equilibrium yield of products cannot be changed by changing pressure.

    15.69

    1. Endothermic. Bond breaking is always an endothermic process.
    2. The equilibrium constant increases. For an endothermic reaction, heat is a “reactant.” An increase in temperature and heat favors the forward reaction and the value of K c increases.
    3. The forward rate constant increases by a larger amount than the reverse rate constant. If K c = k f /k r and the value of K c increases, the value of k f must increase by a greater amount than the value of k r .

    15.70  False. When the temperature of an exothermic reaction increases, the rate constants of both the forward and the reverse reactions increase, but the value of the reverse rate constant increases by a greater amount.

    Additional Exercises

    15.71

    1. Because both the forward and reverse processes are elementary steps, we can write the rate laws directly from the chemical equation.

      rate f = k f [CO][Cl 2 ] = rate r = k r [COCl][Cl]

      k f k r = [ COCl ] [ Cl ] [ CO ] [ Cl 2 ] = K

      K c = k f k r = 1.4 × 10 28 M 1 s 1 9.3 × 10 10 M 1 s 1 = 1.5 × 10 39

      For a homogeneous equilibrium in the gas phase, we usually write K in terms of partial pressures. In this exercise, concentrations are more convenient because the rate constants are expressed in terms of molarity. For this reaction, the value of K is the same regardless of how it is expressed, because there is no change in the moles of gas in going from reactants to products.

    2. Because the K is quite small, reactants are much more plentiful than products at equilibrium.

    15.72  2 A(g) ⇌ B(g), K c = 1

    [ B ] [ A ] 2 = 1, [B] = [A] 2 and [A] = [B] 1/2


    15.73  CH 4 (g) + H 2 O(g) → CO(g) + 3 H 2 (g)

    K p = P CO × P H 2 3 P CH 4 × P H 2 O ; P = g RT MM V ; T = 1000 K

    P CO = 8.62 g 28 .01 g/mol × 0.08206 L-atm mol-K × 1000 K 5 .00 L = 5.0507 = 5.05 atm

    P H 2 = 2.60 g 2 .016 g/mol × 0.08206 L-atm mol-K × 1000 K 5 .00 L = 21.1663 = 21.2 atm

    P CH 4 = 43.0 g 16 .04 g/mol × 0.08206 L-atm mol-K × 1000 K 5 .00 L = 43.9973 = 44.0 atm

    P H 2 O = 48.4 g 18 .02 g/mol × 0.08206 L-atm mol-K × 1000 K 5 .00 L = 44.0811 = 44.1 atm

    K p = ( 5.0507 ) ( 21.1663 ) 3 ( 43.9973 ) ( 44.0811 ) = 24.6949 = 24.7

    K p = K c (RT) Δ n , K c = K p /(RT) Δ n ; Δn = 4 – 2 = 2

    K c = (24.6949)/[(0.08206)(1000)] 2 = 3.6673 × 10 −3 = 3.67 × 10 −3

    15.74 [ SO 2 Cl 2 ] = 2.00 mol 2.00 L = 1.00 M

    The change in [SO 2 Cl 2 ] = 0.56(1.00 M ) = 0.56 M

    SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) K c = [ SO 2 ] [ Cl 2 ] [ SO 2 Cl 2 ]
    initial 1.00 M 0 0
    change –0.56 M +0.56 M +0.56 M
    equil. 0.44 M +0.56 M +0.56 M
    1. K c = ( 0.56 ) 2 0.44 = 0.7127 = 0.71
    2. K p = K c (RT) Δ n ; Δn = 2 – 1 = 1;  K p = (0.7127)(0.08206)(303) = 17.7214 = 18
    3. Increase. There are more moles of gas in the products, so increasing the container volume will shift equilibrium toward products.
    4. [ SO 2 Cl 2 ] = 2.00 mol 15.00 L = 0.13333 = 0.133 M . Let x equal the change in [SO 2 Cl 2 ].

      The equilibrium concentrations are: [SO 2 Cl 2 ] = (0.13333 – x); [SO 2 ] = [Cl 2 ] = x

      K c = ( x ) 2 ( 0.13333 x ) = 0.7127 ; solving the quadratic, x = 0.1148 = 0.11 M

      (We expect the decomposition to be greater than 56%, so we must use the quadratic formula to solve for x.)

      % decomposition = (0.1148/0.1333) × 100 = 86%; the increase in volume does shift the equilibrium toward products.


    15.75

    1. Exothermic. The values of K c in the table decrease as temperature increases; the reverse reaction is favored. This is the case if heat is a “product” of the reaction.
    2. Δ H rxn o = Σ n Δ H f o (products) Σ n Δ H f o (reactants) . Be careful with coefficients, states, and signs.

      Δ H rxn o = 2 Δ H f o NH 3 ( g ) 3 Δ H f o H 2 ( g ) Δ H f o N 2 ( g )

      = 2(–46.19 kJ) – 3(0 kJ) – 0 kJ = –92.38 kJ

      Yes, the calculated value of ΔH is negative, which agrees with the prediction from part (a) that the reaction is exothermic.

    3. [NH 3 ] = 0.025 mol/1.00 L = 0.025 M
      N 2 + 3 H 2 2 NH 3 K c = [ NH 3 ] 2 [ N 2 ][H 2 ] 3 = 0.058
      initial 0 M 0 M 0.0250 M
      change +x M +3x M –2x M
      equil. x M 3x M (0.0250 – 2x) M

      K c = 0.058 = ( 0.0250 2 x) 2 ( x ) ( 3 x) 3 = ( 0.0250 2 x) 2 9 x 4 . Take the square root of both sides .

      0.24083 = 0.0250 2 x 3 x 2 ; 0.72249 x 2 + 2x 0.0250 = 0 ; x = 0 .01244 = 0 .012 M

      [NH 3 ] at equilibrium = 0.025 – 2(0.01244) = 0.000120 = 1.2 × 10 −4 M or, [NH 3 ] at equilibrium = 0.025 – 2(0.012) = 0.001 M

      In this case, intermediate rounding changes the result by an order of magnitude. However, by either method, a very small concentration of NH 3 (g) is present at equilibrium.

    15.76

    1. K p = P Br 2 × P NO 2 P NOBr 2 ; P = gRT MM × V ; T = 100 ° C + 273 = 373 K

      P Br 2 = 4.19 g 159 .8 g/mol × 0.08206 L-atm mol-K × 373 5.00 L = 0.16051 = 0.161 atm

      P NO = 3.08 g 30 .01 g/mol × 0.08206 L-atm mol-K × 373 5.00 L = 0.62828 = 0.628 atm

      P NOBr = 3.22 g NOBr 109 .9 g/mol × 0.08206 L-atm mol-K × 373 5.00 L = 0.17936 = 0.179 atm

      K p = ( 0.16051 ) ( 0.62828 ) 2 ( 0.17936 ) 2 = 1.9695 = 1.97 K p = K c (RT) Δ n , Δ n = 3 2 = 1

      K c = K p /RT = 1.9695/(0.08206)(373) = 0.064345 = 0.0643

    2. P t = P Br 2 + P NO + P NOBr = 0.16051 + 0.62828 + 0.17936 = 0.96815 = 0.968 atm

    1. All NO and Br 2 present at equilibrium came from the decomposition of the original NOBr. The mass of original NOBr is the sum of the masses of all compounds at equilibrium.

      Original g NOBr = 4.19 gBr 2 + 3.08 g NO + 3.22 g NOBr = 10.49 g

    15.77

    1. A(g) 2B(g)
      initial 0.75 atm 0
      change –0.39 atm +0.78 atm
      equil. 0.36 atm 0.78 atm

      P t = P A + P B = 0.36 atm + 0.78 atm = 1.14 atm

    2. K p = ( P B ) 2 P A = ( 0.78 ) 2 0.36 = 1.690 = 1.7
    3. Increasing the volume of the flask favors the reaction with more moles of gas. Doing the reaction in a larger flask maximizes the yield of B.

    15.78

    1. K p = P NH 3 2 P N 2 × P H 2 3 = 4.34 × 10 3 ; T = 300 ° C + 273 = 573 K

      P NH 3 = gRT MM × V = 1.05 g 17.03 g/mol × 0.08206 L-atm mol-K × 573 K 1.00 L = 2.899 = 2.90 atm

      N 2 (g)+ 3 H 2 (g) 2 NH 3 (g)
      initial 0 atm 0 atm ?
      change x 3x –2x
      equil. x atm 3x atm 2.899 atm

      (Remember, only the change line reflects the stoichiometry of the reaction.)

      K p = ( 2.899 ) 2 ( x) (3x) 3 = 4.34 × 10 3 ; 27 x 4 = ( 2.899 ) 2 4.34 × 10 3 ; x 4 = 71.725

      x = 2.910 = 2.91 atm = P N 2 ; P H 2 = 3 x = 8.730 = 8.73 atm

      g N 2 = M M × PV RT = 28.02 g N 2 mol N 2 × mol-K 0 .08206 L-atm × 2.910 atm × 1.00 L 573 K = 1.73 g N 2

      g H 2 = 2 .016 g H 2 mol H 2 × mol-K 0 .08206 L-atm × 8.730 atm × 1.00 L 573 K = 0.374 g H 2

    2. The initial P NH 3 = 2.899 atm + 2(2 .910 atm) = 8.719 = 8.72 atm

      g NH 3 = 17.03 g NH 3 mol NH 3 × mol-K 0 .08206 L-atm × 8.719 atm × 1.00 L 573 K = 3.16 g NH 3

    3. P t = P N 2 + P H 2 + P NH 3 = 2.910 atm + 8 .730 atm + 2 .899 atm = 14.54 atm

    15.79

    2IBr  ⇌ I 2 + Br 2
    initial 0.025 atm 0 0
    change –2x x x
    equil. (0.025 – 2 x) atm x x

    K p = 8.5 × 10 3 = P I 2 × P Br 2 P IBr 2 = x 2 ( 0.025 2 x ) 2 ; taking the square root of both sides

    x 0.025 2 x = ( 8.5 × 10 3 ) 1 / 2 = 0.0922 ; x = 0.0922 ( 0.025 2 x)

    x + 0.184 x = 0.002305; 1.184 x = 0.002305; x = 0.001947 = 1.9 × 10 −3

    At equilibrium, P I 2 = P Br 2 = x = 1.9 × 10 3 atm

    P IBr at equilibrium = 0.025 – 2(1.947 × 10 −3 ) = 0.02111 = 0.021 atm

    Check. K p = (0.001947) 2 /(0.02111) 2 = 8.5 × 10 −3 ; the calculated concentrations are self-consistent.

    15.80

    1. K p = 0.052; K p = K c (RT) Δ n ; Δn = 2 – 0 = 2; K c = K p /(RT) 2

      K c = 0.052/[0.08206)(333)] 2 = 6.964 × 10 −5 = 7.0 × 10 −5

    2. PH 3 BCl 3 is a solid and its concentration is taken as a constant, C.

      [ BCl 3 ] = 0.0500 g BCl 3 1.500 L × mol BCl 3 117.17 g BCl 3 = 2.8449 × 10 4 = 2.84 × 10 4 M BCl 3

      PH 3 BCl 3 PH 3 + BCl 3
      initial C 0 M 2.84 × 10 −4 M
      change +x M +x M
      equil. C +x M (2.84 × 10 −4 + x) M

    K c = [ PH 3 ] [ BCl 3 ] ; 6.964 × 10 5 = x ( 2.84 × 10 4 + x ) ; x 2 + 2.84 × 10 4 x 6.964 × 10 5 = 0 x = 2.84 × 10 4 ± [ ( 2.84 × 10 4 ) 2 4 ( 1 ) ( 6.964 × 10 5 ) ] 1 / 2 2 ( 1 ) = 0.008204 = 8.2 × 10 3 M PH 3

    Check. K c = (8.2 × 10 −3 ) ( 2.84 × 10 −4 + 8.2 × 10 −3 ) = 7.0 × 10− 5 .

    15.81 K p = P NH 3 × P H 2 S ; P t = 0.614 atm

    If the equilibrium amounts of NH 3 and H 2 S are solely due to the decomposition of NH 4 HS(s), the equilibrium pressures of the two gases are equal, and each is 1/2 of the total pressure.

    P NH 3 = P H 2 S = 0.614 atm/2 = 0.307 atm

    K p = (0.307) 2 = 0.0943


    15.82 Initial P SO 3 = gRT MM V = 0 .831 g 80 .07 g/mol × 0 .08206 L-atm mol-K × 1100 K 1.00 L = 0.9368 = 0.937 atm

    2 SO 3 2 SO 2 + O 2
    initial 0.9368 atm 0 0
    change –2 x +2x +x
    equil. 0.9368–2 x 2x x
    [equil.] 0.2104 atm 0.7264 atm 0.3632 atm

    P t = (0.9368 – 2x) + 2 x + x; 0.9368 + x = 1.300 atm; x = 1.300 – 0.9368 = 0.3632 = 0.363 atm

    K p = P SO 2 2 × P O 2 P SO 3 2 = ( 0.7264 ) 2 ( 0.3632 ) ( 0.2104 ) 2 = 4.3292 = 4.33

    K p = K c (RT) Δ n ; Δn = 3 – 2 = 1; K p = K c (RT)

    K c = K p /RT = 4.3292/[(0.08206)(1100)] = 0.04796 = 0.0480

    15.83  In general, the reaction quotient is of the form Q = P NOCl 2 P NO 2 × P Cl 2 .

    1. Q = ( 0.11 ) 2 ( 0.15 ) 2 ( 0.31 ) = 1.7

      Q > K p . The mixture is not at equilibrium. It will shift to the left and produce more reactants as it moves toward equilibrium.

    2. Q = ( 0.050 ) 2 (0 .12) 2 ( 0.10 ) = 1.7

      Q > K p . The mixture is not at equilibrium. It will shift to the left and produce more reactants as it moves toward equilibrium.

    3. Q = ( 5.10 × 10 3 ) 2 ( 0.15 ) 2 ( 0.20 ) = 5.8 × 10 3

      Q < K p . The mixture is not at equilibrium. It will shift to the right and produce more products as it moves toward equilibrium.

    15.84 K c = [ CO 2 ] = 0.0108 ; [ CO 2 ] = g CO 2 44 .01 g/mol × 1 10.0 L

    In each case, calculate [CO 2 ] and determine the position of the equilibrium.

    1. [ CO 2 ] = 4.25 g 44 .01 g/mol × 1 10.0 L = 9.657 × 10 3 = 9.66 × 10 3 M

      Q = 9.66 × 10 −3 < K c . The reaction proceeds to the right to achieve equilibrium and the amount of CaCO 3 (s) decreases.

    2. [ CO 2 ] = 5.66 g CO 2 44.01 g/mol × 1 10.0 L = 0.0129 M

      Q = 0.0129 > K c . The reaction proceeds to the left to achieve equilibrium and the amount of CaCO 3 (s) increases.

    3. 6.48 g CO 2 means [CO 2 ] > 0.0129 M ; Q > 0.0129 > K c , the amount of CaCO 3 increases.

    15.85

    CO 2 (g) + H 2 (g) CO(g) + H 2 O(g)
    initial 1.50 mol 1.50 mol 0 0
    change −x −x +x +x
    equil. (1.50 − x)mol (1.50 − x)mol x x

    Since Δn = 0, the volume terms cancel and we can use moles in place of molarity in the K expression.

    K c = 0.802 = [ CO ] [ H 2 O ] [ CO 2 ] [ H 2 O ] = x 2 ( 1.50 x ) 2

    Take the square root of both sides.

    (0.802) 1/2 = x/(1.50 – x); 0.8955(1.50 – x) = x

    1.3433 = 1.8955x, x = 0.7087 = 0.709 mol

    [CO] = [H 2 O] = 0.7087 mol/3.00 L = 0.236 M

    [CO 2 ] = [H 2 ] = (1.50 – 0.709)mol/3.00 L = 0.264 M

    15.86

    1. [CO 2 ] = 25 .0 g CO 2 44 .01 g/mol × 1 3.00 L = 0.18935 = 0.189 M
      C(s) + CO 2 (g) 2CO(g)
      initial excess 0.189 0
      change −x −x +2x
      equil. 0.189− x +2x

      K c = 1.9 = [CO] 2 [CO 2 ] = ( 2 x ) 2 0.189 x ; 4x 2 = 1.9(0.189 − x); 4x 2 + 1.9x – 0.36 = 0.

      Solve the quadratic for x.

      x = 1.9 ± ( 1.9 ) 2 4 ( 4 ) ( 0.36 ) 2 ( 4 ) = 0.14505 = 0.15 M

      [CO] = 2x = 2(0.14505) = 0.2901 = 0.29 M

      0 .2901 mol CO L × 3.00 L × 28 .01 g CO mol = 24.38 = 24 g CO

    2. The amount of C(s) consumed is related to x. Change M to mol to g C.

      0 .14505 mol L × 3.00 L × 12.01 g = 5.226 = 5.2 g C consumed

    3. A smaller vessel at the same temperature increases the total pressure of the mixture. The equilibrium shifts to form fewer total moles of gas, which favors reactants. The yield of CO product will be smaller in a smaller vessel.
    4. The two K c values are 0.133 at 298 K and 1.9 at 1000 K. The reaction is endothermic, because K is larger at higher temperature.

    15.87 K p = P CO 2 P CO = 6.0 × 10 2

    If P CO is 150 torr, P CO 2 can never exceed 760 – 150 = 610 torr. Then Q = 610/150 = 4.1. Because this is far less than K, the reaction will shift in the direction of more product. Reduction will therefore occur.

    15.88  The anecdote tells us that increasing the volume of the reaction container, the furnace, had no effect on the amount of unreacted CO(g), the amount of CO(g) expelled. This is true for reactions that have the same number of moles of gaseous products and reactants, as this one does. It also means that K p = K c .

    15.89

    1. CCl 4 (g)  ⇌ C(s) + 2 Cl 2 (g)
      initial 2.00 atm 0 atm
      change –x atm +2 x atm
      equil. (2.00 – x) atm 2 x atm

      K p = 0.76 = P Cl 2 2 P CCl 4 = ( 2x) 2 ( 2 .00 x)

      1.52 – 0.76x = 4x 2 ; 4x 2 + 0.76x – 1.52 = 0

      Using the quadratic formula, a = 4, b = 0.76, c = –1.52

      x = 0.76 ± ( 0.76 ) 2 4 ( 4 ) ( 1.52 ) 2 ( 4 ) = 0.76 + 4.99 8 = 0.5287 = 0.53 atm

      Fraction CCl 4 reacted = x atm 2 .00 atm = 0 .5287 2 .00 = 0.264 = 26 %

    2. P Cl 2 = 2 x = 2 ( 0.5287 ) = 1.06 atm

      P CCl 4 = 2.00 x = 2.00 0.5287 = 1.47 atm

    15.90

    1. Q = P PCl 5 P PCl 3 × P Cl 2 = ( 0.20 ) ( 0.50 ) ( 0.50 ) = 0.80

      0.80 (Q) > 0.0870 (K), the reaction proceeds to the left.

    2. PCl 3 (g) + Cl 2 (g) PCl 5 (g)
      initial 0.50 atm 0.50 atm 0.20 atm
      change +x atm +x atm –x atm
      equil. (0.50 + x) atm (0.50 + x) atm (0.20 – x) atm

      (Because the reaction proceeds to the left, P PCl 5 must decrease and P PCl 3 and P Cl 2 must increase.)

      K p = 0.0870 = ( 0.20 x) ( 0.50 + x) (0 .50 + x) ; 0.0870 = (0 .20 x) ( 0.250 + 1.00 x + x 2 )

      0.0870(0.250 + 1.00 x + x 2 ) = 0.20 – x; –0.17825 + 1.0870 x + 0.0870 x 2 = 0


      x = 1.0870 ± ( 1.0870 ) 2 4 ( 0.0870 ) ( 0.17825 ) 2 ( 0 .0870) = 1.0870 + 1.1152 0.174 = 0.162

      P PCl 3 = ( 0.50 + 0.162 ) atm = 0.662 P Cl 2 = ( 0.50 + 0.162 ) atm = 0.662 atm

      P PCl 5 = ( 0.20 0.162 ) atm = 0.038 atm

      To two decimal places, the pressures are 0.66, 0.66, and 0.04 atm, respectively. When substituting into the K p expression, pressures to three decimal places yield a result much closer to 0.0870.

    1. Increasing the volume of the container favors the process where more moles of gas are produced, so the reverse reaction is favored and the equilibrium shifts to the left; the mole fraction of Cl 2 increases.
    1. For an exothermic reaction, increasing the temperature decreases the value of K; more reactants and fewer products are present at equilibrium and the mole fraction of Cl 2 increases.

    15.91 Analyze/Plan. Calculate the equilibrium pressures of H 2 , I 2 , and HI; use them to calculate K p .   Set up a new equilibrium table and calculate new equilibrium pressures. Solve.

    P n = RT V = 0 .08206 L-atm mol-K × 731 K 5.00 = 11 .997 atm mol

    P H 2 = P I 2 = 0.112 mol × 11 .997 atm mol = 1.344 = 1.34 atm

    P HI = 0.775 mol × 11.997 atm mol = 9.298 = 9.30 atm

    H 2 (g) + I 2 (g) ⇌ 2HI(g); K p = P HI 2 P H 2 × P I 2 = ( 9.298 ) 2 ( 1.344 ) 2 = 47.861 = 47.9

    P HI ( added) = 0.200 mol × 11 .997 atm mol = 2.3994 = 2.40 atm

    H 2 (g) + I 2 (g) 2HI(g)
    initial 1.34 atm 1.34 atm 9.30 atm + 2.40 atm
    change +x atm +x atm –2x atm
    equil. (1.34 + x) atm (1.34 + x) atm (11.70 – 2x) atm

    K p = 47.86 = ( 11.70 2 x) 2 ( 1.34 + x) 2 . Take the square root of both sides:

    6.918 = 11.70 2 x 1.34 + x ; 9.270 + 6.918 x = 11.70 2 x; 8 .918 x = 2 .430 ; x = 0.27248 = 0.272

    P H 2 = P I 2 = 1.34 + 0.272 = 1.612 = 1.61 atm; P HI = 11.70 2 ( 0.272 ) = 11.156 = 11.16 atm

    Check. ( 11.156 ) 2 ( 1.612 ) 2 = 47.89 = 47.9


    15.92

    1. Because the volume of the vessel = 1.00 L, mol = M . The reaction will proceed to the left to establish equilibrium.
      A(g) + 2B(g) 2 C(g)
      initial 0 M 0 M 1.00 M
      change +x M +2 x M –2 x M
      equil. x M 2 x M (1.00 – 2 x) M

      At equilibrium, [C] = (1.00 – 2x) M , [B] = 2x M .

    2. x must be less than 0.50 M (so that [C], 1.00 –2 x, is not less than zero).
    3. K c = [ C ] 2 [ A ] [ B ] 2 ; ( 1.00 2 x) 2 ( x) (2x) 2 = 0.25

      1.00 – 4x + 4x 2 = 0.25(4x) 3 ; x 3 – 4x 2 + 4x – 1 = 0

    4. image
    5. From the plot, x ≈ 0.383 M

      [A] = x = 0.383 M ; [B] = 2 x = 0.766 M

      [C] = 1.00 – 2x = 0.234 M

      Using the K c expression as a check:

      K c = 0.25 ; ( 0.234 ) 2 ( 0.383 ) ( 0.766 ) 2 = 0.24 ; the estimated values are reasonable .

    15.93 K p = P O 2 × P CO 2 P CO 2 2 1 × 10 13 ; P O 2 = ( 0.03 ) ( 1 atm) = 0.03 atm

    P CO = ( 0.002 ) ( 1 atm) = 0.002 atm; P CO 2 = ( 0.12 ) ( 1 atm) = 0.12 atm

    Q = ( 0.03 ) ( 0.002 ) 2 ( 0.12 ) 2 = 8.3 × 10 6 = 8 × 10 6

    Because Q > K p , the system will shift to the left to attain equilibrium. Thus, a catalyst that promoted the attainment of equilibrium would result in a lower CO content in the exhaust.


    15.94

    1. At 700 K: K c = k f k r = 1.8 × 10 3 M 1 s 1 6.3 × 10 2 M 1 s 1 = 0.02857 = 0.029
    2. At 800 K: K c = k f k r = 0.097 M 1 s 1 2.6 M 1 s 1 = 0.03731 = 0.037

      The value of K c increases when the temperature increases. The reaction is endothermic.

    Integrative Exercises

    15.95  Calculate the initial [IO 4− ], and then construct an equilibrium table to determine [H 4 IO 6− ] at equilibrium.

    M c × V c = M d × L d ; 0.905 M × 25.0 mL 500.0 mL = M d = 0.04525 = 0.0453 M IO 4

    IO 4 (aq) + 2H 2 O(l) ⇌ H 4 IO 6 (aq)
    initial 0.0453 M 0
    change −x +x
    equil. 0.0453 − x +x

    K c = 3.5 × 10 2 = [ H 4 IO 6 ] [ IO 4 ] = x ( 0.0453 x )

    Because K c is relatively large and [IO 4 ] is relatively small, we cannot assume x is small relative to 0.0453.

    0.035(0.04525 – x) = x; 0.001584 – 0.035 x = x; 0.001584 = 1.035 x

    x = 0.001584/1.035 = 0.001530 = 0.0015 M H 4 IO 6 at equilibrium

    15.96

    1. CoO(s) + H 2 (g) ⇌ Co(s) + H 2 O(g)        K 1 = 67

      CO(g) + H 2 O(g) ⇌ H 2 (g) + CO 2 (g)        1/K 2 = 1/0.14

      CoO(s)+H 2 (g)+CO(g)+H 2 O(g) ⇌ Co(s)+H 2 O(g)+H 2 (g)+CO 2 (g) K c = K 1 /K 2

      CoO(s) + CO(g) ⇌ Co(s) + CO 2 (g) K c = 4.8 × 10 2

    2. Based on the results of part (a), CO(g) is a stronger reducing agent than H 2 (g) at 823 K. In the first reaction above, H 2 (g) reduces CoO(s) and the value of K is 67. In the fourth reaction, CO(g) reduces CoO(s) and the value of K is 480. The much larger equilibrium constant for the fourth reaction indicates that products are more favored than the products in the first reaction.
    3. 5.00 g CoO(s) × 1 mol CoO(s) 74.932 g CoO(s) = 0.066727 = 0.0667 mol CoO(s)

      n = PV RT = 1.00 atm 0 .08206 L-atm/mol-K × 2 .5 × 10 1 L 298 K = 0 .010223 = 0 .010 mol CO


    1. CoO(s) + CO(g)  ⇌ Co(s) + CO 2 (g)
      initial 0.0667 mol 0.010 mol 0 mol 0 mol
      change –x mol –x mol x mol x mol
      equil. 0.0667 – x mol 0.0102 – x mol x mol x mol

      K c = 67 0.14 = [ CO 2 ] [CO(g)] = x 0.0102 x ; 478 .57(0 .010 x) = x

      4.7857 – 478.57 x = x; 4.7857 = 479.57 x; x = 0.009979 = 0.010 mol

      At equilibrium, mol CoO(s) = 0.0667 – 0.009979 = 0.05672 = 0.057 mol CoO(s)

      g CoO(s) = 0.05672 mol CoO × 74.932 g/mol = 4.250 = 4.3 g CoO(s)

      [Note that the equilibrium concentration of CO is essentially zero. With an equilibrium constant of 4.8 × 10 2 , the reaction “goes to completion” and CO is the limiting reactant.]

    15.97  Consider the energy profile for an exothermic reaction.

    image

    The activation energy in the forward direction, E af , equals E u and the activation energy in the reverse reaction, E ar , equals E u – ΔH. (The same is true for an endothermic reaction because the sign of ΔH is the positive and E ar < E af .) For the reaction in question,

    K = k f k r = A f e E af / RT A r e E ar / RT

    Because the ln form of the Arrhenius equation is easier to manipulate, we will consider ln K.

    ln K = ln ( k f k r ) = ln k f ln k r = E af RT + ln A f [ E ar RT + ln A r ]

    Substituting E u for E af and (E u – ΔH) for E ar

    ln K = E u RT + ln A f [ ( E u Δ H) RT + ln A r ] ; ln K = E u + ( E u Δ H) RT + ln A f ln A r

    ln K = Δ H RT + ln A f A r


    For the catalyzed reaction, E cat < E u and E af = E cat , E ar = E cat – ΔH. The catalyst does not change the value of ΔH.

    ln K cat = E cat + ( E cat Δ H) RT + ln A f ln A r ln K cat = Δ H RT + ln A f A r

    Thus, assuming A f and A r are not changed by the catalyst, ln K = ln K cat and K = K cat .

    15.98

    1. For the reaction SO 2 (l) ⇌ SO 2 (g), K p = P SO 2 . From the phase diagram, as T increases P SO 2 and K p increase. For an endothermic reaction, K increases as T increases. The phase diagram tells us that the vaporization of SO 2 (l) is an endothermic process.
    2. Read P SO 2 from the liquid-gas line on the phase diagram at 0 o C and 100 o C. Note that the pressure axis of the phase diagram is logarithmic with respect to pressure, but linear with respect to logP. In terms of logP, the axis labels would be –1, 0, 1 and 2. The logP values at 0 o C and 100 o C are approximately 0.25 and 1.4. The values of P SO 2 and K p are 10 0.25 and 10 1.4 , 1.8, and 25, respectively.
    3. It is not possible to calculate an equilibrium constant between the gas and liquid phases in the supercritical region, because they do not exist separately in this region. That is, the gas and liquid phases are indistinguishable in the supercritical region.
    4. Gases are most ideal at high temperature and low pressure. The red dot at slightly greater than 240 o C is the point where SO 2 (g) most closely approaches ideal behavior.
    5. The point near 15 o C is the one at the lowest temperature, but it is also at low pressure. In general, the closer the pressure and temperature conditions are to the point of phase transition, the less ideal the behavior of the gas (because it is nearly a liquid). This describes the point near 115 o C and 20 atm, which is at relatively high pressure and near the liquid-gas line.

    15.99

    1. The larger rate constant is k r . K c = k f /k r and the value of K c is much less than one (3.1 × 10 −4 ). The value of k r must be larger than the value of k f .
    2. K c = k f k r ; k r = k f K c = 0.27 s 1 3.1 × 10 4 = 870.97 = 8.7 × 10 2
    3. The forward reaction is bond breaking, which is always endothermic.
    4. The reverse rate constant k r will increase. An increase in temperature causes both k r and k f to increase. For an endothermic reaction, we expect the value of K c to increase with an increase in temperature, so the increase in k f will be greater than the increase in k r .

    15.100

    1. H 2 O(l) ⇌ H 2 O(g); K p = P H 2 O
    2. At 30°C, the vapor pressure of H 2 O(l) is 31.82 torr. K P = P H 2 O = 31.82 torr

      K p = 31.82 torr × 1 atm/760 torr = 0.041868 = 0.04187 atm


    1. From part (b), the value of K p is the vapor pressure of the liquid at that temperature. By definition, vapor pressure = atmospheric pressure = 1 atm at the normal boiling point. K p = 1 atm

    15.101

    1. VSEPR indicates that each O atom has four electron domains about it and thus adopts tetrahedral geometry. One O atom has two covalent bonds to H and two hydrogen bonds to H atoms on the second water molecule. The O atom on the second water molecule has two nonbonding electron pairs. The water dimer is not symmetrical.
      image
    2. Hydrogen-bonding is the intermolecular interaction involved in water dimer formation.
    3. Water dimer formation is exothermic, because the value of K decreases as temperature increases.

    15.102  The O 2 -binding reaction occurs in aqueous solution, so we will write a K c expression. The amount of O 2 (g) will appear as a pressure. By convention, reactions which involve gaseous and aqueous substances have mixed equilibrium expressions written in terms of both pressures and molar concentrations.

    K c = [Hb-(O 2 ) 4 ] P O 2 4 × [ Hb ]

    The P50 value is the partial pressure at which 50% of the hemoglobin is saturated with O 2 (g). At this partial pressure, the concentrations of O 2 -bound hemoglobin and free hemoglobin are equal, [Hb – (O 2 ) 4 ] = [Hb]. Substitute the two P50 values into the K c expression and compare the values for fetal and adult hemoglobin.

    at P50, K cF = 1 P O 2 4 = 1 19 4 = 7.7 × 10 6 ; K cA = 1 P O 2 4 = 1 26.8 4 = 1.94 × 10 6

    Comparing the two values, K cF /K cA = 7.7 × 10− 6 /1.94 × 10− 6 ≈ 4. The equilibrium constant for O 2 -binding by fetal hemoglobin is approximately four times that by adult hemoglobin.


     

    16 Acid–Base Equilibria

    Visualizing Concepts

    16.1 Analyze .  From the structures decide which reactant fits the description of a Brønsted–Lowry (B-L) acid, a B-L base, a Lewis acid, and a Lewis base. Plan . A B-L acid is an H + donor, and a B-L base is an H + acceptor. A Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor. Solve .

    1. HCl is a B-L acid, because it donates H + during reaction. NH 3 is a B-L base, because it accepts H + during reaction.
    2. By virtue of its unshared electron pair, NH 3 is the electron pair donor and a Lewis base. HCl is the electron pair acceptor and a Lewis acid.

    16.2 Plan .  The stronger the acid, the greater the extent of ionization. The stronger the acid, the weaker its conjugate base. In an acid–base reaction, equilibrium will favor the side with the weaker acid and base. Solve .

    1. HY is stronger than HX. Starting with six HY molecules, four are dissociated; of six HX molecules, only two are dissociated. Because it is dissociated to a greater extent, HY is the stronger acid.
    2. If HY is the stronger acid, Y is the weaker base and X is the stronger base.
    3. HX and Y , the reactants, are the weaker acid and base. Equilibrium lies to the left, and K c < 1.

    16.3

    1. True. Solution A is the color of methyl orange in an acidic solution.
    2. False. Methyl orange turns yellow at a pH slightly greater than 4, so solution B could be at any pH greater than 4.
    3. True. The basic color of any indicator occurs at higher pH than the acidic color does.

    16.4 Analyze/Plan. The pH reading on the meter will identify the solution. Use the definition of pH or pOH to calculate the concentration of the solution. Solve .

    1. KOH(aq). The pH meter reads 12.08, indicating that the solution is strongly basic. KOH(aq) is the only choice that is a base.
    2. pOH = 14.00 – pH; pOH = 14.00 – 12.08 = 1.92

      pOH = –log [OH ]; [OH ] = 10 −pOH ; [OH ] = 10 −1.92 = 0.01202 = 0.012 M

    3. The pH scale for aqueous solutions is based on the value of K w for water. K w is an equilibrium constant whose value depends on temperature. If temperature is significantly different from 25 o C, the value of K w is different from 1 × 10 −14 and relationships like {pH + pOH = 14} do not hold true.

    16.5 Plan .  Strong acids are completely ionized. The acid that is least ionized is the weakest and has the smallest K a value. At equal concentrations, the weakest acid has the smallest [H + ] and highest pH. Solve .

    1. HY is a strong acid. There are no neutral HY molecules in the solution, only H + cations and Y anions.
    2. HX has the smallest K a value. It has the most neutral acid molecules and fewest ions.
    3. HX has the fewest H + ions and, therefore, the highest pH.

    16.6 Analyze/Plan. We are shown a plot of concentration solution versus [H + ]. Use the definition of acids and bases along with relationships between total solute concentration and [H + ] for strong or weak acids and bases to answer the questions. Solve .

    1. The substance is a strong acid. It is an acid because [H + ] increases as solute concentration increases. It is a strong acid because [H + ] is directly proportional to solute concentration. This is not true for weak acids.
    2. Yes. A strong acid is completely ionized in aqueous solution, so [H + ] = solute concentration = 0.18 M . pH = –log[H + ] = –log(0.18) = 0.74.
    3. No. In pure water, there is a finite concentration of H + (aq) and [OH ], owing to the autoionization of water.

    16.7  Statement (e) is true. Only line C shows the trend in percent ionization of a weak acid when initial acid concentration increases. Statement (f) is false because the value of K a , and thus the strength of a weak acid, does not change with acid concentration.

    16.8 Analyze/Plan . Write the formula of each molecule and compare them to the entries in Tables 16.2 and 16.4. Select the molecule that fits the definition of an acid and the one that fits the definition of a base. Solve .

    1. Molecule A acts as a base. It is hydroxyl amine, NH 2 OH, an entry in Table 16.4. Molecule A is an H + acceptor because of the nonbonded electron pair on the N atom of the amine (–NH 2 ) group, not because it contains an –OH group. The presence of an –OH group in an organic molecule does not mean that the molecule is a base.
    2. Molecule B acts as an acid; it is formic acid, HCOOH. It is similar to CH 3 COOH, an entry in Table 16.2. The H atom bonded to O is ionizable and HCOOH is an H + donor. In general, organic molecules that contain a carboxyl (–COOH) group are acids.
    3. Molecule C acts as neither an acid nor a base; it is methanol, CH 3 OH. In organic molecules, the –OH functional group is an alcohol. The H atom bonded to O is not ionizable, and the –OH group does not dissociate in aqueous solution. An alcohol is neither an acid nor a base.

    16.9

    1. Basic. Because of the amine group (N, with a lone electron pair, bound to two C atoms and an H atom), we expect phenylephrine solution to be basic. [Note that –OH bound to a benzene ring is very weakly acidic (K a for phenol is 1.3 × 10 −10 ), but probably not acidic enough to dominate the acid/base properties of the molecule.]

    1. Phenylephrine is a neutral molecule, whereas phenylephrine hydrochloride is a salt. It is the product when HCl reacts with phenylephrine. The cation in the salt has an additional H atom bound to the N of phenylephrine. The anion is chloride.
    1. Acidic. The cation of phenylephrine hydrochloride is the conjugate acid of phenylephrine, so a solution of the salt is acidic. (Chloride anion is a negligible base.)

    16.10  Diagram C best represents an aqueous solution of NaF; it contains mostly Na + and F , along with a few HF molecules and OH ions. The HF and OH are present because F is a weak Brønsted–Lowry base; it accepts H + from a water molecule, producing HF and OH . The solution is basic because it contains OH .

    16.11 Plan .  Evaluate the structures to determine if the molecules are binary acids or oxyacids. Consider the trends in acid strength for both classes of acids. Solve .

    1. If X is the same atom on both molecules, the molecule (b) is more acidic. The carboxylate anion, the conjugate base of this carboxylic acid, is stabilized by resonance, whereas the conjugate base of (a) is not resonance-stabilized. Stabilization of the conjugate base causes the ionization equilibrium to favor products, and (b) is the stronger acid.
    2. Increasing the electronegativity of X increases the strength of both acids. As X becomes more electronegative and attracts more electron density, the O–H bond becomes weaker and more polar. This increases the likelihood of ionization, which increases acid strength. An electronegative X group also stabilizes the anionic conjugate bases by delocalizing the negative charge. This causes the ionization equilibrium to favor products, and the values of K a to increase.

    16.12 Analyze/Plan . Consider the definitions of Lewis, Arrhenius, and Brønsted-Lowry acids and bases. Solve .

    1. PCl 4 + , Lewis acid; Cl , Lewis base. There are no hydrogen atoms or ions in any of the reactants or products. Only the Lewis definition applies.
    2. NH 3 , Lewis base; BF 3 , Lewis acid. No hydrogen ions are transferred in the reaction. Only the Lewis definition applies. Also, from Lewis structures, we know that NH 3 has a nonbonded electron pair on N, and BF 3 is electron deficient.
    3. [Al(H 2 O) 6 ] 3+ , Brønsted-Lowry acid; H 2 O, Brønsted-Lowry base. This is an H + transfer reaction and fits the Brønsted-Lowry definition. Because the reaction occurs in aqueous solution, [Al(H 2 O) 6 ] 3+ also fits the definition of an Arrhenius acid.

    Arrhenius and Brønsted–Lowry Acids and Bases (Sections 16.1 and 16.2)

    16.13  HCl(g) + NH 3 (g)   →   NH 4 + Cl (s). HCl is the B-L (Brønsted–Lowry) acid; it donates an H + to NH 3 to form NH 4 + . NH 3 is the B-L base; it accepts the H + from HCl.

    16.14  Statement (e) is false. One type of compound that contains an –OH group and is not (usually) a Brønsted-Lowry base is an alcohol.


    16.15 Analyze/Plan .  Follow the logic in Sample Exercise 16.1. A conjugate base has one less H + than its conjugate acid. A conjugate acid has one more H + than its conjugate base. Solve .

      1. IO 3
      2. NH 3
      1. OH
      2. H 3 PO 4

    16.16  A conjugate base has one less H + than its conjugate acid. A conjugate acid has one more H + than its conjugate base.

      1. HCOO
      2. PO 4 3–
      1. HSO 4
      2. CH 3 NH 3 +

    16.17 Analyze/Plan .  Use the definitions of B-L acids and bases, and conjugate acids and bases to make the designations. Evaluate the changes going from reactant to product to inform your choices. Solve .

    B-L Acid + B-L Base Conjugate Acid + Conjugate Base
    (a) NH 4 + (aq) CN (aq) HCN(aq) NH 3 (aq)
    (b) H 2 O(l) (CH 3 ) 3 N(aq) (CH 3 ) 3 NH + (aq) OH (aq)
    (c) HCOOH(aq) PO 4 3– (aq) HPO 4 2– (aq) HCOO (aq)

    16.18

    B-L Acid + B-L Base Conjugate Acid + Conjugate Base
    (a) HBrO(aq) H 2 O(l) H 3 O + (aq) BrO–(aq)
    (b) HSO 4 (aq) HCO 3 (aq) H 2 CO 3 (aq) SO 4 2– (aq)
    (c) H 3 O + (aq) HSO 3 (aq) H 2 SO 3 (aq) H 2 O(l)

    16.19 Analyze/Plan .  Follow the logic in Sample Exercise 16.2. Solve .

    1. Acid: HSO 3 (aq) + H 2 O(l) SO 3 2– (aq) + H 3 O + (aq)
      B-L acid B-L base conj. base conj. acid
      Base: HSO 3 (aq) + H 2 O(l) H 2 SO 3 (aq) + OH (aq)
      B-L base B-L acid conj. acid conj. base
    2. H 2 SO 3 is the conjugate acid of HSO 3 .

      SO 3 2– is the conjugate base of HSO 3 .

    16.20

    1. H 2 C 6 H 7 O 5 (aq) + H 2 O(l) ⇌ H 3 C 6 H 7 O 5 (aq) + OH (aq)
    2. H 2 C 6 H 7 O 5 (aq) + H 2 O(l) ⇌ HC 6 H 7 O 5 2– (aq) + H 3 O + (aq)
    3. H 3 C 6 H 7 O 5 is the conjugate acid of H 2 C 6 H 7 O 5 .

      HC 6 H 7 O 5 2– is the conjugate base of H 2 C 6 H 7 O 5 .

    16.21 Analyze/Plan . Based on the chemical formula, decide whether the base is strong, weak, or negligible. Is it the conjugate of a strong acid (negligible base), weak acid (weak base), or negligible acid (strong base)? Also check Figure 16.4. To write the formula of the conjugate acid, add a single H and increase the particle charge by one. Solve .

    1. CH 3 COO , weak base; CH 3 COOH, weak acid
    2. HCO 3 , weak base; H 2 CO 3 , weak acid
    3. O 2– , strong base; OH , negligible acid

    1. Cl , negligible base; HCl, strong acid
    1. NH 3 , weak base; NH 4 + , weak acid

    16.22 Analyze/Plan . Based on the chemical formula, decide whether the acid is strong, weak, or negligible. Is it one of the known seven strong acids (Section 16.5)? Also check Figure 16.4. To write the formula of the conjugate base, remove a single H and decrease the particle charge by one. Solve .

    1. HCOOH, weak acid; HCOO , weak base
    2. H 2 , negligible acid; H , strong base
    3. CH 4 , negligible acid; CH 3 , strong base
    4. HF, weak acid; F , weak base
    5. NH 4 + , weak acid; NH 3 , weak base

    16.23 Analyze/Plan .  Based on the chemical formula, determine the strength of acids and bases by checking the known strong acids (Section 16.5). Recall the paradigm “The stronger the acid, the weaker its conjugate base, and vice versa.” Solve .

    1. HBr. It is one of the seven strong acids (Section 16.5).
    2. F . HCl is a stronger acid than HF, so F is the stronger conjugate base.

    16.24

    1. HClO 3 . It is one of the seven strong acids (Section 16.5). Also, in a series of oxyacids with the same central atom (Cl), the acid with more O atoms is stronger (Section 16.10).
    2. HS . H 2 SO 4 is a stronger acid than H 2 S, so HS is the stronger conjugate base. In fact, because H 2 SO 4 is one of the seven strong acids, HSO 4 is a negligible base.

    16.25 Analyze/Plan .  Acid–base equilibria favor formation of the weaker acid and base. Compare the relative strengths of the substances acting as acids on opposite sides of the reaction arrow. (Bases can also be compared; the conclusion should be the same.) Solve .

    Base + Acid Conjugate Acid + Conjugate Base
    (a) O 2– (aq) + H 2 O(l) OH (aq) + OH (aq)
    H 2 O is a stronger acid than OH , so the equilibrium lies to the right.
    (b) HS (aq) + CH 3 COOH(aq) H 2 S(aq) + CH 3 COO (aq)
    CH 3 COOH is a stronger acid than H 2 S, so the equilibrium lies to the right.
    (c) NO 2 (aq) + H 2 O(l) HNO 2 (aq) + OH (aq)
    HNO 2 is a stronger acid than H 2 O, so the equilibrium lies to the left.

    16.26

    Base + Acid Conjugate Acid + Conjugate Base
    (a) OH (aq) + NH 4 + (aq) H 2 O(l) + NH 3 (aq)
    OH is a stronger base than NH 3 (Figure 16.4), so the equilibrium lies to the right.
    (b) CH 3 COO (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O(l)
    H 3 O + is a stronger acid than CH 3 COOH (Figure 16.4), so the equilibrium lies to the right.
    (c) F –(aq) + HCO 3 (aq) CO 3 2– (aq) + HF(aq)
    CO 3 2– is a stronger base than F , so the equilibrium lies to the left.

    Autoionization of Water (Section 16.3)

    16.27  Statement (ii) is correct. In pure water, the only source of H + is the autoionization reaction, which produces equal concentrations of H + and OH . As the temperature of water changes, the value of K w changes, and the pH at which [H + ] = [OH ] changes.

    16.28

    1. H 2 O(l)  ⇌  H + (aq) + OH (aq)
    2. K w = [H + ][OH ]
    3. Statement (iii) is true. If a solution is basic, it contains more OH than H + .

    16.29 Analyze/Plan .  Follow the logic in Sample Exercise 16.5. In pure water at 25 °C, [H + ] = [OH ] = 1 × 10 −7 M . If [H + ] > 1 × 10 −7 M , the solution is acidic; if [H + ] < 1 × 10 −7 M , the solution is basic. Solve .

    1. [ H + ] = K w [OH ] = 1.0 × 10 14 4.5 × 10 4 M = 2.2 × 10 11 M < 1 × 10 7 M ; basic
    2. [ H + ] = K w [OH ] = 1.0 × 10 14 8.8 × 10 9 M = 1.1 × 10 6 M > 1 × 10 7 M ; acidic
    3. [OH ] = 100[H + ]; K w = [H + ] × 100[H + ] = 100[H + ] 2 ;

      [H + ] = (K w /100) 1/2 = 1.0 × 10 −8 M < 1 × 10 −7 M ; basic

    16.30  In pure water at 25 °C, [H + ] = [OH ] = 1 × 10 −7 M . If [OH ] > 1 × 10 −7 M , the solution is basic; if [OH ] < 1 × 10 −7 M , the solution is acidic.

    1. [ OH ] = K w [H + ] = 1.0 × 10 14 0.0505 M = 1.98 × 10 13 M < 1 × 1 0 7 M ; acidic
    2. [ OH ] = K w [H + ] = 1.0 × 10 14 2.5 × 10 10 M = 4.0 × 10 5 M > 1 × 10 7 M ; basic
    3. [H + ] = 1000[OH ]; K w = 1000[OH ][OH ] = 1000[OH ] 2

      [OH ] = (K w /1000) 1/2 = 3.2 × 10 −9 M < 1 × 10 −7 M ; acidic

    16.31 Analyze/Plan .  Follow the logic in Sample Exercise 16.4. Note that the value of the equilibrium constant (in this case, K w ) changes with temperature. Solve .

    At 0 °C, K w = 1.2 × 10 −15 = [H + ][OH ]

    In pure water, [H + ] = [OH ]; 1.2 × 10 −15 = [H + ] 2 ; [H + ] = (1.2 × 10 −15 ) 1/2

    [H + ] = [OH ] = 3.5 × 10 −8 M

    16.32  K w = [D + ][OD–]; for pure D 2 O, [D + ] = [OD–]; 8.9 × 10 −16 = [D + ] 2 ;

    [D + ] = [OD ] = 3.0 × 10 −8 M


    The pH Scale (Section 16.4)

    16.33 Analyze/Plan .  A change of one pH unit (in either direction) is:

    Δ pH = pH 2 pH 1 = ( log[H + ] 2 log [H + ] 1 ) = log [H + ] 2 [H + ] 1 = ± 1. The antilog of +1 is 10; the antilog of –1 is 1 × 10 −1 . Thus, a ΔpH of one unit represents an increase or decrease in [H + ] by a factor of 10. Solve .

    1. ΔpH = ±2.00 is a change of 10 2.00 ; [H + ] changes by a factor of 100.
    2. ΔpH = ±0.5 is a change of 10 0.50 ; [H + ] changes by a factor of 3.2.

    16.34  [H + ] A = 250 [H + ] B . From Solution 16.33, Δ pH = log [H + ] B [H + ] A

    Δ pH = log [H + ] B 250 [H + ] B = log ( 1 250 ) = 2.40

    The pH of solution A is 2.40 pH units lower than the pH of solution B, because [H + ] A is 250 times greater than [H + ] B . The greater the [H + ], the lower the pH of the solution.

    16.35 Analyze/Plan .  At 25 °C, [H + ][OH ] = 1 × 10 −14 ; pH + pOH = 14. Use these relationships to complete the table. If pH < 7, the solution is acidic; if pH > 7, the solution is basic. Solve .

    [H + ] [OH ] pH pOH acidic or basic
    7.5 × 10 −3 M 1.3 × 10 −12 M 2.12 11.88 acidic
    2.8 × 10 −5 M 3.6 × 10 −10 M 4.56 9.44 acidic
    5.6 × 10 −9 M 1.8 × 10 −6 M 8.25 5.75 basic
    5.0 × 10 −9 M 2.0 × 10 −6 M 8.30 5.70 basic

    Check. pH + pOH = 14; [H + ][OH ] = 1 × 10 −14

    16.36

    pH pOH [H + ] [OH ] acidic or basic
    5.25 8.75 5.6 × 10 −6 M 1.8 × 10 −9 M acidic
    11.98 2.02 1.1 × 10 −12 M 9.6 × 10 −3 M basic
    9.36 4.64 4.4 × 10 −10 M 2.3 × 10 −5 M basic
    12.93 1.07 1.2 × 10 −13 M 8.5 × 10 −2 M basic

    16.37 Analyze/Plan .  Based on the pH and a new value of the equilibrium constant K w , calculate equilibrium concentrations of H + (aq) and OH (aq). The definition of pH remains pH = –log[H + ]. Solve .

    pH = 7.40; [H + ] = 10 −pH = 10 −7.40 = 4.0 × 10 −8 M

    K w = 2.4 × 10 −14 = [H + ][OH ]; [OH ] = 2.4 × 10 −14 / [H + ]

    [OH ] = 2.4 × 10 −14 / 4.0 × 10 −8 = 6.0 × 10 −7 M ; pOH = –log(6.0 × 10 −7 ) = 6.22

    Alternately, pH + pOH = pK w . At 37 °C, pH + pOH = –log(2.4 × 10 −14 )

    pH + pOH = 13.62; pOH = 13.62 – 7.40 = 6.22

    [OH ] = 10 −pOH = 10 −6.22 = 6.0 × 10 −7 M


    16.38  The pH ranges from 5.2 to 5.6; pOH ranges from (14.0–5.2 =) 8.8 to (14.0–5.6 =) 8.4.

    [H + ] = 10 −pH , [OH ] = 10 −pOH

    [H + ] = 10 −5.2 = 6.31 × 10 −6 = 6 × 10 −6 M ; [H + ] = 10 −5.6 = 2.51 × 10 −6 = 3 × 10 −6 M

    The range of [H + ] is 6 × 10 −6 M to 3 × 10 −6 M .

    [OH ] = 10 −8.8 = 1.58 × 10 −9 = 2 × 10 −9 M ; [OH ] = 10 −8.4 = 3.98 × 10 −9 = 4 × 10 −9 M

    The range of [OH ] is 2 × 10 −9 M to 4 × 10 −9 M .

    (The pH has one decimal place, so concentrations are reported to 1 sig fig.)

    16.39 Analyze/Plan. We are given the behavior of an unknown solution in two indicators. Use information from Figure 16.8 to answer the questions about the solution. Solve .

    1. Acidic. Both indicators change color at pH less than 7.
    2. The range of possible integer pH values for the solution is 4 to 6. Methyl orange changes color near pH 4 and bromthymol (or bromothymol) blue changes near pH 6.
    3. Methyl red changes color slightly above pH 5. It would help narrow the range of possible pH values.

    16.40

    1. Acidic. Bromthymol (or bromothymol) blue, which changes color at a lower pH than phenolphthalein, is yellow below pH 6, so the solution is acidic.
    2. (ii), a maximum pH. The solution is the lower-pH color for both indicators, so we know only that the maximum pH is 6.
    3. From Figure 16.8, methyl violet, thymol blue, methyl orange, and methyl red would help determine the pH of the solution more precisely. These indicators change colors at pH values from approximately 1 to 5. [One strategy would be to start at the low end of the pH range with methyl violet and work up.472]

    Strong Acids and Bases (Section 16.5)

    16.41

    1. True.
    2. True.
    3. False. A 1.0 M strong acid solution has a pH of 0.

    16.42

    1. True.
    2. True.
    3. False. Base strength should not be confused with solubility. Base strength describes the tendency of a dissolved molecule [formula unit for ionic compounds such as Mg(OH) 2 ] to dissociate into cations and hydroxide ions. Mg(OH) 2 is a strong base because each Mg(OH) 2 unit that dissolves also dissociates into Mg 2+ (aq) and OH (aq). Mg(OH) 2 is not very soluble, so relatively few Mg(OH) 2 units dissolve when the solid compound is added to water.

    16.43 Analyze/Plan .  Follow the logic in Sample Exercise 16.8. Strong acids are completely ionized, so [H + ] = original acid concentration and pH = –log[H + ]. For the solutions obtained by dilution, use the “dilution” formula, M 1 V 1 = M 2 V 2 , to calculate molarity of the acid. Solve .

    1. 8.5 × 10 −3 M HBr = 8.5 × 10 −3 M H + ; pH = –log (8.5 × 10 −3 ) = 2.07
    2. 1.52 g HNO 3 0.575 L soln × 1 mol HNO 3 63.02 g HNO 3 = 0.041947 = 0.0419 M HNO 3

      [H + ] = 0.0419 M ; pH = –log (0.041947) = 1.377

    3. M c × V c = M d × V d ; 0.250 M × 0.00500 L = ? M × 0.0500 L

      M d = 0.250 M × 0.00500 L 0.0500 L = 0.0250 M HCl

      [H + ] = 0.0250 M ; pH = –log (0.0250) = 1.602

    4. [ H + ] total = mol H + from HBr + mol H + from HCl total L solution

      [ H + ] total = ( 0.100 M HBr × 0 .0100 L) + ( 0.200 M × 0.0200 L) 0.0300 L

      [ H + ] total = 1.00 × 10 3 mol H + + 4.00 × 10 3 mol H + 0.0300 L = 0.1667 = 0.167 M

      pH = –log (0.1667 M ) = 0.778

    16.44  For a strong acid, which is completely ionized, [H + ] = the initial acid concentration.

    1. 0.0167 M HNO 3 = 0.0167 M H + ; pH = –log (0.0167) = 1.777
    2. 0.225 g HClO 3 2.00 L soln × 1 mol HClO 3 84.46 g HClO 3 = 1.332 × 10 3 = 1.33 × 10 3 M HClO 3

      [H + ] = 1.33 × 10 −3 M ; pH = –log (1.332 × 10 −3 ) = 2.875

    3. M c × V c = M d × V d ; 0.500 L = 500 mL

      1.00 M HCl × 15.00 mL HCl = M d HCl × 500 mL HCl

      M d HCl = 1.00 M × 1 5.00 mL 500 mL = 3.00 × 10 2 M HCl = 3.00 × 10 2 M H +

      pH = –log (3.00 × 10 −2 ) = 1.523

    4. [ H + ] total = mol H + from HCl + mol H + from HI total L solution ; mol = M × L

      [ H + ] total = ( 0.020 M HCl × 0 .0500 L) + (0 .010 M HI × 0 .125 L) 0.175 L

      [ H + ] total = 1.0 × 10 3 m o l H + + 1.25 × 10 3 mol H + 0.175 L = 0.01286 = 0.013 M

      pH = –log (0.01286) = 1.89


    16.45 Analyze/Plan .  Follow the logic in Sample Exercise 16.9. Strong bases dissociate completely upon dissolving. pOH = –log[OH ]; pH = 14 – pOH. Solve .

    1. Pay attention to the formula of the base to get [OH ].

      [OH ] = 2[Sr(OH) 2 ] = 2(1.5 × 10 −3 M ) = 3.0 × 10 −3 M OH

      pOH = –log (3.0 × 10 −3 ) = 2.52; pH = 14 – pOH = 11.48

    2. mol/LiOH = g LiOH/molar mass LiOH. [OH ] = [LiOH].

      2.250 g LiOH 0.2500 L soln × 1 mol LiOH 23.948 g LiOH = 0.37581 = 0.3758 M LiOH = [ OH ]

      pOH = –log (0.37581) = 0.4250; pH = 14 – pOH = 13.5750

    3. Use the dilution formula to get the [NaOH] = [OH ].

      M c × V c = M d × V d ; 0.175 M × 0.00100 L = ? M × 2.00 L

      M d = 0.175 M × 0.00100 L 2.00 L = 8.75 × 10 5 M NaOH = [ OH ]

      pOH = –log (8.75 × 10 −5 ) = 4.058; pH = 14 – pOH = 9.942

    4. Consider total mol OH from KOH and Ca(OH) 2 as well as total solution volume.

      [ OH ] total = mol OH from KOH + mol OH from Ca (OH) 2 total L soln

      [ OH ] total = ( 0.105 M × 0.00500 L) + 2(9 .5 × 10 2 M × 0.0150 L) 0.0200 L

      [ OH ] total = 0.525 × 10 3 mol OH + 2.85 × 10 3 mol OH 0.0200 L = 0.16875 = 0.17 M

      pOH = –log (0.16875) = 0.77; pH = 14 – pOH = 13.23

      (9.5 × 10 −2 M has 2 sig figs, so the [OH ] has 2 sig figs and pH and pOH have 2 decimal places.)

    16.46  For a strong base, which is completely dissociated, [OH ] = the initial base concentration. Then, pOH = –log [OH ] and pH = 14 – pOH.

    1. 0.182 M KOH = 0.182 M OH ; pOH = –log (0.182) = 0.740; pH = 14 – 740 = 13.260
    2. 3.165 g KOH 0.5000 L × 1 mol KOH 56.106 g KOH = 0.112822 = 0.1128 M = [ OH ]

      pOH = –log (0.112822) = 0.9476; pH = 14 – pOH = 13.0524

    3. M c × V c = M d × V d

      0.0105 M Ca(OH) 2 × 10.0 mL = M d Ca(OH) 2 × 500 mL

      M d Ca(OH) 2 = 0.0105 M Ca(OH) 2 × 10.0 mL 500.0 mL = 2.10 × 10 4 M Ca(OH) 2

      Ca(OH) 2 (aq)  →  Ca 2+ (aq) + 2OH (aq)

      [OH ] = 2[Ca(OH) 2 ] = 2(2.10 × 10 −4 M ) = 4.20 × 10 −4 M

      pOH = –log (4.20 × 10 −4 ) = 3.377; pH = 14 – pOH = 10.623


    1. [ OH ] total = mol OH from NaOH + mol OH from Ba(OH) 2 total L solution

      ( 8.2 × 10 3 M × 0.0400 L) + 2(0 .015 M × 0.0200 L) 0.0600 L

      [ OH ] total = 3 .28 × 10 4 mol OH + 6.0 × 10 4 mol OH 0.0600 L = 0.01547 = 0.015 M OH

      pOH = –log (0.01547) = 1.81; pH = 14 – 1.81 = 12.19

    16.47 Analyze/Plan .  pH   →   pOH   →   [OH ] = [NaOH]. Solve .

    pOH = 14 – pH = 14.00 – 11.50 = 2.50

    pOH = 2.50 = –log[OH ]; [OH ] = 10 −2.50 = 3.2 × 10 −3 M

    [OH ] = [NaOH] = 3.2 × 10 −3 M

    16.48  pOH = 14 – pH = 14.00 – 10.05 = 3.95

    pOH = 3.95 = –log[OH ]; [OH ] = 10 −3.95 = 1.122 × 10 −4 M = 1.1 × 10 −4 M

    [OH ] = 2[Ca(OH) 2 ]; [Ca(OH) 2 ] = [OH ] / 2 = 1.122 × 10 −4 M /2 = 5.6 × 10 −5 M

    Weak Acids (Section 16.6)

    16.49 Analyze/Plan .  Remember that K a = [products]/[reactants]. If H 2 O(l) appears in the equilibrium reaction, it will not appear in the K a expression, because it is a pure liquid. Solve .

    1. HBrO 2 (aq) H + ( aq) + BrO 2 ( aq); K a = [ H + ] [ BrO 2 ] [ HBrO 2 ]

      HBrO 2 (aq) + H 2 O(l) H 3 O + ( aq ) + BrO 2 ( aq ) ; K a = [ H 3 O + ] [ BrO 2 ] [ HBrO 2 ]

    2. C 2 H 5 COOH(aq) ⇌ H + ( aq) + C 2 H 5 COO ( aq); K a = [ H + ] [ C 2 H 5 COO ] [ C 2 H 5 COOH ]

      C 2 H 5 COOH(aq) + H 2 O(l) ⇌ H 3 O + ( aq ) + C 2 H 5 COO ( aq); K a = [ H 3 O + ] [ C 2 H 5 COO ] [ C 2 H 5 COOH ]

    16.50

    1. C 6 H 5 COOH(aq) ⇌ H + ( aq) + C 6 H 5 COO ( aq); K a = [ H + ] [ C 6 H 5 COO ] [ HC 6 H 5 COOH ]

      C 6 H 5 COOH(aq) + H 2 O(l) ⇌ H 3 O + ( aq) + C 6 H 5 COO ( aq); K a = [ H 3 O + ] [ C 6 H 5 COO ] [ HC 6 H 5 COOH ]

    2. HCO 3 (aq) ⇌ H + ( aq) + CO 3 2 ( aq); K a = [ H + ] [ CO 3 2 ] [ HCO 3 ]

      HCO 3 (aq) + H 2 O(l) ⇌ H 3 O + ( aq) + CO 3 2 ( aq); K a = [ H 3 O + ] [ CO 3 2 ] [ HCO 3 ]


    16.51 Analyze/Plan .  Follow the logic in Sample Exercise 16.10. Solve .

    CH 3 CH(OH)COH ⇌ H + ( aq) + CH 3 CH(OH)COO ( aq); K a = [ H + ] [ CH 3 CH(OH)COO ] [ CH 3 CH(OH)COOH ]

    [H + ] = [CH 3 CH(OH)COO ] = 10 −2.44 = 3.63 × 10 −3 = 3.6 × 10 −3 M

    [CH 3 CH(OH)COOH] = 0.10 – 3.63 × 10 −3 = 0.0964 = 0.096 M

    K a = ( 3.63 × 10 3 ) 2 ( 0.0964 ) = 1.4 × 10 4

    16.52 C 6 H 5 CH 2 COOH(aq) H + ( aq) + C 6 H 5 CH 2 CO O ( aq); K a = [ H + ] [ C 6 H 5 CH 2 COO ] [ C 6 H 5 CH 2 COOH ]

    [H + ] = [C 6 H 5 CH 2 COO ] = 10 −2.68 = 2.09 × 10 −3 = 2.1 × 10 −3 M

    [C 6 H 5 CH 2 COOH] = 0.085 – 2.09 × 10 −3 = 0.0829 = 0.083 M

    K a = ( 2.09 × 10 3 ) 2 0.0829 = 5.3 × 10 5

    16.53 Analyze/Plan .  Write the equilibrium reaction and the K a expression. Use percent ionization to get equilibrium concentration of [H + ], and by stoichiometry, [X ] and [HX]. Calculate K a Solve .

    [H + ] = 0.110 × [CH 2 ClCOOH] initial = 0.0110 M

    CH 2 ClCOOH(aq) H + (aq) + CH 2 ClCOO (aq)
    initial 0.100 M 0 0
    equil. 0.089 M 0.0110 M 0.0110 M

    K a = [ H + ] [ CH 2 ClCOO ] [ CH 2 ClCOOH ] = ( 0.0110 ) 2 0.089 = 1.4 × 10 3

    16.54  [H + ] = 0.132 × [BrCH 2 COOH] initial = 0.0132 M

    BrCH 2 COOH(aq) H + (aq) + BrCH 2 COO (aq)
    initial 0.100 M 0 0
    equil. 0.087 0.0132 M 0.0132 M

    K a = [ H + ] [ BrCH 2 COO ] [ BrCH 2 COOH ] = ( 0.0132 ) 2 0.087 = 2.0 × 10 3

    16.55 Analyze/Plan .  Write the equilibrium reaction and the K a expression.

    [H + ] = 10 −pH = [CH 3 COO ]; [CH 3 COOH] = x – [H + ].

    Substitute into the K a expression and solve for x. Solve .

    [H + ] = 10 −pH = 10 −2.90 = 1.26 × 10 −3 = 1.3 × 10 −3 M

    K a = 1.8 × 10 5 = [ H + ] [ CH 3 COO ] [ C H 3 COOH ] = ( 1.26 × 10 3 ) 2 ( x 1.26 × 10 3 )

    1.8 × 10 −5 (x – 1.26 × 10 −3 ) = (1.26 × 10 −3 ) 2 ;

    1.8 × 10 −5 x = 1.585 × 10 −6 + 2.266 × 10 −8 = 1.608 × 10 −6 ;

    x = 0.08931 = 0.089 M CH 3 COOH


    16.56  [H + ] = 10 −pH = 10 −3.65 = 2.239 × 10 −4 = 2.2 × 10 −4 M

    K a = 6.8 × 10 4 = [ H + ] [ F ] [ HF ] = ( 2.239 × 10 4 ) 2 x 2.239 × 10 4

    6.8 × 10 −4 (x – 2.239 × 10 −4 ) = (2.239 × 10 −4 ) 2 ;

    6.8 × 10 −4 x = 1.522 × 10 −7 + 0.501 × 10 −7 = 2.024 × 10 −7

    x = 2.976 × 10 −4 = 3.0 × 10 −4 M HF

    16.57 Analyze/Plan .  Follow the logic in Sample Exercise 16.12. Write K a , construct the equilibrium table, solve for x = [H + ], and then get equilibrium [C 6 H 5 COO ] and [C 6 H 5 COOH] by substituting [H + ] for x. Solve .

    C 6 H 5 COOH(aq) H + (aq) + C 6 H 5 COO (aq)
    initial 0.050 M 0 0
    equil. (0.050 – x) M x M x M

    K a = [ H + ] [ C 6 H 5 COO ] [ C 6 H 5 COOH ] = x 2 ( 0.050 x) x 2 0.050 = 6.3 × 10 5

    x 2 = 0.050 (6.3 × 10 −5 ); x = 1.8 × 10 −3 M = [H + ] = [H 3 O + ] = [C 6 H 5 COO ]

    [C 6 H 5 COOH] = 0.050 – 0.0018 = 0.048 M

    Check . 1.8 × 10 3 M H + 0.050 M C 6 H 5 COOH × 100 = 3.6 % ionization; the approximation is valid

    16.58

    HClO 2 (aq) H + (aq) + ClO 2 (aq)
    initial 0.0125 M 0 0
    equil. (0.0125 – x) M x M x M

    K a = [ H + ] [ ClO 2 ] [ HClO 2 ] = x 2 ( 0.0125 x) x 2 0.0125 = 1.1 × 10 2

    Assuming x is small relative to 0.0125, x 2 = 0.0125(0.011); x = 1.2 × 10 −2 M .

    Clearly, x is not small relative to 0.0125, so we must solve the quadratic formula for [H + ].

    x 2 = 0.011 (0.0125 – x); x 2 + 0.011x – 1.38 × 10 −4 = 0

    x = 0.011 ± ( 0.011 ) 2 4 ( 1 ) ( 1.38 × 10 4 ) 2 ( 1 ) = 0.007452 = 0.0075 M ;

    [H + ] = [H 3 O + ] = [ClO 2 ] = 0.0075 M ; [HClO 2 ] = 0.0125 – 0.0075 = 0.005045 = 5.0 × 10 −3 M

    Check . K a = ( 7.5 × 10 3 ) 2 5.0 × 10 3 = 0.011 ; our results agree

    16.59 Analyze/Plan .  Follow the logic in Sample Exercise 16.12. Solve .

    1. C 2 H 5 COOH(aq) H + (aq) + C 2 H 5 COO (aq)
      initial 0.095 M 0 0
      equil. (0.095 – x) M x M x M

      K a = [ H + ] [ C 2 H 5 COO ] [ C 2 H 5 COOH ] = x 2 ( 0.095 x) x 2 0.095 = 1.3 × 10 5

      x 2 = 0.095(1.3 × 10 −5 ); x = 1.111 × 10 −3 = 1.1 × 10 −3 M H + ; pH = 2.95


    1. Check . 1.1 × 10 3 M H + 0.095 M C 2 H 5 COO H × 100 = 1.2 % ionization; the approximation is valid

    1. K a = [ H + ] [ CrO 4 2 ] [ HCrO 4 ] = x 2 ( 0.100 x ) x 2 0.100 = 3.0 × 10 7

      x 2 = 0.100(3.0 × 10 −7 ); x = 1.732 × 10 −4 = 1.7 × 10 −4 M H +

      pH = –log(1.732 × 10 −4 ) = 3.7614 = 3.76

      Check . 1.7 × 10 4 M H + 0.100 M HCrO 4 × 100 = 0.17 % ionization; the approximation is valid

    1. Follow the logic in Sample Exercise 16.15. pOH = –log[OH ]; pH = 14 – pOH
      C 5 H 5 N(aq) + H 2 O(l) C 5 H 5 NH + (aq) + OH (aq)
      initial 0.120 M 0 0
      equil. (0.120 – x) M x M x M

      K b = [ C 5 H 5 NH + ] [ OH ] [ C 5 H 5 N] = x 2 ( 0.120 x) x 2 0.120 = 1.7 × 10 9

      x 2 = 0.120(1.7 × 10 −9 ); x = 1.428 × 10 −5 = 1.4 × 10 −5 M OH ; pH = 9.15

      Check . 1.4 × 10 5 M OH 0.120 M C 5 H 5 N × 100 = 0.012 % ionization; the approximation is valid

    16.60

    1. HOCl(aq) H + (aq) + OCl (aq)
      initial 0.095 M 0 0
      equil. (0.095 – x) M x M x M

      K a = [ H + ] [ OCl ] [ HOCl] = x 2 ( 0.095 x) x 2 0.095 = 3.0 × 10 8

      x 2 = 0.095 (3.0 × 10 −8 ); x = [H + ] = 5.3 × 10 −5 M , pH = 4.27

      Check . 5.3 × 10 5 M H + 0.095 M HOCl × 100 = 0.056 % ionization

      The approximation is nearly valid. To 2 sig figs, the quadratic formula gives the same [H + ].

    2. H 2 NNH 2 (aq) + H 2 O(l) H 2 NNH 3 + (aq) + OH (aq)
      initial 0.0085 M 0 0
      equil. (0.0085 – x) M x M x M

      K b = [ H 2 NNH 3 + ] [ OH ] [ H 2 NNH 2 ] = x 2 ( 0.0085 x) x 2 0.0085 = 1.3 × 10 6

      x 2 = 0.0085 (1.3 × 10 −6 ); x = [OH ] = 1.051 × 10 −4 = 1.1 × 10 −4 M


    1. Clearly, 1.1 × 10 −4 M OH is not small compared to 8.5 × 10 −3 M H 2 NNH 2 , and we must solve the quadratic.

      x 2 = 1.3 × 10 −6 (0.0085 – x); x 2 + 1.3 × 10 −6 x – 1.105 × 10 −8 = 0

      x = 1.3 × 10 6 ± ( 1.3 × 10 6 ) 2 4 ( 1 ) ( 1.105 × 10 8 ) 2 ( 1 ) = 1.0447 × 10 4 = 1.0 × 10 4 M O H

      pOH = 3.981 = 3.98; pH = 14 – pOH = 14 – 3.981 = 10.019 = 10.02

      Check . Although this solution has more than 12% ionization, the difference in [OH ] between the estimate and the quadratic is not great.

    1. HONH 2 (aq) + H 2 O(l) HONH 3 + (aq) + OH (aq)
      initial 0.165 M 0 0
      equil. (0.165 – x) M x M x M

      K b = [ HONH 3 + ] [ OH ] [ HONH 2 ] = x 2 ( 0.165 x) x 2 0.165 = 1.1 × 10 8

      x 2 = 0.165 (1.1 × 10 −8 ); x = [OH ] = 4.3 × 10 −5 M , pH = 9.63

      Check . 4.3 × 10 5 M OH 0.165 M HONH 2 × 100 = 0.026 % ionization; the approximation is valid

    16.61 Analyze/Plan .  K a = 10 −pKa . Follow the logic in Sample Exercise 16.13. Solve .

    Let [H + ] = [NC 7 H 4 SO 3 ] = z. K a = antilog (–2.32) = 4.79 × 10 −3 = 4.8 × 10 −3 .

    z 2 0.10 z = 4.79 × 10 3 . As K a is relatively large, solve the quadratic .

    z 2 + 4.79 × 10 −3 z – 4.79 × 10 −4 = 0

    z = 4.79 × 10 3 ± ( 4.79 × 10 3 ) 2 4 ( 1 ) ( 4.79 × 10 4 ) 2 ( 1 ) = 4.79 × 10 3 ± 1.937 × 10 3 2

    z = 1.96 × 10 −2 = 2.0 × 10 −2 M H + ; pH = –log (1.96 × 10 −2 ) = 1.71

    16.62  Calculate the initial concentration of HC 9 H 7 O 4 .

    2 tablets × 500 mg tablet × 1 g 1000 mg × 1 mol HC 9 H 7 O 4 180.2 g HC 9 H 7 O 4 = 0.005549 = 0.00555 mol HC 9 H 7 O 4

    0.005549 mol HC 9 H 7 O 4 0.250 L = 0.02220 = 0.0222 M HC 9 H 7 O 4

    HC 9 H 7 O 4 (aq) C 9 H 7 O 4 + H + (aq)
    initial 0.0222 M 0 M 0 M
    equil. (0.0222 – x) x M x M

    K a = 3.3 × 10 4 = [ H + ] [ C 9 H 7 O 4 ] [ HC 9 H 7 O 4 ] = x 2 ( 0.0222 x)


    Assuming x is small compared to 0.0222,

    x 2 = 0.0222 (3.3 × 10 −4 ); x = [H + ] = 2.7 × 10 −3 M

    2.7 × 10 3 M H + 0.0222 M HC 9 H 7 O 4 × 100 = 12 % ionization; the approximation is not valid

    Using the quadratic formula, x 2 + 3.3 × 10 −4 x –7.325 × 10 −6 = 0

    x = 3.3 × 10 4 ± ( 3.3 × 10 4 ) 2 4 ( 1 ) ( 7.325 × 10 6 ) 2 ( 1 ) = 3.3 × 10 4 ± 2.941 × 10 5 2

    x = 2.547 × 10 −3 = 2.5 × 10 −3 M H + ; pH = –log(2.547 × 10 −3 ) = 2.594 = 2.59

    16.63 Analyze/Plan .  Follow the logic in Sample Exercise 16.12 and 16.13. Solve .

    1. HN 3 (aq) H + (aq) + N 3 (aq)
      initial 0.400 M 0 0
      equil. (0.400 – x) M x M x M

      K a = [ H + ] [ N 3 ] [ HN 3 ] = 1.9 × 10 5 ; x 2 ( 0.400 x) x 2 0.400 = 1.9 × 10 5

      x = 0.00276 = 2.8 × 10 3 M = [ H + ] ; % ionization = 2.76 × 10 3 0.400 × 100 = 0.69 %

    2. 1.9 × 10 5 x 2 0.100 ; x = 0.00138 = 1.4 × 10 3 M H +

      % ionization = 1.38 × 10 3 M H + 0.100 M HN 3 × 100 = 1.4 %

    3. 1.9 × 10 5 x 2 0.0400 ; x = 8.72 × 10 4 = 8.7 × 10 4 M H +

      % ionization = 8.72 × 10 4 M H + 0.0400 M HN 3 × 100 = 2.2 %

      Check . Notice that a tenfold dilution [part (a) versus part (c)] leads to a slightly more than threefold increase in percent ionization.

    16.64

    1. C 2 H 5 COOH(aq) ⇌ H + (aq) + C 2 H 5 COO (aq)

      K a = 1.3 × 10 5 = [ H + ] [ C 2 H 5 COO ] [ C 2 H 5 COOH ] = x 2 0.250 x

      x 2 ≈ 0.250 (1.3 × 10 −5 ); x = 1.803 × 10 −3 = 1.8 × 10 −3 M H +

      % ionization = 1.803 × 10 3 M H + 0.250 M C 2 H 5 COOH × 100 = 0.721 %

    2. x 2 0.0800 1.3 × 10 5 ; x = 1.020 × 10 3 = 1.0 × 10 3 M H +

      % ionization = 1.020 × 10 3 M H + 0.0800 M C 2 H 5 COOH × 100 = 1.27 %


    1. x 2 0.0200 1.3 × 10 5 ; x = 5.099 × 10 4 = 5.1 × 10 4 M H +

      % ionization = 5.099 × 10 4 M H + 0.0200 M C 2 H 5 COOH × 100 = 2.55 %

    16.65 Analyze/Plan .  Follow the logic in Sample Exercise 16.14. Citric acid is a triprotic acid with three K a values that do not differ by more than 10 3 . We must consider all three steps. Also, C 6 H 5 O 7 3– is only produced in step 3. Solve .

    H 3 C 6 H 5 O 7 (aq) H + (aq) + H 2 C 6 H 5 O 7 (aq) K a1 = 7 .4 × 10 4

    H 2 C 6 H 5 O 7 (aq) H + (aq) + HC 6 H 5 O 7 2 (aq) K a2 = 1.7 × 10 5

    HC 6 H 5 O 7 2 (aq) H + (aq) + C 6 H 5 O 7 3 (aq) K a3 = 4 .0 × 10 7

    1. To calculate the pH of a 0.040 M solution, assume initially that only the first ionization is important:
      H 3 C 6 H 5 O 7 (aq) H + (aq) + H 2 C 6 H 5 O 7 (aq)
      initial 0.040 M 0 0
      equil. (0.040 – x) M x M x M

      K a1 = [ H + ] [ H 2 C 6 H 5 O 7 ] [ H 3 C 6 H 5 O 7 ] = x 2 ( 0.040 x) = 7.4 × 10 4

      x 2 = (0.040 – x)(7.4 × 10 −4 ); x 2 ≈ (0.040)(7.4 × 10 −4 ); x = 0.00544 = 5.4 × 10 −3 M

      As this value for x is rather large in relation to 0.040, a better approximation for x can be obtained by substituting this first estimate into the expression for x 2 , and then solving again for x:

      x 2 = (0.040 – x) (7.4 × 10 −4 ) = (0.040 – 5.44 × 10 −3 ) (7.4 × 10 −4 )

      x 2 = 2.557 × 10 −5 ; x = 5.057 × 10 −3 = 5.1 × 10 −3 M

      (This is the same result obtained from the quadratic formula.)

      The correction to the value of x, though not large, is significant. Does the second ionization produce a significant additional concentration of H + ?

      H 2 C 6 H 5 O 7 (aq) H + (aq) + HC 6 H 5 O 7 2– (aq)
      initial 5.1 × 10 −3 M 5.1 × 10 −3 M 0
      equil. (5.1 × 10 −3 – y) (5.1 × 10 −3 + y) y

      K a2 = [ H + ] [ HC 6 H 5 O 7 2 ] [ H 2 C 6 H 5 O 7 ] = 1.7 × 10 5 ; ( 5.1 × 10 3 + y) (y) ( 5.1 × 10 3 y) = 1.7 × 10 5

      Assume that y is small relative to 5.1 × 10 −3 ; that is, that additional ionization of H 2 C 6 H 5 O 7 is small, then

      ( 5.1 × 10 3 ) y ( 5.1 × 10 3 ) = 1.7 × 10 5 M ; y = 1.7 × 10 5 M

      This value is indeed small compared to 5.1 × 10 −3 M ; [H + ] and pH are determined by the first ionization step. pH = –log(5.057 × 10 −3 ) = 2.30.


    1. Yes. We started the calculation by assuming that only the first step made a significant contribution to [H + ] and pH. Calculation proved this assumption to be true. Next, we assumed [H + ] from the first ionization was small relative to 0.040 M citric acid; this assumption was not valid. Finally, we assumed that additional ionization of H 2 C 6 H 5 O 7 was small, which was true.
    1. The concentration of citrate ion, [C 6 H 5 O 7 3– ], is much less than [H + ]. Because the second ionization does not contribute significantly to [H + ], we know that [HC 6 H 5 O 7 2– ] is less than [H + ]. The third ionization is even less extensive, so [C 6 H 5 O 7 3– ] is much less than [H + ].

    16.66  H 2 C 4 H 4 O 6 (aq) ⇌ H + (aq) + HC 4 H 4 O 6 (aq) K a1 = 1.0 × 10 −3

    HC 4 H 4 O 6 (aq) ⇌ H + (aq) + C 4 H 4 O 6 2– (aq)  K a2 = 4.6 × 10 −5

    Begin by calculating the [H + ] from the first ionization. The equilibrium concentrations are [H + ] = [HC 4 H 4 O 6 ] = x, [H 2 C 4 H 4 O 6 ] = 0.25 – x.

    K a1 = [ H + ] [ HC 4 H 4 O 6 ] [ H 2 C 4 H 4 O 6 ] = x 2 0.25 x ; x 2 + 1.0 × 10 3 x 2.5 × 10 4 = 0

    Using the quadratic formula, x = 1.532 × 10 −2 = 0.015 M H + from the first ionization. Next, calculate the H + contribution from the second ionization.

    HC 4 H 4 O 6 (aq) H + (aq) + C 4 H 4 O 6 2– (aq)
    initial 0.015 0.015 0
    equil. (0.015 – y) (0.015 + y) y

    K a2 = ( 0.015 + y) (y) ( 0.015 y ) = 4.6 × 10 5 ; assuming y is small compared to 0 .015,

    y = 4.6 × 10 −5 M C 4 H 4 O 6 2– (aq)

    This approximation is reasonable, because 4.6 × 10 −5 is only 0.3% of 0.015.

    [H + ] = 0.015 M (first ionization) + 4.6 × 10 −5 (second ionization)

    Because 4.6 × 10 −5 is 0.3% of 0.015 M , it can be safely ignored when calculating total [H + ].

    pH = –log(0.01532) = 1.18148 = 1.181

    Assumptions:

    1. The ionization can be treated as a series of steps (valid by Hess’s law).
    2. The extent of ionization in the second step (y) is small relative to that from the first step (valid for this acid and initial concentration). This assumption was used twice, to calculate the value of y from K a2 and to calculate total [H + ] and pH.

    Weak Bases (Section 16.7)

    16.67

    1. HONH 3 +
    2. When hydroxylamine acts as a base, the nitrogen atom accepts a proton.
    3. 14 e , 7 e pairs
      image

    1. In neutral hydroxylamine, both O and N have zero formal charges. Nitrogen is less electronegative than oxygen and more likely to share a lone pair of electrons with an incoming (and electron-deficient) H + . The resulting cation with the +1 formal charge on N is more stable than the one with the +1 formal charge on O.

    16.68

    1. Analyze/Plan . To determine relative strength, compare the K b values of the two bases. Solve .

      K b for OCl = K w K a for HClO = 1.0 × 10 14 3.0 × 10 8 = 3.3 × 10 7

      K b for hydroxylamine is 1.1 × 10 −8 . OCl is a stronger base than hydroxylamine.

    2. When OCl acts as a base, the O atom is the proton acceptor.
    3. 14 e , 7 e pairs
      image

      In OCl , the –1 formal charge is on O. H + attaches to the atom with the negative formal charge.

    16.69 Analyze/Plan .  Remember that K b = [products]/[reactants]. If H 2 O(l) appears in the equilibrium reaction, it will not appear in the K b expression, because it is a pure liquid. Solve .

    1. (CH 3 ) 2 NH(aq) + H 2 O(l) ⇌ (CH 3 ) 2 NH 2 + ( aq) + OH ( aq); K b = [ ( CH 3 ) 2 NH 2 + ] [ OH ] [ (CH 3 ) 2 NH ]
    2. CO 3 2– (aq) + H 2 O(l) ⇌ HCO 3 (aq) + OH (aq); K b = [ HCO 3 ] [ OH ] [ CO 3 2 ]
    3. HCOO (aq) + H 2 O(l) ⇌ HCOOH(aq) + OH ( aq ) ; K b = [ HCOOH ] [ OH ] [ HCOO ]

    16.70

    1. C 3 H 7 NH 2 (aq) + H 2 O(l) ⇌ C 3 H 7 NH 3 + ( aq) + OH ( aq); K b = [ C 3 H 7 NH 3 + ] [ OH ] [ C 3 H 7 NH 2 ]
    2. HPO 4 2– (aq) + H 2 O(l) ⇌ H 2 PO 4 ( aq) + OH (aq); K b = [ H 2 PO 4 ] [ OH ] [ HPO 4 2 ]
    3. C 6 H 5 CO 2 (aq) + H 2 O(l) ⇌ C 6 H 5 CO 2 H(aq) + OH ( aq); K b = [ C 6 H 5 CO 2 H ] [ OH ] [ C 6 H 5 CO 2 ]

    16.71 Analyze/Plan .  Follow the logic in Sample Exercise 16.15. Solve .

    C 2 H 5 NH 2 (aq) + H 2 O(l) C 2 H 5 NH 3 + (aq) + OH (aq)
    initial 0.075 M 0 0
    equil. (0.075 – x) M x M x M

    K b = [ C 2 H 5 NH 3 + ] [ OH ] [ C 2 H 5 NH 2 ] = (x) (x) (0 .075 x) = x 2 0.075 = 6.4 × 10 4

    x 2 = 0.075 (6.4 × 10 −4 ); x = [OH ] = 6.9 × 10 −3 M ; pH = 11.84


    Check . 6.9 × 10 3 M OH 0.075 M C 2 H 5 NH 2 × 100 = 9.2 % ionization; the assumption is not valid

    To obtain a more precise result, the K b expression is rewritten in standard quadratic form and solved via the quadratic formula.

    x 2 0.075 x = 6.4 × 10 4 ; x 2 + 6.4 × 10 4 x 4.8 × 10 5 = 0

    x = b ± b 2 4 ac 2 a = 6.4 × 10 4 ± ( 6.4 × 10 4 ) 2 4 ( 1 ) ( 4.8 × 10 5 ) 2

    x = 6.62 × 10 −3 = 6.6 × 10 −3 M OH ; pOH = 2.18; pH = 14.00 – pOH = 11.82

    Note that the pH values obtained using the two algebraic techniques are very similar.

    16.72

    BrO (aq) + H 2 O(l) HOBr(aq) + OH (aq)
    initial 0.724 M 0 0
    equil. (0.724 – x) M x M x M

    K b = [ HOBr ] [ OH ] [ BrO ] = x 2 0.724 x x 2 0.724 = 4.0 × 10 6

    x 2 = 0.724 (4.0 × 10 −6 ); x = [OH ] = 1.70 × 10 −3 = 1.7 × 10 −3 M ; pH = 11.23

    Check . 1.7 × 10 3 M OH 0.724 M BrO × 100 = 0.24 % hydrolysis; the approximation is valid

    16.73 Analyze/Plan .  Based on the pH and initial concentration of base, calculate all equilibrium concentrations. pH → pOH → [OH ] at equilibrium. Construct the equilibrium table and calculate other equilibrium concentrations. Substitute into the K b expression and calculate K b . Solve .

    1. [OH ] = 10 −pOH ; pOH = 14 – pH = 14.00 – 11.33 = 2.67

      [OH ] = 10 −2.67 = 2.138 × 10 −3 = 2.1 × 10 −3 M

      C 10 H 15 ON(aq) + H 2 O(l) C 10 H 15 ONH + (aq) + OH (aq)
      initial 0.035 M 0 0
      equil. 0.033 M 2.1 × 10 −3 M 2.1 × 10 −3 M
    2. K b = [ C 10 H 15 ONH + ] [ OH ] [ C 10 H 15 ON ] = ( 2.138 × 10 3 ) 2 ( 0.03286 ) = 1.4 × 10 4

    16.74

    1. pOH = 14.00 – 9.95 = 4.05; [OH ] = 10 −4.05 = 8.91 × 10 −5 = 8.9 × 10 −5 M
      C 18 H 21 NO 3 (aq) + H 2 O(l) C 18 H 21 NO 3 H + (aq) + OH (aq)
      initial 0.0050 M 0 0
      equil. (0.0050 – 8.9 × 10 −5 ) 8.9 × 10 −5 M 8.9 × 10 −5 M

      K b = [C 18 H 21 NO 3 H + ] [ OH ] [ C 18 H 21 NO 3 ] = ( 8.91 × 10 5 ) 2 ( 0.0050 8.91 × 10 5 ) = 1.62 × 10 6 = 1.6 × 10 6

    2. pK b = –log (K b ) = –log (1.62 × 10 −6 ) = 5.79

    The K a – K b Relationship; Acid–Base Properties of Salts (Sections 16.8 and 16.9)

    16.75 Analyze/Plan . Refer to Equation 16.6 and Sample Exercise 16.17. Solve .

    1. C 6 H 5 OH(aq) + H 2 O(l) ⇌ H 3 O + (aq) + C 6 H 5 O (aq)
    2. K b = K w /K a = 1.0 × 10 −14 / 1.3 × 10 −10 = 7.7 × 10 −5
    3. Phenol is a stronger acid than water. The benchmark for acid strength in water is 1.0 × 10 −14 . All acids listed in Table D.1 of Appendix D are stronger acids than water. (The one notable exception is K a2 for H 2 S. HS , the product of the first ionization of H 2 S, has a K a value of 1 × 10 −19 .)

    16.76  The stronger a base, the weaker its conjugate acid. From the K a values in Table 16.3, place the conjugate acids of these oxyanions in order of increasing K a value, increasing acid strength, and decreasing conjugate base strength. Use K a2 for H 2 SO 4 , H 2 CO 3 , and H 2 SO 3 and K a3 for H 3 PO 4 .

    In order of increasing K a value and acid strength: HPO 4 2– < HCO 3 < HSO 3 < HSO 4

    In order of decreasing base strength: PO 4 3– > CO 3 2– > SO 3 2– > SO 4 2–

    16.77 Analyze/Plan .  Based on K a , determine relative strengths of the acids and their conjugate bases. The greater the magnitude of K a , the stronger the acid and the weaker the conjugate base. K b (conjugate base) = K w /K a . Solve .

    1. Acetic acid is stronger, because it has the larger K a value.
    2. Hypochlorite ion is the stronger base because the weaker acid, hypochlorous acid, has the stronger conjugate base.
    3. K b for CH 3 COO = K w /K a for CH 3 COOH = 1.0 × 10 −14 /1.8 × 10 −5 = 5.6 × 10 −10

      K b for ClO = K w /K a for HClO = 1 × 10 −14 /3.0 × 10 −8 = 3.3 × 10 −7

      Note that K b for ClO is greater than K b for CH 3 COO .

    16.78

    1. Ammonia is the stronger base because it has the larger K b value.
    2. Hydroxylammonium is the stronger acid because the weaker base, hydroxylamine, has the stronger conjugate acid.
    3. K a for NH 4 + = K w /K b for NH 3 = 1.0 × 10 −14 /1.8 × 10 −5 = 5.6 × 10 −10

      K a for HONH 3 + = K w /K b for HONH 2 = 1.0 × 10 −14 /1.1 × 10 −8 = 9.1 × 10 −7

      Note that K a for HONH 3 + is larger than K a for NH 4 + .

    16.79 Analyze .  When the solute in an aqueous solution is a salt, evaluate the acid/base properties of the component ions.

    1. Plan .  NaBrO is a soluble salt and, thus, a strong electrolyte. When it is dissolved in H 2 O, it dissociates completely into Na + and BrO . [NaBrO] = [Na + ] = [BrO ] = 0.10 M . Na + is the conjugate acid of the strong base NaOH and, thus, does not influence the pH of the solution. BrO , on the other hand, is the conjugate base of the weak acid HBrO and does influence the pH of the solution. Like any other weak base, it hydrolyzes water to produce OH (aq). Solve the equilibrium problem to determine [OH ]. Solv e .

    1. BrO (aq) + H 2 O(l) HBrO(aq) + OH (aq)
      initial 0.10 M 0 0
      equil. (0.10 – x) M x M x M

      K b for BrO = [ HBrO ] [ OH ] [ BrO ] = K w K a for HBrO = 1 × 10 14 2.5 × 10 9 = 4.00 × 10 6 = 4.0 × 10 6

      4.00 × 10 6 = ( x) (x) (0 .10 x) ; assume the percent of BrO that hydrolyzes is small

      x 2 = 0.10 (4.00 × 10 −6 ); x = [OH ] = 6.32 × 10 −4 = 6.3 × 10 −4 M

      pOH = 3.20; pH = 14 – 3.20 = 10.80

    1. Plan .  NaHS(aq)   →   Na + (aq) + HS (aq)

      HS is the conjugate base of H 2 S and its hydrolysis reaction will determine the [OH ] and pH of the solution [see similar explanation for NaBrO in part (a)]. We will assume the process HS (aq) ⇌ H + (aq) + S (aq) will not significantly affect the [OH ] in solution because K a2 for H 2 S is so small. Solve the equilibrium problem for [OH ]. Solve .

      HS (aq) + H 2 O(l) H 2 S(aq) + OH (aq)
      initial 0.080 M 0 0
      equil. (0.080 – x) M x x

      K b = [ H 2 S ] [ OH ] [ HS ] = K w K a for H 2 S = 1.0 × 10 14 9.5 × 10 8 = 1.053 × 10 7 = 1.1 × 10 7

      1.053 × 10 7 = x 2 ( 0.080 x) ; x 2 = 0.080 ( 1.053 × 10 7 ) ; x = 9.177 × 10 5 = 9.2 × 10 5 M OH

      (Assume x is small compared to 0.080); pOH = 4.04; pH = 14 – 4.04 = 9.96

      Check . 9.2 × 10 5 M OH 0.080 M HS × 100 = 0.12 % hydrolysis; the approximation is valid

    1. Plan .  For the two salts present, Na + and Ca 2+ are negligible acids. NO 2 is the conjugate base of HNO 2 and will determine the pH of the solution. Solve .

      Calculate total [NO 2 ] present initially.

      [NO 2 ] total = [NO 2 ] from NaNO 2 + [NO 2 ] from Ca(NO 2 ) 2

      [NO 2 ] total = 0.10 M + 2(0.20 M ) = 0.50 M

      The hydrolysis equilibrium is:

      NO 2 (aq) + H 2 O(l) HNO 2 (aq) + OH (aq)
      initial 0.50 M 0 0
      equil. (0.50 – x) M x M x M

      K b = [ HNO 2 ] [ OH ] [ NO 2 ] = K w K a for HNO 2 = 1.0 × 10 14 4.5 × 10 4 = 2.22 × 10 11 = 2.2 × 10 11

      2.2 × 10 11 = x 2 ( 0.50 x) x 2 0.50 ; x 2 = 0.50 ( 2.22 × 10 11 )

      x = 3.33 × 10 −6 = 3.3 × 10 −6 M OH ; pOH = 5.48; pH = 14 – 5.48 = 8.52


    16.80

    1. Proceeding as in Solution 16.79(a):
      F (aq) + H 2 O(l) HF(aq) + OH (aq)
      initial 0.105 M 0 M 0 M
      equil. (0.105 – x) M x M x M

      K b for F = [ HF ] [ OH ] [ F ] = K w K a for HF = 1.0 × 10 14 6.8 × 10 4 = 1.47 × 10 11 = 1.5 × 10 11

      1.5 × 10 11 = ( x) (x) (0 .105 x) ; assume the amount of F that hydrolyzes is small

      x 2 = 0.105(1.47 × 10 −11 ); x = [OH ] = 1.243 × 10 −6 = 1.2 × 10 −6 M

      pOH = 5.91; pH = 14 – 5.91 = 8.09

    2. Na 2 S(aq) → S 2– (aq) + 2Na + (aq)

      S 2– (aq) + H 2 O(l) ⇌ HS (aq) + OH (aq)

      As in part (a), [OH ] = [HS ] = x; [S 2– ] = 0.035 M

      K b = [ HS ] [ OH ] [ S 2 ] = K w K a for HS = 1.0 × 10 14 1 × 10 19 = 1 × 10 5

      Because K b >> 1, this equilibrium lies far to the right and [OH ] = [HS ] = 0.035 M . K b for HS = 1.05 × 10 −7 ; [OH ] produced by further hydrolysis of HS amounts to 6.1 × 10 −5 M . The second hydrolysis step does not make a significant contribution to the total [OH ] and pH.

      [OH ] = 0.035 M ; pOH = 1.46, pH = 12.54

    3. As in Solution 16.79(c), calculate [CH 3 COO ].

      [CH 3 COO ] t = [CH 3 COO ] from NaCH 3 COO + [CH 3 COO ] from Ba(CH 3 COO) 2

      [CH 3 COO ] t = 0.045 M + 2(0.055 M ) = 0.155 M

      The hydrolysis equilibrium is:

      CH 3 COO (aq) + H 2 O(l)  ⇌ CH 3 COOH(aq) + OH (aq)

      K b = [ CH 3 COOH ] [ OH ] [ CH 3 COO ] = K w K a for CH 3 COOH = 1.0 × 10 14 1.8 × 10 5 = 5.56 × 10 10 = 5.6 × 10 10

      [OH ] = [CH 3 COOH] = x; [CH 3 COO ] = 0.155 – x

      K b = 5.56 × 10 10 = x 2 ( 0.155 x) ; assume x is small compared to 0 .155 M

      x 2 = 0.155 (5.56 × 10 −10 ); x = [OH ] = 9.280 × 10 −6 = 9.3 × 10 −6

      pH = 14 + log (9.280 × 10 −6 ) = 8.97

    16.81 Analyze/Plan .  The salt dissociates to form Na + and CH 3 COO . Na + is a negligible base; the hydrolysis equilibrium of CH 3 COO determines the pH of the solution.

    Solve . The hydrolysis equilibrium is:

    CH 3 COO (aq) + H 2 O(l)  ⇌ CH 3 COOH(aq) + OH (aq)

    K b = [ CH 3 COOH ] [ OH ] [ CH 3 COO ] = K w K a for CH 3 COOH = 1.0 × 10 14 1.8 × 10 5 = 5.56 × 10 10 = 5.6 × 10 10


    [CH 3 COO ] = x; [OH ] = [CH 3 COOH] = 10 −pOH

    pOH = 14.00 – pH = 14.00 – 9.70 = 4.30; [OH ] = 10 −4.30 = 5.012 × 10 −5 = 5.0 × 10 −5

    K b = 5.56 × 10 10 = ( 5.012 × 10 5 ) 2 x ; x = 4 .518 = 4 .5 M NaCH 3 COO

    Note that no assumption was required in this calculation.

    16.82

    1. C 5 H 5 NH + (aq) + H 2 O(l) ⇌ C 5 H 5 N(aq) + H 3 O + (aq)
    2. K a = [ C 5 H 5 N ] [ H + ] [ C 5 H 5 NH + ] = K w K b for C 5 H 5 N = 1.0 × 10 14 1.7 × 10 9 = 5.882 × 10 6 = 5.9 × 10 6
    3. [C 5 H 5 NH + ] = x; [C 5 H 5 N] = [H + ] = 10 −pH = 10 −2.95 = 1.122 × 10 −3 = 1.1 × 10 −3 M

      K a = 5.882 × 10 6 = ( 1.122 × 10 3 ) 2 x ; x = 0 .2140 = 0 .21 M C 5 H 5 NH +

      Note that no assumption was required in this calculation.

    16.83 Analyze/Plan .  Based on the formula of a salt, predict whether an aqueous solution will be acidic, basic, or neutral. Evaluate the acid–base properties of both ions and determine the overall effect on solution pH. Solve .

    (a) acidic; NH 4 + is a weak acid, Br is negligible.
    (b) acidic; Fe 3+ is a highly charged metal cation and a Lewis acid; Cl is negligible.
    (c) basic; CO 3 2– is the conjugate base of HCO 3 ; Na + is negligible.
    (d) neutral; both K + and ClO 4 are negligible.
    (e) acidic; HC 2 O 4 is amphoteric, but K a for the acid dissociation (6.4 × 10 −5 ) is much greater than K b for the base hydrolysis
    (1.0 × 10 −14 / 5.9 × 10 −2 = 1.7 × 10 −13 ).

    16.84

    (a) acidic; Al 3+ is a highly charged metal cation and a Lewis acid; Cl is negligible.
    (b) neutral; both Na + and Br are negligible.
    (c) basic; ClO is the conjugate base of HClO; Na + is negligible.
    (d) acidic; CH 3 NH 3 + is the conjugate acid of CH 3 NH 2 ; NO 3 is negligible.
    (e) basic; SO 3 2– is the conjugate base of H 2 SO 3 ; Na + is negligible.

    16.85 Plan .  Estimate pH using relative base strength and then calculate to confirm prediction. NaCl is a neutral salt, so it is not the unknown. The unknown is a relatively weak base, because a pH of 8.08 is not very basic. Because F is a weaker base than OCl , the unknown is probably NaF. Calculate K b for the unknown from the data provided. Solve .

    [OH ] = 10 −pOH ; pOH = 14.00 – pH = 14.00 – 8.08 = 5.92

    [OH ] = 10 −5.92 = 1.202 × 10 −6 = 1.2 × 10 −6 M = [HX]

    [NaX] = [X ] = 0.050 mol salt/0.500 L = 0.10 M

    K b = [ OH ] [ HX ] [ X ] = ( 1.202 × 10 6 ) 2 ( 0.10 1.2 × 10 6 ) = ( 1.202 × 10 6 ) 2 0.10 = 1.4 × 10 11

    K b for F = K w /K a for HF = 1.0 × 10 −14 /6.8 × 10 −4 = 1.5 × 10 −11

    The unknown is NaF.


    16.86 Plan .  Estimate pH of salt solution by evaluating the ions in the salts. Calculate to confirm if necessary. Solve .

    KBr: salt of strong acid and strong base, neutral solution. The unknown is probably KBr. Check the others to be sure.
    NH 4 Cl: salt of a weak base and a strong acid, acidic solution
    KCN: salt of a strong base and a weak acid, basic solution
    K 2 CO 3 : salt of a strong base and a weak acid (HCO 3 ), basic solution

    Only KBr fits the acid–base properties of the unknown.

    Acid–Base Character and Chemical Structure (Section 16.10)

    16.87

    1. HNO 3 is a stronger acid than HNO 2 because it has one more nonprotonated oxygen atom and, thus, a higher oxidation number on N.
    2. H 2 S is a stronger acid than H 2 O. For binary hydrides, acid strength increases going down a family.
    3. H 2 SO 4 is a stronger acid than H 2 SeO 4 . For oxyacids, the greater the electronegativity of the central atom, the stronger the acid.
    4. CCl 3 COOH is stronger than CH 3 COOH because the electronegative Cl atoms withdraw electron density from other parts of the molecule, which weakens the O–H bond and makes H + easier to remove. Also, the electronegative Cl delocalizes negative charge on the carboxylate anion. This stabilizes the conjugate base, favoring products in the ionization equilibrium and increasing K a .

    16.88

    1. HCl is a stronger acid than HF. For binary hydrides, acid strength increases going down a column.
    2. H 3 PO 4 is a stronger acid than H 3 AsO 4 . For oxyacids, the more electronegative the central atom, the stronger the acid.
    3. HBrO 3 is a stronger acid than HBrO 2 because it has one more nonprotonated oxygen and a higher oxidation number on Br.
    4. H 2 C 2 O 4 is a stronger acid than HC 2 O 4 . The first ionization of a polyprotic acid is always stronger because H + is more tightly held by an anion.
    5. C 6 H 5 COOH is stronger than C 6 H 5 OH. The conjugate base of benzoic acid, C 6 H 5 COO , is stabilized by resonance, whereas the conjugate base of phenol, C 6 H 5 O , is not. C 6 H 5 COOH has greater tendency to form its conjugate base and is the stronger acid.

    16.89

    (a) BrO (HClO is the stronger acid because Cl is more electronegative than Br, so BrO is the stronger base.)
    (b) BrO (HBrO 2 has more nonprotonated O atoms and is the stronger acid, so BrO is the stronger base.)
    (c) HPO 4 2– (larger negative charge, greater attraction for H + )

    16.90

    (a) NO 2 (HNO 3 is the stronger acid because it has more nonprotonated O atoms, so NO 2 is the stronger base.)
    (b) PO 4 3– (K a for HAsO 4 2– is greater than K a for HPO 4 2– , so K b for PO 4 3– is greater and PO 4 3– is the stronger base. Note that P is more electronegative than As and H 3 PO 4 is a stronger acid than H 3 AsO 4 , which could lead to the conclusion that AsO 4 3– is the stronger base. As in all cases, the measurement of base strength, K b , supercedes the prediction. Chemistry is an experimental science.
    (c) CO 3 2– (The more negative the anion, the stronger the attraction for H + .)

    16.91

    1. True.
    2. False. In a series of acids that have the same central atom, acid strength increases with the number of nonprotonated oxygen atoms bonded to the central atom.
    3. False. H 2 Te is a stronger acid than H 2 S because the H–Te bond is longer, weaker, and more easily ionized than the H–S bond. Binary hydride acid strength increase going down a family.

    16.92

    1. True.
    2. False. For oxyacids with the same structure but different central atom, the acid strength increases as the electronegativity of the central atom increases.
    3. False. HF is a weak acid, weaker than the other hydrogen halides, primarily because the H–F bond energy is exceptionally high.

    Lewis Acids and Bases (Section 16.11)

    16.93  NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH (aq)

    Ammonia, NH 3 , acts as an Arrhenius base because it increases the concentration of hydroxide ion, OH , in aqueous solution. It acts like a Brønsted-Lowry base because it is a proton, H + , acceptor. It acts like a Lewis base because it is an electron pair donor. If a substance is an Arrhenius base, it must also be a Brønsted–Lowry base and a Lewis base.

    16.94

    1. F (aq) + H 2 O(l) ⇌ HF(aq) + OH (aq)
    2. Basic. The F ion is the conjugate base of the weak acid HF.
    3. The F ion acts as a Lewis base, donating an electron pair to water.

    16.95 Analyze/Plan .  Identify each reactant as an electron pair donor (Lewis base) or electron pair acceptor (Lewis acid). Remember that a Brønsted–Lowry acid is necessarily a Lewis acid, and a Brønsted–Lowry base is necessarily a Lewis base (Solution 16.93). Solve .

    Lewis Acid Lewis Base
    (a) Fe(ClO 4 ) 3 or Fe 3+ H 2 O
    (b) H 2 O CN
    (c) BF 3 (CH 3 ) 3 N
    (d) HIO NH 2

    16.96

    Lewis Acid Lewis Base
    (a) HNO 2 (or H + ) OH
    (b) FeBr 3 (Fe 3+ ) Br
    (c) Zn 2+ NH 3
    (d) SO 2 H 2 O

    16.97

    1. Cu 2+ , higher cation charge
    2. Fe 3+ , higher cation charge
    3. Al 3+ , smaller cation radius, same charge

    16.98

    1. ZnBr 2 , smaller cation radius, same charge
    2. Cu(NO 3 ) 2 , higher cation charge
    3. NiBr 2 , smaller cation radius, same charge

    Additional Exercises

    16.99

    1. Correct.
    2. Incorrect. A Brønsted–Lowry acid must have ionizable hydrogen. Lewis acids are electron pair acceptors, but need not have ionizable hydrogen.
    3. Correct.
    4. Incorrect. K + is a negligible Lewis acid because it is the conjugate of strong base KOH. Its relatively large ionic radius and low positive charge render it a poor attractor of electron pairs.
    5. Correct.

    16.100  Calculate moles OH , calculate moles H + , determine which is in excess after neutralization, calculate pH

    0.300 g Ca(OH) 2 × 1 mol Ca(OH) 2 74.093 g Ca(OH) 2 × 2 mol OH 1 mol Ca(OH) 2 = 8.0979 × 10 3 = 8.10 × 10 3 M OH

    1.40 M HNO 3 × 0.0500 L = 0.0700 mol H + ; H + is in excess

    0.0700 mol H + – 0.00810 mol OH = 0.0619 mol H + remain

    0.0619 mol H + /0.0750 L = 0.82536 = 0.825 M H + ; pH = –log (0.82536) = 0.03356 = 0.0834

    16.101

    H 3 C 6 H 5 O 7 + CH 3 NH 2 CH 3 NH 3 + + H 2 C 6 H 5 O 7
    citric acid methylamine odorless salt

    H 3 C 6 H 5 O 7 ⇌ H + + H 2 C 6 H 5 O 7 K a1 = 7.4 × 10− 4

    CH 3 NH 2 + H 2 O ⇌ CH 3 NH 3 + + OH K b = 4.4 × 10− 4

    H + + OH ⇌ H 2 O 1/K w = 1/1.0 × 10− 14

    H 3 C 6 H 5 O 7 + CH 3 NH 2 + H 2 O + H + + OH ⇌ H 2 C 6 H 5 O 7 + CH 3 NH 3 + + H + + OH + H 2 O

    H 3 C 6 H 5 O 7 + CH 3 NH 2 ⇌ H 2 C 6 H 5 O 7 + CH 3 NH 3 +

    K = K a 1 × K b K w = ( 7.4 × 10 4 ) ( 4.4 × 10 4 ) 1.0 × 10 14 = 3.256 × 10 7 = 3.3 × 10 7


    16.102  Statements (a), (d), and (f) are true. Statements (b), (c), and (e) are the opposites of the true statements.

    16.103 Analyze/Plan . Brønsted-Lowry acids are H + donors, Brønsted-Lowry bases are H + acceptors. Examine the structures of the molecules and ions for acidic or basic functional groups. Solve .

    1. Bicarbonate ion, HCO 3 , both acid and base. Bicarbonate ion is amphiprotic; it can act like an H + donor or acceptor.
    2. Prozac, base . Prozac contains an –NH– (amine) group. The nonbonded electron pair on N causes it to be a hydrogen ion acceptor.
    3. PABA, both acid and base. The molecule contains both a –COOH (carboxylic acid) group and an –NH 2 (amine) group. It can act like an H + donor and acceptor.
    4. TNT, neither acid nor base. TNT contains –NO 2 (nitro) groups. The N atoms in these groups do not have a nonbonded electron pair and are neither acidic nor basic.
    5. N-Methylpyridinium ion, neither acid nor base. The N atom in this ion forms four covalent bonds and does not have a nonbonded electron pair. It cannot act as an H + acceptor.

    16.104  Upon dissolving, Li 2 O dissociates to form Li + and O 2– . According to Equation 16.22, O 2– is completely protonated in aqueous solution.

    Li 2 O(s) + H 2 O(l)   →   2Li + (aq) + 2OH (aq)

    Thus, initial [Li 2 O] = [O 2 ]; [OH ] = 2[O 2– ] = 2[Li 2 O]

    [ Li 2 O ] = mol Li 2 O L solution = 2.50 g Li 2 O × 1 mol Li 2 O 29.88 g Li 2 O × 1 1.500 L = 0.0558 = 0.0558 M

    [OH ] = 0.11156 = 0.112 M ; pOH = 0.9525 = 0.953; pH = 14.00 – pOH = 13.0475 = 13.048

    16.105

    1. The conjugate base of benzoic acid is benzoate anion, C 6 H 5 COO . The conjugate acid of aniline is anilinium cation, C 6 H 5 NH 3 + .
    2. To compare relative acidity, compare the K a values for benzoic acid and anilinium ion.

      K a for C 6 H 5 NH 3 + = K w K b for C 6 H 5 NH 2 = 1.0 × 10 14 4.3 × 10 10 = 2.3256 × 10 5 = 2.3 × 10 5

      K a for C 6 H 5 COOH, 6.3 × 10 −5 , is greater than K a for C 6 H 5 NH 3 + , 2.3 × 10 −5 . The 0.10 M solution of benzoic acid will be somewhat more acidic.

    3. C 6 H 5 COOH ⇌ H + + C 6 H 5 COO K a = 6.3 × 10 −5

      C 6 H 5 NH 2 + H 2 O ⇌ C 6 H 5 NH 3 + + OH K b = 4.3 × 10 −10

      H + + OH ⇌ H 2 O 1/K w = 1/1.0 × 10− 14

      C 6 H 5 COOH+C 6 H 5 NH 2 +H 2 O+H + +OH ⇌ C 6 H 5 COO +C 6 H 5 NH 3 + +H + +OH +H 2 O

      C 6 H 5 COOH + C 6 H 5 NH 2 ⇌ C 6 H 5 COO + C 6 H 5 NH 3 +

    K = K a × K b K w = ( 6.3 × 10 5 ) ( 4.3 × 10 10 ) 1.0 × 10 14 = 2.7090 = 2.7


    16.106  Assume T = 25 °C. If [OH ] =2.5 × 10 −9 M , pOH = 8.60 and pH = 5.40. This does not make sense (!) because NaOH is a strong base. Usually, we assume that [H + ] and [OH ] from the autoionization of water do not contribute to the overall [H + ] and [OH ]. However, for acid or base solute concentrations less than 1 × 10 −6 M , the autoionization of water produces significant [H + ] and [OH ] and we must consider it when calculating pH.

    H 2 O(l) [H + ] + [OH ]
    initial C 0 2.5 × 10 −9 M
    equil. C x (x + 2.5 × 10 −9 ) M

    K w = 1.0 × 10 −14 = [H + ][OH ] = (x)(x + 2.5 × 10 −9 ); x 2 + 2.5 × 10 −9 x – 1.0 × 10 −14 = 0

    From the quadratic formula, x = 2.5 × 10 9 ± ( 2.5 × 10 9 ) 2 4 ( 1 ) ( 1 × 10 14 ) 2 ( 1 ) = 9.876 × 10 8 = 9.9 × 10 8 M H +

    [H + ] = 9.9 × 10 −8 M ; [OH ] = (9.876 × 10 −8 + 2.5 × 10 −9 ) = 1.013 × 10 −7 = 1.0 × 10 −7 M

    pH = 7.0054 = 7.01

    Check : [9.876 × 10 −8 ][1.013 × 10 −7 ] = 1.0 × 10 −14 . Now our answer makes sense. The very small concentration of OH from the solute raises the solution pH to slightly more than 7.

    16.107

    1. False. H 2 C 2 O 4 has no capacity to accept H + .
    2. True.
    3. True. HC 2 O 4 can act like either a Brønsted–Lowry acid or base. It is a stronger acid than base (K a2 > K b1 ), so a solution of the salt will be acidic.

    16.108

    H 2 Suc(aq) H + (aq) + HSuc (aq) K a1 = 6.9 × 10 −5
    HSuc (aq) H + (aq) + Suc 2– (aq) K a2 = 2.5 × 10 −6
    1. Calculating the [H + ] from the first ionization. The equilibrium concentrations are [H + ] = [HSuc ] = x; [H 2 Suc] = 0.32 – x.

      K a1 = [ H + ] [ HSuc ] [ H 2 Suc ] = x 2 0.32 x ; assume x is small relative to 0 .32

      x 2 = ( 0.32 ) ( 6.9 × 10 5 ) ; x = 0.0046989 = 0.0047 M H + ; pH = 2.328 = 2.33

    2. To calculate [Suc 2– ], consider the second ionization equilibrium. Initial [HSuc ] and [H + ] are 0.0047 M , from the first ionization. Then,

      [Suc 2– ] = y; [H + ] = 0.0047 + y; [HSuc ] = 0.0047 – y.

      K a2 = ( 0.0047 + y) (y) ( 0.0047 y ) = 2.5 × 10 6 ; assuming y is small compared to 0 .0047,

      y = 2.5 × 10 −6 M Suc 2– . This assumption is reasonable, because 2.5 × 10 −6 is only 0.05% of 0.0047.


    1. The assumption in part (a) is reasonable. From part (b), [Suc 2– ] and [H + ] from the second ionization are 2.5 × 10 −6 M. This is only 0.05% of 0.0047 M , [H + ] from the first ionization. Only the first dissociation is relevant for calculating pH.
    1. Acidic. HSuc can act like either a Brønsted–Lowry acid or base. It is a stronger acid than base (K a2 > K b1 ), so a solution of the salt will be acidic.

    16.109

    1. K b = K w /K a ; pK b = 14 – pK a ; pK b = 14 – 4.84 = 9.16
    2. K a for butyric acid (buCOOH) is 10 −4.84 = 1.4454 × 10 −5 = 1.4 × 10 −5

      K a = [H + ][buCOO ] [buCOOH] ; [ H + ] = [ buCOO ] = x ; [ buCOOH ] = 0.050 x

      1.4454 × 10 5 = x 2 0.050 x ; assume x is small relative to 0 .050

      x 2 = 7.227 × 10 −7 ; x = [H + ] = 8.501 × 10 −4 = 8.5 × 10 −4 M H + ; pH = 3.07

      (This represents 1.7% ionization, so the approximation is valid.)

    3. K b for butyrate anion (buCOO ) is 10 −9.16 = 6.918 = 6.918 × 10 −10 = 6.9 × 10 −10

      K b = [ OH ] [ buCOOH ] [ buCOO ] ; [ OH ] = [ buCOOH ] = x ; [ buCOO ] = 0.050 x

      6.918 × 10 10 = x 2 0.050 x ; assume x is small relative to 0 .050

      x 2 = 3.459 × 10 −11 ; x = [OH ] = 5.881 × 10 −6 = 5.9 × 10 −6 M OH

      pOH = 5.23; pH = 8.77

    16.110 Analyze/Plan .  Evaluate the acid–base properties of the cation and anion to determine whether a solution of the salt will be acidic, basic, or neutral. Solve .

    1. NH 4 NO 3 : NH 4 + , weak conjugate acid of NH 3 ; NO 3 , negligible conjugate base of HNO 3 ; acidic solution
    2. NaNO 3 : Na + , negligible conjugate acid of NaOH; NO 3 , negligible conjugate base of HNO 3 ; neutral solution
    3. CH 3 COONH 4 : NH 4 + , weak conjugate acid of NH 3 , K a = K w /1.8 × 10 −5 = 5.6 × 10 −10 ; CH 3 COO , weak conjugate base of CH 3 COOH, K b = K w /1.8 × 10 −5 = 5.6 × 10 −10 ; neutral solution (K a for the cation and K b for the anion are accidentally equal, producing a neutral solution)
    4. NaF: Na + , negligible conjugate acid of NaOH; F , weak conjugate base of HF, K b = K w /6.8 × 10 −4 = 1.5 × 10 −11 ; basic solution
    5. CH 3 COONa: Na + , negligible; CH 3 COO , weak base, K b = 5.6 × 10 −10 ; basic solution

    In order of increasing acidity (and decreasing pH):

    0.1 M CH 3 COONa > 0.1 M NaF > 0.1 M CH 3 COONH 4 = 0.1 M NaNO 3 > 0.1 M NH 4 NO 3 ; (v) > (iv) > (iii) ~ (ii) > (i)

    (iv) and (v) are both bases, and (v) has the greater K b value and higher pH. (ii) and (iii) are both neutral and (i) is acidic.


    16.111  Calculate K b for A and then K a for HA.

    A = [NaA] = [A ] = 0.25 M ; [OH ] = 10 −pOH ; pOH = 14.00 – pH = 14.00 – 9.29 = 4.71

    [OH ] = 10 −4.71 = 1.950 × 10 −5 = 2.0 × 10 −5 M = [HX]

    K b = [ OH ] [ HA ] [ A ] = ( 1.950 × 10 5 ) 2 ( 0.25 1.950 × 10 5 ) = ( 1.950 × 10 5 ) 2 0.25 = 1.521 × 10 9 = 1.5 × 10 9

    K a for HA = K w /K b for A = 1.0 × 10 −14 /1.521 × 10 −9 = 6.576 × 10 −6 = 6.6 × 10 −6

    16.112  The value of pK a2 can only be choice (iii).

    Calculate K a1 , assuming that only the first ionization determines pH.

    [H 2 A] = 0.10 M ; [H + ] = [HA ] = 10 −pH = 10 −3.30 = 5.0119 × 10 −4 = 5.0 × 10 −4

    K a1 = [ H + ] [ A ] [ HA ] = ( 5.0119 × 10 4 ) 2 ( 0.10 5.0119 × 10 4 ) = ( 5.0119 × 10 4 ) 2 0.0995 = 2.524 × 10 6 = 2.5 × 10 6

    pK a1 = 5.60. pK a2 must be greater than pK a1 , which eliminates choices (i) and (ii).

    If a solution of the salt NaHA is acidic, then K a2 > K b1 and pK a2 < pK b1 .

    pK b1 = 14.00 – pK a1 = 14.00 – 5.60 = 8.84. This eliminates choice (iv).

    16.113  Call each compound in the neutral form Q.

    Then, Q(aq) + H 2 O(l) ⇌ QH + (aq) + OH . K b = [QH + ] [OH ]/[Q]

    The ratio in question is [QH + ]/[Q], which equals K b /[OH ] for each compound.

    At pH = 2.5, pOH = 11.5, [OH ] = antilog (–11.5) = 3.16 × 10 −12 = 3 × 10 −12 M . Now calculate K b /[OH ] for each compound:

    Nicotine [QH + ] [Q] = 7 × 10 7 / 3.16 × 10 12 = 2 × 10 5

    Caffeine [QH + ] [Q] = 4 × 10 14 / 3.16 × 10 12 = 1 × 10 2

    Strychnine [QH + ] [Q] = 1 × 10 6 / 3.16 × 10 12 = 3 × 10 5

    Quinine [QH + ] [Q] = 1.1 × 10 6 / 3.16 × 10 12 = 3.5 × 10 5

    For all the compounds except caffeine, the protonated form has a much higher concentration than the neutral form. However, for caffeine, a very weak base, the neutral form dominates.

    16.114

    1. Consider the formation of the zwitterion as a series of steps (Hess’s law).

      NH 2 CH 2 COOH + H 2 O NH 2 CH 2 COO + H 3 O + K a

      NH 2 CH 2 COOH + H 2 O + NH 3 CH 2 COOH + OH K b

      H 3 O + + OH ⇌ 2H 2 O           1/K w

      NH 2 CH 2 COOH + NH 3 CH 2 COO K a × K b K w

      K = K a × K b K w = ( 4.3 × 10 3 ) ( 6.0 × 10 5 ) 1.0 × 10 14 = 2.6 × 10 7


    1. As glycine exists as the zwitterion in aqueous solution, the pH is determined by the following equilibrium.

      + NH 3 CH 2 COO + H 2 O NH 2 CH 2 COO + H 3 O +

      K a = [ NH 2 CH 2 COO ] [ H 3 O + ] [ + NH 3 CH 2 COO ] = K w K b = 1.0 × 10 14 6.0 × 10 5 = 1.67 × 10 10 = 1.7 × 10 10

      x = [ H 3 O + ] = [ NH 2 CH 2 COO ] ; K a = 1.67 × 10 10 = ( x) (x) (0 .050 x) x 2 0.050

      x = [H 3 O + ] = 2.89 × 10 −6 = 2.9 × 10 −6 M ; pH = 5.54

    1. In strongly basic solution (pH 13), the –NH 3 + group would be deprotonated, so glycine would be in the form H 2 NCH 2 CO 2 . In a strongly acidic (pH 1) solution, the –CO 2 function would be protonated, so glycine would exist as + H 3 NCH 2 COOH.

    16.115  Answer (c) is correct. Water itself is amphiprotic; it can act like an H + donor or acceptor. The autoionization equilibrium, Equation 16.12, is also unique in that it describes water acting as an acid and a base simultaneously. The value of K w is the value of K a for water acting like an acid and K b for water acting like a base. Thus, the pK b of water is 14 (and the pK a of water is 14.)

    Integrative Exercises

    16.116  At 25 °C, [H + ] = [OH ] = 1.0 × 10 −7 M

    1.0 × 10 7 mol H + 1 L H 2 O × 0.0010 L × 6 .022 × 10 23 H + ions mol H + = 6.0 × 10 13 H + ions

    16.117 Analyze .  Based on the mass % and density of concentrated HCl, calculate volume of concentrated solution required to produce 10.0 L of HCl with pH = 2.05. Plan . Calculate molarity of concentrated solution from density and mass %. Calculate molarity of dilute solution from pH. Use the dilution formula to calculate volume (mL) of concentrated solution required. Solve .

    1.18 g conc . soln . mL conc . soln . × 36.0 g HCl 100 g conc . soln . × 1000 mL 1 L × 1 mol HCl 36.46 g HCl = 11.651 mol HCl/L = 11.7 M HCl/L

    For the dilute HCl solution, [H + ] = 10 −pH = 10 −2.05 = 8.913 × 10 −3 = 8.9 × 10 −3 M HCl

    M c × L c = M d × M d ; 11.651 × L c = 8.913 × 10 −3 M × 10.0 L;

    L c = 7.650 × 10 3 ; 7.650 × 10 3 L × 1000 mL 1 L = 7.65 = 7.7 mL conc . HCl

    16.118  [H + ] = 10 −pH = 10 −2 = 1 × 10 −2 M H + ; 1 × 10 −2 M × 0.400 L = 4.0 × 10 −3 = 4 × 10 −3 mol H +

    HCl(aq) + HCO 3 (aq) → Cl (aq) + H 2 O(l) + CO 2 (g)

    4 × 10 3 mol H + = 4 × 10 3 mol HCO 3 × 84.01 g NaHCO 3 1 mol HCO 3 = 0.336 = 0.3 g NaHCO 3


    16.119 Analyze .  If pH were directly related to CO 2 concentration, this exercise would be simple. Unfortunately, we must solve the equilibrium problem for the diprotic acid H 2 CO 3 to calculate [H + ] and pH. We are given ppm CO 2 in the atmosphere at two different times and the pH that corresponds to one of these CO 2 levels. We are asked to find pH at the other atmospheric CO 2 level.

    Plan .  Assume all dissolved CO 2 is present as H 2 CO 3 (aq) (Sample Exercise 16.14).

    pH → [H + ] → [H 2 CO 3 ]. Although H 2 CO 3 is a diprotic acid, the two K a values differ by more than 10 3 , so we can ignore the second ionization when calculating [H 2 CO 3 ]. Change 380 ppm CO 2 to pressure and calculate the Henry’s law constant for CO 2 . Calculate the dissolved [CO 2 ] = [H 2 CO 3 ] at 315 ppm and then solve the K a1 expression for [H + ] and pH. Solve .

    1. H 2 CO 3 (aq) ⇌ H + (aq) + HCO 3 (aq)

      K a1 = 4.3 × 10 7 = [ H + ] [ HCO 3 ] [ H 2 CO 3 ] ; [ H + ] = 10 5.4 = 3.98 × 10 6 = 4 × 10 6 M

      [H + ] = [HCO 3 ]; [H 2 CO 3 ] = x – 4 × 10 −6

      4.3 × 10 7 = ( 3.98 × 10 6 ) 2 ( x 3.98 × 10 6 ) ; 4.3 × 10 7 x = 1.585 × 10 11 + 1.711 × 10 12

      x = 1.756 × 10 −11 /4.3 × 10 −7 = 4.084 × 10 −5 = 4 × 10 −5 M H 2 CO 3

      380 ppm = 380 mol CO 2 /1 × 10 6 mol air = 0.000380 mol % CO 2

      Because of the properties of gases, mol % = pressure %. P CO 2 = 0.000380 atm . According to Equation 13.4, S CO 2 = kP CO 2 ;

      4.084 × 10 −5 mol/L = k(3.80 × 10 −4 atm); k = 0.1075 = 0.1 mol/L -atm .

      Forty years ago, S CO 2 = 0.1075 mol L-atm × 3.15 × 10 4 atm = 3.385 × 10 5 = 3 × 10 5 M

      Now solve K a1 for [H + ] at this [H 2 CO 3 ]. [H + ] = x.

      We cannot assume x is small, because [H 2 CO 3 ] is so low.

      4.3 × 10 −7 = x 2 /(3.385 × 10 −5 – x); x 2 + 4.3 × 10 −7 x – 1.456 × 10 −11 = 0

      x = 4.3 × 10 7 ± ( 4.3 × 10 7 ) 2 4 ( 1.456 × 10 11 ) 2 = 4.3 × 10 7 + 7.644 × 10 6 2 = 3.607 × 10 6 = 4 × 10 6 M H + ; [H + ] = 4 × 10 −6 M ; pH = 5.443 = 5.4

      (Note that, to the precision that the pH data is reported, the change in atmospheric CO 2 leads to no change in pH.)

    2. From part (a), [H 2 CO 3 ] today = 4.084 × 10 −5 M

      4.084 × 10 5 mol H 2 CO 3 1 L × 20.0 L = 8.168 × 10 4 = 8 × 10 4 mol CO 2

      V = nRT P = 8.168 × 10 4 mol × 298 K 1 .0 atm × 0.08206 L-atm mol-K = 0.01997 = 0.02 L = 20 mL


    16.120

    1. K w = [H + ][OH ] = 5.48 × 10 −14 . In pure water, [H + ] = [OH ].

      [H + ] 2 = 5.48 × 10 −14 [H + ] = 2.34 × 10 −7 M ; pH = 6.63

    2. The value of K w increases with increasing temperature, so the sign of ΔH is positive. The autoionization of water is endothermic.

    16.121

    1. 24 valence e , 12 e pairs
      image

      The formal charges on all atoms are zero. Structures with multiple bonds lead to nonzero formal charges. There are three electron domains about Al. The electron-domain geometry and molecular structure are trigonal planar.

    2. The Al atom in AlCl 3 has an incomplete octet and is electron deficient. It “needs” to accept another electron pair, to act like a Lewis acid.
    3. image

      Both the Al and N atoms in the product have tetrahedral geometry.

    4. The Lewis theory is most appropriate. H + and AlCl 3 are both electron pair acceptors, Lewis acids.

    16.122 Plan .  Use acid ionization equilibrium to calculate the total moles of particles in solution. Use density to calculate kg solvent. From the molality ( m ) of the solution, calculate ΔT b and T b . Solve .

    HSO 4 (aq) H + (aq) + SO 4 2– (aq)
    initial 0.10 M 0 0
    equil. 0.10 – x M x M x M
    0.071 M 0.029 M 0.029 M

    K a = 1.2 × 10 2 = [ H + ] [ SO 4 2 ] [ HSO 4 ] = x 2 0.10 x ; K a is relatively large, so use the quadratic.

    x 2 + 0.012 x 0.0012 = 0 ; x = 0.012 ± ( 0.012 ) 2 4 ( 1 ) ( 0.0012 ) 2 ; x = 0.029 M H + , SO 4 2

    Total ion concentration = 0.10 M Na + + 0.071 M HSO 4 + 0.029 M H + + 0.029 M SO 4 2– = 0.229 = 0.23 M

    Assume 100.0 mL of solution. 1.002 g/mL × 100.0 mL = 100.2 g solution.

    0.10 M NaHSO 4 × 0.1000 L = 0.010 mol NaHSO 4 × 120.1 g NaHSO 4 mol NaHSO 4 = 1.201 = 1.2 g NaHSO 4

    100.2 g soln – 1.201 g NaHSO 4 = 99.0 g = 0.099 kg H 2 O

    m = mol ions kg H 2 O = 0.229 M × 0.1000 L 0.0990 kg = 0.231 = 0.23 m ions

    ΔT b = K b ( m ) = 0.52 °C/ m × (0.23 m ) = +0.12 °C; T b = 100.0 + 0.12 = 100.1 °C


    16.123

    Rx 1: ΔH = D(H–F) + 2D(H–O) – 3D(H–O) = D(H–F) – D(H–O)
    ΔH = 567 kJ – 463 kJ = 104 kJ
    Rx 2: ΔH = D(H–Cl) + 2D(H–O) – 3D(H–O) = D(H–Cl) – D(H–O)
    ΔH = 431 kJ – 463 kJ = –32 kJ

    The reaction involving HCl is exothermic, whereas the reaction involving HF is endothermic, owing to the smaller bond dissociation enthalpy of H–Cl. HCl is a stronger acid than HF, and the enthalpy of ionization for HCl is exothermic, whereas that of HF is endothermic. This is consistent with the trend in acid strength for binary acids with heavy atoms (X) in the same family. That is, the longer and weaker the H–X bond, the stronger the acid (and the more exothermic the ionization reaction).

    16.124  Calculate M of the solution from osmotic pressure and K b using the equilibrium expression for the hydrolysis of cocaine. Let Coc = cocaine and CocH + be the conjugate acid of cocaine.

    Π = M RT; M = Π / RT = 52.7 torr 288 K × 1 atm 760 torr × mol-K 0 .08206 L-atm = 0.002934 = 2.93 × 10 3 M Coc

    pH = 8.53; pOH = 14 – pH = 5.47; [OH ] = 10 −5.47 = 3.39 × 10 −6 = 3.4 × 10 −6 M

    Coc(aq) + H 2 O(l) CocH + (aq) + OH (aq)
    initial 2.93 × 10 −3 M 0 0
    equil. (2.93 × 10 −3 – 3.4 × 10 −6 ) M 3.4 × 10 −6 M 3.4 × 10 −6 M

    K b = [ CocH + ] [ OH ] [ Coc ] = ( 3.39 × 10 6 ) 2 ( 2.934 × 10 3 3.39 × 10 6 ) = 3.9 × 10 9

    Note that % hydrolysis is small in this solution, so 3.39 × 10 −6 M is small compared to 2.93 × 10 −3 M and can be ignored in the denominator of the calculation.

    16.125

    1. rate = k[IO 3 ][SO 3 2– ][H + ]
    2. ΔpH = pH 2 – pH 1 = 3.50 – 5.00 = –1.50

      ΔpH = –log [H + ] 2 – (–log [H + ] 1 ); –ΔpH = log [H + ] 2 – log [H + ] 1

      –ΔpH = log [H + ] 2 / [H + ] 1 ; [H + ] 2 / [H + ] 1 = 10 Δ pH

      [H + ] 2 /[H + ] 1 = 10 1.50 = 31.6 = 32. The rate will increase by a factor of 32 if [H + ] increases by a factor of 32. The reaction goes faster at lower pH.

    3. As H + does not appear in the overall reaction, it is either a catalyst or an intermediate. An intermediate is produced and then consumed during a reaction, so its contribution to the rate law can usually be written in terms of concentrations of other reactants (Sample Exercise 14.15). A catalyst is present at the beginning and end of a reaction and can appear in the rate law if it participates in the rate-determining step (Solution 14.78). This reaction is pH dependent because H + is a homogeneous catalyst that participates in the rate-determining step499.

    16.126

    1. (i) HCO 3 (aq) ⇌ H + (aq) + CO 3 2 ( aq) K 1 = K a2 for H 2 CO 3 = 5.6 × 10 11
      H + (aq) + OH–(aq) ⇌ H 2 O(l) K 2 = 1 /K w = 1 × 10 14
      HCO 3 ( aq) + OH (aq) CO 3 2 (aq) + H 2 O(l) K = K 1 × K 2 = 5.6 × 10 3
      (ii) NH 4 + ( aq) H + (aq) + NH 3 ( aq) K 1 = K a for NH 4 + = 5.6 × 10 10
      CO 3 2 (aq) + H + ( aq) HCO 3 (aq) K 2 = 1 /K a2 for H 2 CO 3 = 1.8 × 10 10
      NH 4 + ( aq) + CO 3 2 (aq) HCO 3 (aq) + NH 3 (aq) K = K 1 × K 2 = 10
    2. Both (i) and (ii) have K > 1, although K = 10 is not much greater than 1. Both could be written with a single arrow. (This is true in general when a strong acid or strong base, H + (aq) or OH (aq), is a reactant.)

     

    17 Additional Aspects of Aqueous Equilibria

    Visualizing Concepts

    17.1 Analyze .  Given diagrams showing equilibrium mixtures of HX and X with different compositions, decide which has the highest pH. HX is a weak acid and X is its conjugate base. Plan . Evaluate the contents of the boxes. Use acid–base equilibrium principles to relate [H + ] to box composition. Solve .

    Use the following acid ionization equilibrium to describe the mixtures:

    HX(aq) ⇌ H + (aq) + X (aq). Each box has 4 HX molecules, but differing amounts of X ions. The greater the amount of X (conjugate base), for the same amount of HX (weak acid), the lower the amount of H + and the higher the pH. The middle box, with most X , has least H + and highest pH.

    17.2

    1. The yellow solution has the higher pH. According to Figure 16.8, methyl orange is yellow above pH 4.5 and red (really pink) below pH 3.5. The beaker on the left has a pH greater than 4.5, and the one on the right has pH less than 3.5. (By calculation, pH of left beaker = 4.7, pH of right beaker = 2.9.) The right beaker, with lower pH and greater [H + ], is pure acetic acid. The left beaker contains equal amounts of the weak acid and its conjugate base, acetic acid and acetate ion. Adding the ”common-ion” acetate (in the form of sodium acetate) shifts the acid ionization equilibrium to the left, decreases [H + ], and raises pH.
    2. When small amounts of NaOH are added, the left beaker is better able to maintain its pH. For solutions of the same weak acid, pH depends on the ratio of conjugate base to conjugate acid. Small additions of base (or acid) have the least effect when this ratio is close to one. The left beaker is a buffer because it contains a weak conjugate acid/conjugate base pair and resists rapid pH change upon addition of small amounts of strong base or acid.

    17.3   Statement (b) is correct, [HA] > [A ]. Buffers prepared from weak acids (HA) and their conjugate bases (A , usually in the form of a salt) have pH values in a range of approximately 2 pH units, centered around pK a for the weak acid. If concentration of the weak acid is greater than concentration of the conjugate base, pH < pK a . If concentration of the conjugate base is greater than concentration of the weak acid, pH > pK a . This is generally true for buffers containing a weak conjugate acid/conjugate base (CA/CB) pair.

    [CA] > [CB], pH of buffer < pK a of CA

    [CA] < [CB], pH of buffer > pK a of CA


    17.4 Analyze/Plan .  When strong acid is added to a buffer, it reacts with conjugate base (CB) to produce conjugate acid (CA). [CA] increases and [CB] decreases. The opposite happens when strong base is added to a buffer, [CB] increases and [CA] decreases. Match these situations to the drawings. Solve .

    The buffer begins with equal concentrations of HX and X .

    1. After addition of strong acid, [HX] will increase and [X ] will decrease. Drawing (3) fits this description.
    2. Adding of strong base causes [HX] to decrease and [X ] to increase. Drawing (1) matches the description.
    3. Drawing (2) shows both [HX] and [X ] to be smaller than the initial concentrations shown on the left. This situation cannot be achieved by adding strong acid or strong base to the original buffer.

    17.5 Analyze/Plan .  Consider the reaction HA + OH → A + H 2 O. What are the major species present in solution at the listed stages of the titration? Which diagram represents these species? Solve .

    1. Before addition of NaOH , the solution is mostly HA. The only A is produced by the ionization equilibrium of HA and is too small to appear in the diagram. This situation is shown in diagram (iii), which contains only HA.
    2. After addition of NaOH but before the equivalence point , some, but not all, HA has been converted to A . The solution contains a mixture of HA and A ; this is shown in diagram (i).
    3. At the equivalence point , all HA has been converted to A , with no excess HA or OH present. This is shown in diagram (iv).
    4. After the equivalence point , the same amount of A as at the equivalence point is present, plus some excess OH . This is diagram (ii).

    17.6 Analyze/Plan .  In each case, the first substance is in the buret, and the second is in the flask. If acid is in the flask, the initial pH is low; with base in the flask, the pH starts high. Strong acids have lower pH than weak acids; strong bases have higher pH than weak bases. Polyprotic acids and bases have more than one ”jump” in pH. Solve .

    1. Strong base in flask, pH starts high, ends low as acid is added. Only diagram (ii) fits this description.
    2. Weak acid in flask, pH starts low, but not extremely low. Diagrams (i), (iii), and (iv) all start at low pH and get higher. Diagram (i) has very low initial pH, and likely has strong acid in the flask. Diagram (iv) has two pH jumps, so it has a polyprotic acid in the flask. Diagram (iii) best fits the profile of adding a strong base to a weak acid.
    3. Strong acid in the flask, pH starts very low, diagram (i).
    4. Polyprotic acid, more than one pH jump, diagram (iv).

    17.7 Analyze . Given two titration curves where 0.10 M NaOH is the titrant, decide which represents the more concentrated acid, and which the stronger acid.

    Plan . For equal volumes of acid, concentration is related to volume of titrant (0.10 M NaOH) at the equivalence points. To determine K a , pH = pK a half-way to the equivalence point.

    Solve .

    1. Both acids have one ionizable hydrogen, because there is one “jump” in each titration curve. For equal volumes of acid, and the same titrant, the more concentrated acid requires a greater volume of titrant to reach equivalence. The equivalence point of the blue curve is at 25 mL NaOH and of the red curve is at 35 mL NaOH. The red acid is more concentrated.
    2. According to the Henderson–Hasselbach equation, pH = pK a + log [ conj . base] [conj . acid] .

      At half-way to the equivalence point, [conj. acid] = [conj. base] and pH = K a of the conjugate acid. For the blue curve, half-way is 12.5 mL NaOH. The pH at this volume is approximately 7.0. For the red curve, half-way is 17.5 mL NaOH. The pH at this volume is approximately 4.2. A pK a of 7 corresponds to K a of 1 × 10– 7 , whereas pK a of 4.2 corresponds to K a of 6 × 10– 5 . The red acid has the larger K a value.

    Note that the stronger acid, the one with the larger K a value, has a larger change in pH (jump) at the equivalence point. Also note that initial acid pH was not a definitive measure of acid strength, because the acids have different starting concentrations. Both K a values and concentration contribute to solution pH.

    17.8 Analyze/Plan . The beaker of saturated Cd(OH) 2 (aq) contains undissolved Cd(OH) 2 (s), Cd 2+ (aq), and OH (aq). Decide how amounts of each of these three components change when HCl(aq) is added. Solve.

    When HCl(aq) is added, it reacts with OH (aq) to form H 2 O(l) and Cl (aq). (Both have been omitted from the figure.) When OH (aq) is removed from solution, more Cd(OH) 2 (s) dissolves to replace it; [Cd 2+ (aq)] increases, [OH (aq)] decreases and the amount of undissolved Cd(OH) 2 (s) decreases. In the resulting solution, [Cd 2+ (aq)] is greater than [OH (aq)] and there is less undissolved solid on the bottom of the beaker. Beaker A accurately represents the solution after equilibrium is reestablished.

    17.9 Analyze/Plan .  Common anions or cations decrease the solubility of salts. Ions that participate in acid–base or complex ion equilibria increase solubility. Solve .

    1. CO 2 3– from BaCO 3 reacts with H + from HNO 3 , causing solubility of BaCO 3 to increase with increasing HNO 3 concentration. This behavior matches the right diagram.
    2. Extra CO 2 3– from Na 2 CO 3 decreases the solubility of BaCO 3 . Solubility of BaCO 3 decreases as [Na 2 CO 3 ] increases. This behavior matches the left diagram.
    3. NaNO 3 has no common ions, nor does it enter into acid–base or complex ion equilibria with Ba 2+ or CO 3 2– ; it does not affect the solubility of BaCO 3 . This behavior is shown in the center diagram.

    17.10 Analyze/Plan . Calculate the molarity of the solution assuming all Ca(OH) 2 (s) dissolves. Use this concentration along with the K sp expression for Ca(OH) 2 to answer the questions. Solve .

    1. [ Ca(OH ) 2 ] = 0.370 g Ca(OH ) 2 0.500 L soln × 1 mol Ca(OH) 2 74.093 g Ca(OH) 2 = 0.00998745 = 0.00999 M

      [Ca 2+ ] = 0.00999 M ; [OH ] = 2(0.00998745) = 0.0199749 = 0.0200 M;

      K sp = [Ca 2+ ][OH ] 2 . Calculate the reaction quotient using the calculated molarities. If it is equal to or greater than K sp , the resulting solution is saturated. Q = (0.0098745)(0.0199749) 2 = 3.99 × 10 –6 . Q < K sp (6.5 × 10 –6 ) and the solution is not saturated.

    2. Consider the beakers individually.
      1. The 50 mL of 1.0 M HCl is more than enough to neutralize 50 mL of

        0.0200 M OH (aq). No precipitate forms.

      2. NaCl does not react with Ca(OH) 2 and the two compounds contain no common ions. No precipitate forms.
      3. CaCl 2 does contain a common ion. Calculate Q for the resulting solution to see if Ca(OH) 2 precipitates. [OH ] in the new solution is 0.00999 M , because it is diluted by a factor of 2. [Ca 2+ ] = (1.0 + 0.00999)/2 = 0.5050 M . Q = (0.5050)(0.00999) 2 = 5.04 × 10 –5 . Q > K sp (6.5 × 10 –6 ) and Ca(OH) 2 precipitates.
      4. A common ion with a different concentration; [Ca 2+ ] = (0.10 + 0.00999)/2 = 0.0550 = 0.055 M . Q = (0.0550)(0.00999) 2 = 5.49 × 10 –6 . Q ≈ K sp (6.5 × 10 –6 ); the solution is very nearly saturated, but no precipitate forms.

    17.11   Statement (c) explains the shape of the graph. Solubility is high initially, at low pH, and then decreases to a minimum as pH increases. This depicts formation of an insoluble hydroxide as [H + ] decreases and [ OH ] increases. Additional base then reacts with the insoluble hydroxide to dissolve it. This curve depicts the behavior of an amphoteric insoluble hydroxide; it dissolves upon addition of either acid or base.

    17.12   According to Figure 17.23, the two precipitating agents are 6 M HCl (first) and H 2 S in 0.2 M HCl (second).

    Cation A = Ag + (precipitates as AgCl)

    Cation B = Cu + (precipitates as CuS, acid insoluble)

    Cation C = Ni 2+ (remains in acidic solution)

    The Common-Ion Effect (Section 17.1)

    17.13   Statement (a) is most correct. The common ion can be either the cation or anion of a salt. The common-ion effect applies to all ions if they are “common” to the salt in question, but not to noncommon ions. Common ions do no affect the equilibrium constant.


    17.14   The added salt is soluble and increases [HB + ] in the solution. For a generic weak base B, K b = [ HB + ] [ OH ] [ B ] .

    1. Stay the same. Addition of a common ion such as HB + does not change the equilibrium constant.
    2. Increase. To maintain the value of the equilibrium constant, addition of HB + requires that [B] also increases.
    3. Decrease. Additional HB + reacts with OH , lowering the pH of the solution.

    17.15 Analyze/Plan .  Follow the logic in Sample Exercise 17.1. Solve .

    1. image

      K a = 1.3 × 10 5 = [ H + ] [ C 2 H 5 COO ] [ C 2 H 5 COOH ] = ( x) (0 .060 + x) ( 0 .085 x)

      Assume x is small compared to 0.060 and 0.085.

      1.3 × 10 5 = 0.060 x 0 .085 ; x = 1.8 × 10 5 = [ H + ] , pH = 4.73

      Check .  Because the extent of ionization of a weak acid or base is suppressed by the presence of a conjugate salt, the 5% rule usually holds true in buffer solutions.

    2. image

      K b = 6.4 × 10 5 = [ OH ] [ (CH 3 ) 3 NH + ] [ (CH 3 ) 3 N ] = ( x) (0 .10 + x) ( 0 .075 x) 0.10 x 0 .075

      x = 4.8 × 10 –5 = [OH ], pOH = 4.32, pH = 14.00 – 4.32 = 9.68

      Check .  In a buffer, if [conj. acid] > [conj. base], pH < pK a of the conj. acid.

      If [conj. acid] < [conj. base], pH > pK a of the conj. acid. In this buffer, pK a of (CH 3 ) 3 NH + is 9.81. [(CH 3 ) 3 NH + ] > [(CH 3 ) 3 N] and pH = 9.68, less than 9.81.

    3. mol = M × L; mol CH 3 COOH = 0.15 M × 0.0500 L = 7.5 × 10 –3 mol

      mol CH 3 COO = 0.20 M × 0.0500 L = 0.010 mol

      image

      [CH 3 COOH(aq) ] = (7.5 × 10 –3 – x) mol/0.1000 L;

      [CH 3 COO (aq)] = (0.010 + x) mol/0.1000 L


    1. K a = 1.8 × 10 5 = [ H + ] [ CH 3 COO ] [ CH 3 COOH ] = ( x) (0 .010 + x)/0 .1000 L ( 0.0075 x)/0 .1000 L x(0 .010) 0 .0075

      x = 1.35 × 10 –5 M = 1.4 × 10 –5 M H + ; pH = 4.87

      Check .  pK a for CH 3 COOH = 4.74. [CH 3 COO ] > [CH 3 COOH], pH of buffer = 4.87, greater than 4.74.

    17.16 Analyze/Plan .  Follow the logic in Sample Exercise 17.1. Solve .

    1. HCOOH is a weak acid, and HCOONa contains the common ion HCOO , the conjugate base of HCOOH. Solve the common-ion equilibrium problem.
      image

      K a = 1.8 × 10 4 = [H + ] [HCOO ] [ HCOOH ] = ( x) (0 .250 + x) ( 0.100 x) 0.250 x 0.100

      x = 7.20 × 10 –5 = 7.2 × 10 –5 M = [H + ], pH = 4.14

      Check .  Because the extent of ionization of a weak acid or base is suppressed by the presence of a conjugate salt, the 5% rule usually holds true in buffer solutions.

    2. C 5 H 5 N is a weak base, and C 5 H 5 NHCl contains the common ion C 5 H 5 NH + , which is the conjugate acid of C 5 H 5 N. Solve the common-ion equilibrium problem.
      image

      K b = 1.7 × 10 9 = [ C 5 H 5 NH + ] [ OH ] [ C 5 H 5 N ] = ( 0.450 + x) (x) ( 0.510 x) 0.450 x 0 .510

      x = 1.927 × 10 –9 = 1.9 × 10 –9 M = [OH ], pOH = 8.715, pH = 14.00 – 8.715 = 5.29

      Check .  In a buffer, if [conj. acid] > [conj. base], pH < pK a of the conj. acid.

      If [conj. acid] < [conj. base], pH > pK a of the conj. acid. In this buffer, pK a of C 5 H 5 NH + is 5.23. [C 5 H 5 NH + ] < [C 5 H 5 N] and pH = 5.29, greater than 5.23.

    3. mol = M × L; mol HF = 0.050 M × 0.055 L = 2.75 × 10 –3 = 2.8 × 10 –3 mol;

      mol F = 0.10 M × 0.125 L = 0.0125 = 0.013 mol

      image

      [HF] = (2.75 × 10 –3 + x)/0.180 L; [F ] = (0.0125 + x)/0.180 L

      Note that the volumes will cancel when substituted into the K a expression.


    1. K a = 6.8 × 10 4 = [ H + ] [ F ] [ HF ] = x(0 .0125 + x)/0 .180 ( 2.75 × 10 3 x)/0 .180 x(0 .0125) 0.00275

      x = 1.50 × 10 –4 = 1.5 × 10 –4 M H + ; pH = 3.83

      Check .  K a for HF = 3.17. [HF] < [F ], pH of buffer = 3.83, greater than 3.17.

    17.17 Analyze/Plan .  We are asked to calculate % ionization of (a) a weak acid and (b) a weak acid in a solution containing a common ion, its conjugate base. Calculate % ionization as in Sample Exercise 16.13. In part (b), the concentration of the common ion is 0.085 M , not x, as in part (a). Solve .

    buCOOH(aq) ⇌ H + (aq) + buCOO (aq) K a = [ H + ] [ buCOO ] [ buCOOH ] = 1.5 × 10 5

    equil (a) 0.0075 – x M x M x M

    equil (b) 0.0075 – x M x M 0.085 + x M

    1. K a = 1.5 × 10 5 = x 2 0.0075 x x 2 0.0075 ; x = [ H + ] = 3.354 × 10 4 = 3.4 × 10 4 M H +

      % ionization = 3.4 × 10 4 M H + 0.0075 M buCOOH × 100 = 4.5 % ionization

    2. K a = 1.5 × 10 5 = (x) (0 .085 + x) 0.0075 x 0 .085 x 0.0075 ; x = 1.3 × 10 6 M H +

      % ionization = 1.3 × 10 6 M H + 0.0075 M buCOOH × 100 = 0.018 % ionization

    Check .  Percent ionization is much smaller when the “common ion” is present.

    17.18       CH 3 CH(OH)COOH ⇌ H + (aq) + CH 3 CH(OH)COO

    equil (a) 0.125 – x M x M x M

    equil (b) 0.125 – x M x M 0.0075 + x M

    K a = [ H + ] [ CH 3 CH ( OH ) COO ] [ CH 3 CH ( OH ) COOH ] = 1.4 × 10 4

    1. K a = 1.4 × 10 4 = x 2 0.125 x x 2 0.125 ; x = [ H + ] = 4.18 × 10 3 M = 4.2 × 10 3 M H +

      % ionization = 4.2 × 10 3 M H + 0.125 M CH 3 CH(OH)COOH × 100 = 3.4 % ionization

    2. K a = 1.4 × 10 4 = (x) (0 .0075 + x) 0.125 x 0 .0075 x 0.125 ; x = 2.3 × 10 3 M H +

      % ionization = 2.3 × 10 3 M H + 0.125 M CH 3 CH ( OH ) COOH × 100 = 1.9 % ionization


    Buffered Solutions (Section 17.2)

    17.19   Only solution (a) is a buffer. CH 3 COOH and CH 3 COONa are a weak conjugate acid/conjugate base pair that acts as a buffer; CH 3 COOH reacts with added base and CH 3 COO reacts with added acid, leaving [H + ] relatively unchanged. Solution (b) contains only a weak acid, which has no capacity to react with added acid. Although solution (c) contains a conjugate acid/conjugate base pair, Cl is a negligible base. In general, the conjugate bases of strong acids are negligible and mixtures of strong acids and their conjugate salts do not act as buffers.

    17.20   Only solution (a) is a buffer. NaOH is a strong base and will react with CH 3 COOH to form CH 3 COONa. As long as CH 3 COOH is present in excess, the resulting solution will contain both the conjugate acid CH 3 COOH(aq) and the conjugate base CH 3 COO (aq), the requirements for a buffer.

    Solution (b) contains a large excess and is essentially a strong base. Solution (c) is essentially a strong acid; HCl determines [H + ] and pH of the solution. Solution (d) contains two salts; one is the conjugate base of a weak acid. It can react with added acid, but no added base.

    17.21 Analyze/Plan .  Follow the logic in Sample Exercise 17.3. Assume that % ionization is small in these buffers (Solutions 17.17 and 17.18). Solve .

    1. K a = [ H + ] [ CH 3 CH ( OH ) COO ] [ CH 3 CH(OH)COOH ] ; [ H + ] = [ K a ] [ CH 3 CH(OH)COOH] [ CH 3 CH(OH)COO ]

      [H + ] = 1.4 × 10 4 ( 0.12 ) ( 0.11 ) ; [H + ] = 1.53 × 10 –4 = 1.5 × 10 –4 M ; pH = 3.82

    2. mol = M × L; total volume = 85 mL + 95 mL = 180 mL

      [ H + ] = K a [CH 3 CH(OH)COOH] [CH 3 CH(OH)COO ] = 1.4 × 10 4 ( 0.13 M × 0.085 L)/0 .180 L ( 0.15 M × 0.095 L)/0 .180 L

      [H + ] = 1.4 × 10 4 ( 0.13 × 0.085 ) ( 0.15 × 0.095 ) ; [H + ] = 1.086 × 10 –4 = 1.1 × 10 –4 M ; pH = 3.96

    17.22   Assume that % ionization is small in these buffers (Solutions 17.17 and 17.18).

    1. The conjugate acid in this buffer is HCO 3 , so use K a2 for H 2 CO 3 , 5.6 × 10 –11

      K a = [ H + ] [ CO 3 2 ] [ HCO 3 ] ; [ H + ] = K a [ HCO 3 ] [ CO 3 2 ] = 5.6 × 10 11 ( 0.105 ) ( 0.125 )

      [H + ] = 4.70 × 10 –11 = 4.7 × 10 –11 M ; pH = 10.33

    2. mol = M × L; total volume = 140 mL = 0.140 L

      [ H + ] = K a ( 0.20 M × 0 .065 L)/0 .140 L ( 0.15 M × 0.075 L)/0 .140 L = 5.6 × 10 11 ( 0.20 × 0.065 ) ( 0.15 × 0.075 )

      [H + ] = 6.47 × 10 –11 = 6.5 × 10 –11 M ; pH = 10.19


    17.23

    1. Analyze/Plan .  Follow the logic in Sample Exercises 17.1 and 17.3. As in Sample Exercise 17.1, start by calculating concentrations of the components. Solve .

      CH 3 COOH(aq) ⇌ H + (aq) + CH 3 COO (aq); K a = 1.8 × 10 5 = [ H + ] [ CH 3 COO ] [ CH 3 COOH ]

      [CH 3 COOH] = 0.150 M

      [ CH 3 COO ] = 20.0 g CH 3 COONa 0.500 L soln × 1 mol CH 3 COONa 82.04 g CH 3 COONa = 0.488 M

      [ H + ] = K a [ CH 3 COOH ] [ CH 3 COO ] = 1.8 × 10 5 ( 0.150 x) ( 0.488 + x) 1.8 × 10 5 (0 .150) ( 0.488 )

      [H + ] = 5.533 × 10 –6 = 5.5 × 10 –6 M , pH = 5.26

    2. Plan .  On the left side of the equation, write all ions present in solution after HCl or NaOH is added to the buffer. Using acid–base properties and relative strengths, decide which ions will combine to form new products. Solve .

      Na + (aq) + CH 3 COO (aq) + H + (aq) + Cl (aq) → CH 3 COOH(aq) + Na + (aq) + Cl (aq)

    3. CH 3 COOH(aq) + Na + (aq) + OH (aq) → CH 3 COO (aq) + H 2 O(l) + Na + (aq)

    17.24   NH 4 + /NH 3 is a basic buffer. Either the hydrolysis of NH 3 or the dissociation of NH 4 + can be used to determine the pH of the buffer. Using the dissociation of NH 4 + leads directly to [H + ] and facilitates use of the Henderson–Hasselbach relationship.

    1. NH 4 + (aq)  ⇌   H + (aq)  +  NH 3 (aq)

      K a = K w K b = 1.0 × 10 14 1.8 × 10 5 = 5.56 × 10 10 = 5.6 × 10 10

      [ NH 3 ] = 1.00 M NH 3

      [ NH 4 + ] = 10.0 g NH 4 Cl 0.250 L × 1 mol NH 4 Cl 53.50 g NH 4 Cl = 0.74766 = 0.748 M NH 4 +

      K a = [ H + ] [ NH 3 ] [ NH 4 + ] ; [ H + ] = K a [ NH 4 + ] [ NH 3 ] = 5.56 × 10 10 ( 0.74766 x) ( 1.00 + x) 5.56 × 10 10 ( 0.74766 ) ( 1.00 )

      [H + ] = 4.1537 × 10 –10 = 4.15 × 10 –10 M , pH = 9.382

    2. NH 3 (aq) + H + (aq) + NO 3 (aq) → NH 4 + (aq) + NO 3 (aq)
    3. NH 4 + (aq) + Cl (aq) + K + (aq) + OH (aq) → NH 3 (aq) + H 2 O(l) + Cl (aq) + K + (aq)

    17.25 Analyze/Plan .  Follow the logic in Sample Exercises 16.12 and 17.5. Solve .

    1. K a = 6.8 × 10 4 = x 2 1.00 x = x 2 1.00 ; x = [ H + ] = 0.02608 = 0.026 M ; pH = 1.58

      There is 2.6% ionization, so the approximation is valid.

    2. In this problem, [F ] is the unknown.

      pH = 3.00, [H + ] = 10 –3.00 = 1.0 × 10 –3 ; [HF] = 1.00 – 0.0010 = 0.999 M

      K a = 6.8 × 10 4 = 1.0 × 10 3 [ F ] 0.999 ; [F ] = 0.6793 = 0.68 M

      0 .6793 mol NaF 1 L × 41.990 g NaF 1 mol NaF × 1.25 L = 35.654 = 36 g NaF


    17.26

    1. C 6 H 5 COOH(aq)  ⇌  H + (aq)  +  C 6 H 5 COO (aq)

      K a = 6.3 × 10 5 = [ H + ] [ C 6 H 5 COO ] [ C 6 H 5 COOH ] ; [C 6 H 5 COOH] = 0.0200 M ;

      [H + ] = [C 6 H 5 COO ] = x

      K a = 6.3 × 10 5 x 2 0.02000 ; x = [ H + ] = 1.123 × 10 3 = 1.1 × 10 3 M ; pH = 2.95

      Note that C 6 H 5 COOH is 5.6% ionized. Solving the quadratic for [H + ] yields (1.0914 × 10 –3 =) 1.1 × 10 –3 M H + , pH = 2.96; this is not a significant difference.

    2. [ H + ] = K a [ C 6 H 5 COOH ] [ C 6 H 5 COO ] ; [ H + ] = 10 4.00 = 1.0 × 10 4 M

      [C 6 H 5 COOH] = 0.0200 M ; calculate [C 6 H 5 COO ]. Because the common ion C 6 H 5 COO reduces % ionization, we assume the 5% approximation is valid.

      [ C 6 H 5 COO ] = K a [ C 6 H 5 COOH ] [ H + ] = 6. 3 × 10 5 ( 0.0200 ) 1.0 × 10 4 = 0.01260 = 0.013 M

      0.0126 mol C 6 H 5 COONa L × 1.50 L × 144.11 g C 6 H 5 COONa 1 mol C 6 H 5 COONa = 2.724 = 2.7 g C 6 H 5 COONa

    17.27 Analyze/Plan .  Follow the logic in Sample Exercises 17.3 and 17.6. Solve .

    1. K a = [ H + ] [ CH 3 COO ] [ CH 3 COOH ] ; [ H + ] = K a [ CH 3 COOH ] [ CH 3 COO ]

      [ H + ] 1 . 8 × 1 0 5 ( 0 . 1 0 ) ( 0 . 1 3 ) = 1 . 3 8 5 × 1 0 5 = 1 . 4 × 1 0 5 M ; p H = 4 . 8 6

    2. CH 3 COOH(aq) + KOH(aq) CH 3 COO (aq) + H 2 O(l) + K + (aq)
      0.10 mol 0.02 mol 0.13 mol
      –0.02 mol –0.02 mol +0.02 mol
      0.08 mol 0    mol 0.15 mol

      [ H + ] = 1.8 × 10 5 ( 0.08 mol/1 .00 L) ( 0.15 mol/1 .00 L) = 9.60 × 10 6 = 1 × 10 5 M ; pH = 5.02 = 5.0

    3. CH 3 COO (aq) + HNO 3 (aq) CH 3 COOH(aq) + NO 3 (aq)
      0.13 mol 0.02 mol 0.10 mol
      –0.02 mol –0.02 mol +0.02 mol
      0.11 mol 0    mol 0.12 mol

      [ H + ] = 1.8 × 10 5 ( 0.12 mol/1 .00 L) ( 0.11 mol/1 .00 L) = 1.96 × 10 5 = 2.0 × 10 5 M ; pH = 4.71


    17.28

    1. K a = [ H + ] [ C 2 H 5 COO ] [ C 2 H 5 COOH ] ; [ H + ] = K a [ C 2 H 5 COOH ] [ C 2 H 5 COO ]

    Because this expression contains a ratio of concentrations, we can ignore total volume and work directly with moles.

    [ H + ] = 1.3 × 10 5 ( 0.15 x) ( 0.10 + x) 1.3 × 10 5 ( 0.15 ) 0.10 = 1.950 × 10 5 = 2.0 × 10 5 M , pH = 4.71

    1. C 2 H 5 COOH(aq) + OH (aq) C 2 H 5 COO (aq) + H 2 O(l)
      0.15 mol 0.01 mol 0.10 mol
      –0.01 mol –0.01 mol +0.01 mol
      0.14 mol 0  mol 0.11 mol

      [ H + ] 1.3 × 10 5 ( 0.14 ) ( 0.11 ) = 1.6545 × 10 5 = 1.7 × 10 5 M ; pH = 4.78

    1. C 2 H 5 COO (aq) + HI(aq) C 2 H 5 COOH(aq) + I (aq)
      0.10 mol 0.01 mol 0.15 mol
      –0.01 mol –0.01 mol +0.01 mol
      0.09 mol 0  mol 0.16 mol

      [ H + ] 1 . 3 × 1 0 5 ( 0 . 1 6 ) ( 0 . 0 9 ) = 2 .3111 × 1 0 5 = 2 × 1 0 5 M ; p H = 4 . 6


    17.31 Analyze .  Given six solutions, decide which two should be used to prepare a pH 3.50 buffer. Calculate the volumes of the two 0.10 M solutions needed to make approximately 1 L of buffer.

    Plan .  A buffer must contain a conjugate acid/conjugate base (CA/CB) pair. By examining the chemical formulas, decide which pairs of solutions could be used to make a buffer. If there is more than one possible pair, calculate pK a for the acids. A buffer is most effective when its pH is within 1 pH unit of pK a for the conjugate acid component. Select the pair with pK a nearest to 3.50. Use Equation 17.9 to calculate the [CB]/[CA] ratio and the volumes of 0.10 M solutions needed to prepare 1 L of buffer. Solve .

    There are three CA/CB pairs:

    HCOOH/HCOONa, pK a = 3.74

    CH 3 COOH/CH 3 COONa, pK a = 4.74

    H 3 PO 4 /NaH 2 PO 4 , pK a = 2.12

    The most appropriate solutions are HCOOH/HCOONa, because pK a for HCOOH is nearest to 3.50.

    pH = pK a + log [CB] [CA] ; 3.50 = 3.7447 + log [HCOONa] [ HCOOH ]

    log [HCOONa] [ HCOOH ] = 0.2447 ; [ HCOONa ] [ HCOOH ] = 0.5692 = 0.57

    Because we are making a total of 1 L of buffer,

    let y = vol HCOONa and (1 – y) = vol HCOOH.

    0.5692 = [HCOONa] [HCOOH] = ( 0.10 M × y)/1L [ 0.10 M × ( 1 y) ] / 1 L ; 0.5692 [ 0.10 ( 1 y)] = 0.10 y;

    0.05692 = 0.15692 y; y = 0.3627 = 0.36 L

    360 mL of 0.10 M HCOONa, 640 mL of 0.10 M HCOOH

    Check .  The pH of the buffer is less than pK a for the conjugate acid, indicating that the amount of CA in the buffer is greater than the amount of CB. This agrees with our result.

    17.32   The solutes listed contain three possible conjugate acid/conjugate base (CA/CB) pairs.

    These are:

    HCOOH/HCOONa, pK a = 3.74

    CH 3 COOH/CH 3 COONa, pK a = 4.74

    HCN/NaCN, pK a = 9.31

    For maximum buffer capacity, pK a should be within 1 pH unit of the buffer. The acetic acid/acetate pair is most appropriate for a buffer with pH 5.00.

    pH = pK a + log [CB] [CA] ; 5.00 = 4.745 + log [CH 3 COONa] [ CH 3 COOH ]

    log [CH 3 COONa] [ CH 3 COOH ] = 0.2553 ; [ CH 3 COONa ] [ CH 3 COOH ] = 1.800 = 1.8


    Because we are making a total of 1 L of buffer,

    let y = vol CH 3 COONa and (1 – y) = vol CH 3 COOH.

    1.800 = [ CH 3 COONa ] [ CH 3 COOH ] = ( 0.10 M × y)/1 .0 L [ 0.10 M × ( 1 y)]/1 .0 L = 0.10 y 0 .10 0.10 y

    1.800(0.10 – 0.10 y) = 0.10 y; 0.1800 = 0.2800 y; y = 0.6429 = 0.64 L

    640 mL of 0.10 M CH 3 COONa, 360 mL of CH 3 COOH

    Check .  pH (buffer) > pK a (CA) and the calculated amount of CB in the buffer is greater than the amount of CA.

    Acid–Base Titrations (Section 17.3)

    17.33

    1. Curve B. The initial pH is lower and the equivalence point region is steeper.
    2. pH at the approximate equivalence point of curve A = 8.0

      pH at the approximate equivalence point of curve B = 7.0

    3. Volume of base required to reach the equivalence point depends only on moles of acid present; it is independent of acid strength. Because acid B requires 40 mL and acid A requires only 30 mL, more moles of acid B are being titrated. For equal volumes of A and B, the concentration of acid B is greater.
    4. pK a of the weak acid is approximately 4.5. In the titration of a weak acid, pH equals pK a of the weak acid at the volume half-way to the equivalence point. On curve A, the equivalence point is at 30 mL, half-way is 15 mL, and the pH there is 4.5.

    17.34

    1. False. The quantity of base required to reach the equivalence point is the same in the two titrations, assuming both acids have the same initial concentrations.
    2. False. The pH is higher initially in the titration of a weak acid.
    3. False. The pH is higher at the equivalence point in the titration of a weak acid.

    17.35

    1. False. The same volume of NaOH(aq) is required to reach the equivalence point of both titrations, because moles of acid to be titrated are the same in both flasks.
    2. True. CH 3 COONa, the salt formed in the titration of CH 3 COOH, produces a basic solution, whereas NaNO 3 , formed in the titration of HNO 3 , produces a neutral solution.
    3. True. Even though the pH values at the equivalence points of the two titrations are different, phenolphthalein changes color over a wide range of pH values and is appropriate for both titrations.

    17.36

    1. False. The pH at the beginning of the titration of the weaker acid, CH 3 COOH, will be higher.
    2. True. Past the equivalence point, the titration curves are very similar (but not identical.
    3. False. According to Figures 17.12 and 17.13, methyl red is suitable for the titration of the strong acid HNO 3 , but not for the titration of the weak acid CH 3 COOH.

    17.37 Analyze .  Given reactants, predict whether pH at the equivalence point of a titration is less than, equal to, or greater than 7.

    Plan .  At the equivalence point of a titration, only product is present in solution; there is no excess of either reactant. Determine the product of each reaction and whether a solution of it is acidic, basic, or neutral. Solve .

    1. NaHCO 3 (aq) + NaOH(aq) → Na 2 CO 3 (aq) + H 2 O(l)

      At the equivalence point, the major species in solution are Na + and CO 3 2– . Na + is negligible and CO 3 2– is the CB of HCO 3 . The solution is basic, above pH 7.

    2. NH 3 (aq) + HCl(aq) → NH 4 Cl(aq)

      At the equivalence point, the major species are NH 4 + and Cl . Cl is negligible and NH 4 + is the CA of NH 3 . The solution is acidic, below pH 7.

    3. KOH(aq) + HBr(aq) → KBr(aq) + H 2 O(l)

      At the equivalence point, the major species are K + and Br–; both are negligible. The solution is at pH 7.

    17.38

    1. HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H 2 O(l)

      At the equivalence point, the major species are Na + and HCOO . Na + is negligible and HCOO is the CB of HCOOH. The solution is basic, above pH 7.

    2. Ca(OH) 2 (aq) + 2 HClO 4 (aq) → Ca(ClO 4 ) 2 (aq) + 2 H 2 O(l)

      At the equivalence point, the major species are Ca 2+ and ClO 4 ; both are negligible. The solution is at pH 7.

    3. C 5 H 5 N(aq) + HNO 3 (aq) → C 5 H 5 NH + NO 3 (aq)

      At the equivalence point, the major species are C 5 H 5 NH + and NO 3 . NO 3 is negligible and C 5 H 5 NH + is the CA of C 5 H 5 N. The solution is acidic, below pH 7.

    17.39   The second color change, from yellow to blue near pH = 8.5, is more suitable for the titration of a weak acid with a strong base. The salt present at the equivalence point of this type of titration produces a slightly basic solution. The second color change of thymol blue is in the correct pH range to show (indicate) the equivalence point.

    17.40

    1. At the equivalence point, moles HA added = moles B initially present = 0.10 M × 0.0300 L = 0.0030 moles HA added.
    2. BH + (aq).
    3. Less than 7. The predominant form at equivalence, BH + , is a weak acid.
    4. Because the pH at the equivalence point will be less than 7, methyl red would be more appropriate.

    17.41 Analyze/Plan .  We are asked to calculate the volume of 0.0850 M NaOH required to titrate various acid solutions to their equivalence point. At the equivalence point, moles base added equals moles acid initially present. Solve the stoichiometry problem, recalling that mol = M × L. In part (c), calculate molarity of HCl from g/L and proceed as outlined earlier. Solve .


    1. 40.0 mL HNO 3 × 0.0900 mol HNO 3 1000 mL soln × 1 mol NaOH 1 mol HNO 3 × 1000 mL soln 0.0850 mol NaOH = 42.353 = 42.4 mL NaOH soln
    2. 35.0 mL CH 3 COOH × 0.0850 M CH 3 COOH 1000 mL soln × 1 mol NaOH 1 mol CH 3 COOH × 1000 mL soln 0.0850 mol NaOH = 35.0 mL NaOH soln
    3. 1.85 g HCl 1 L soln × 1 mol HCl 36.46 g HCl = 0.05074 = 0.0507 M HCl

      50.0 mL HCl × 0 .05074 mol HCl 1000 mL × 1 mol NaOH 1 mol HCl × 1000 mL soln 0.0850 mol NaOH = 29.847 = 29.8 mL NaOH soln

    17.42

    1. 45.0 mL NaOH × 0.0950 mol NaOH 1000 mL soln × 1 mol HCl 1 mol NaOH × 1000 mL soln 0.105 mol HCl = 40.7 mL HCl soln
    2. 22.5 mL NH 3 × 0.118 mol NH 3 1000 mL soln × 1 mol HCl 1 mol NH 3 × 1000 mL soln 0.105 mol HCl = 25.3 mL HCl soln
    3. 125.0 mL × 1.35 g NaOH 1000 mL × 1 mol NaOH 40 .00 g NaOH × 1 mol HCl 1 mol NaOH × 1000 mL soln 0.105 mol HCl = 40.2 mL HCl soln

    17.43 Analyze/Plan .  Follow the logic in Sample Exercise 17.7 for the titration of a strong acid with a strong base. Solve .

    moles H + = M HBr × L HBr = 0.200 M × 0.0200 L = 4.00 × 10 –3 mol

    moles OH = M NaOH × L NaOH = 0.200 M × L NaOH

    mL HBr mL NaOH Total Volume Moles H + Moles OH Molarity Excess Ion pH
    (a) 20.0 15.0 35.0 4.00 × 10 –3 3.00 × 10 –3 0.0286(H + ) 1.544
    (b) 20.0 19.9 39.9 4.00 × 10 –3 3.98 × 10 –3 5 × 10 –4 (H + ) 3.3
    (c) 20.0 20.0 40.0 4.00 × 10 –3 4.00 × 10 –3 1 × 10 –7 (H + ) 7.0
    (d) 20.0 20.1 40.1 4.00 × 10 –3 4.02 × 10 –3 5 × 10 –4 (OH ) 10.7
    (e) 20.0 35.0 55.0 4.00 × 10 –3 7.00 × 10 –3 0.0545(OH ) 12.737

    molarity of excess ion = moles ion/total vol in L

    1. 4.00 × 10 3 mol H + 3.00 × 10 3 mol OH 0.0350 L = 0.0286 M H +
    2. 4.00 × 10 3 mol H + 3.98 × 10 3 mol OH 0.0339 L = 5.01 × 10 4 = 5 × 10 4 M H +

    1. equivalence point, mol H + = mol OH

      NaBr does not hydrolyze, so [H + ] = [OH ] = 1 × 10 –7 M

    1. 4.02 × 10 3 mol OH 4.00 × 10 3 mol H + 0.0401 L = 4.99 × 10 4 = 5 × 10 4 M OH
    1. 7.00 × 10 3 mol OH 4.00 × 10 3 mol H + 0.0550 L = 0.054545 = 0.0545 M OH

    17.44   moles OH = M KOH × L KOH = 0.150 M × 0.0200 L = 3.00 × 10 –3 mol

    moles H + = M HClO 4 × L HClO 4 = 0.125 M × L HClO 4

    mL KOH mL HClO4 Total Volume Moles OH Moles H + Molarity Excess Ion pH
    (a) 20.0 20.0 40.0 3.00 × 10 –3 2.50 × 10 –3 0.013(OH ) 12.10
    (b) 20.0 23.0 43.0 3.00 × 10 –3 2.88 × 10 –3 2.9 × 10 –3 (OH ) 11.46
    (c) 20.0 24.0 44.0 3.00 × 10 –3 3.00 × 10 –3 1.0 × 10 –7 (OH ) 7.00
    (d) 20.0 25.0 45.0 3.00 × 10 –3 3.13 × 10 –3 2.8 × 10 –3 (H + ) 2.56
    (e) 20.0 30.0 50.0 3.00 × 10 –3 3.75 × 10 –3 0.015(H + ) 1.82

    molarity of excess ion = moles ion total vol in L

    1. 3. 00 × 10 3 mol OH 2.50 × 10 3 mol H + 0.0400 L = 0.0125 = 0.013 M OH
    2. 3.00 × 10 3 mol OH 2.875 × 10 3 mol H + 0.0430 L = 2.91 × 10 3 = 2.9 × 10 3 M OH
    3. equivalence point, mol H + = mol OH

      KClO 4 does not hydrolyze, so [H + ] = [OH ] = 1 × 10 –7 M

    4. 3.125 × 10 3 mol H + 3.00 × 10 3 mol OH 0.0450 L = 2.78 × 10 3 = 2.8 × 10 3 M H +
    5. 3.75 × 10 3 mol H + 3.00 × 10 3 mol OH 0.0500 L = 0.0150 = 0.015 M H +

    17.45 Analyze/Plan .  Follow the logic in Sample Exercise 17.8 for the titration of a weak acid with a strong base. Solve .

    1. At 0 mL, only weak acid, CH 3 COOH, is present in solution. Using the acid ionization equilibrium
      CH 3 COOH(aq) H + (aq) + CH 3 COO (aq)
      initial 0.150 M 0 0
      equil. 0.150 – x M x M x M

    1. K a = [ H + ] [ CH 3 COO ] [ CH 3 COOH ] = 1.8 × 10 5 (Appendix D)

      1.8 × 10 5 = x 2 ( 0.150 x) x 2 0.150 ; x 2 = 2.7 × 10 6 ; x = [ H + ] = 0.001643 = 1.6 × 10 3 M ; pH = 2.78

    (b–f) Calculate the moles of each component after the acid–base reaction takes place. Moles CH 3 COOH originally present = M × L = 0.150 M × 0.0350 L = 5.25 × 10 –3 mol. Moles NaOH added = M × L = 0.150 M × y mL.

    NaOH(aq) + CH 3 COOH (aq) → CH 3 COONa(aq) + H 2 O(l)
    (0.150 M × 0.0175 L) =
    (b) before rx 2.625 × 10 –3 mol 5.25 × 10 –3 mol
    after rx 0 2.625 × 10 –3 mol 2.63 × 10 –3 mol
    (0.150 M × 0.0345 L) =
    (c) before rx 5.175 × 10 –3 mol 5.25 × 10 –3 mol
    after rx 0 0.075 × 10 –3 mol 5.18 × 10 –3 mol
    (0.150 M × 0.0350 L) =
    (d) before rx 5.25 × 10 –3 mol 5.25 × 10 –3 mol
    after rx 0 0 5.25 × 10 –3 mol
    (0.150 M × 0.0355 L) =
    (e) before rx 5.325 × 10 –3 mol 5.25 × 10 –3 mol
    after rx 0.075 × 10 –3 mol 0 5.25 × 10 –3 mol
    (0.150 M × 0.0500 L) =
    (f) before rx 7.50 × 10 –3 mol 5.25 × 10 –3 mol
    after rx 2.25 × 10 –3 mol 0 5.25 × 10 –3 mol

    Calculate the molarity of each species ( M = mol/L) and solve the appropriate equilibrium problem in each part.

    1. V T = 35.0 mL CH 3 COOH + 17.5 mL NaOH = 52.5 mL = 0.0525 L

      [ CH 3 COOH ] = 2.625 × 10 3 mol 0.0525 = 0.0500 M

      [ CH 3 COO ] = 2.625 × 10 3 mol 0.0525 = 0.0500 M

      CH 3 COOH(aq)  ⇌  H + (aq)  +  CH 3 COO (aq)

      equil. 0.0500 – x M x M 0.0500 + x M

      K a = [ H + ] [ CH 3 COO ] [ CH 3 COOH ] ; [ H + ] = K a [ CH 3 COOH ] [CH 3 COO ]

      [ H + ] = 1.8 × 10 5 ( 0.0500 x) ( 0.0500 + x) = 1.8 × 10 5 M H + ; pH = 4.74


    1. [ CH 3 COOH ] = 7.5 × 10 5 mol 0.0695 L = 0.001079 = 1.1 × 10 3 M

      [ CH 3 COO ] = 5.175 × 10 3 mol 0.0695 L = 0.07446 = 0.074 M

      [ H + ] = 1.8 × 10 5 ( 1.079 × 10 3 x) ( 0.07446 + x) 2.6 × 10 7 M H + ; pH = 6.58

    1. At the equivalence point, only CH 3 COO is present.

      [ CH 3 COO ] = 5.25 × 10 3 mol 0.0700 L = 0.0750 M

      The pertinent equilibrium is the base hydrolysis of CH 3 COO .

      CH 3 COO (aq)  +  H 2 O(l) CH 3 COOH(aq) + OH (aq)
      initial 0.0750 M 0 0
      equil. 0.0750 – x M x x

      K b = K w K a for CH 3 COOH = 1.0 × 10 14 1.8 × 10 5 = 5.56 × 10 10 = 5.6 × 10 10 = [ CH 3 COOH ] [ OH ] [ CH 3 COO ]

      5.56 × 10 10 = x 2 0.0750 x ; x 2 5.56 × 10 10 (0 .0750); x = 6.458 × 10 6 = 6.5 × 10 6 M OH

      pOH = –log(6.458 × 10 –6 ) = 5.19; pH = 14.00 – pOH = 8.81

    1. After the equivalence point, the excess strong base determines the pOH and pH. The [OH ] from the hydrolysis of CH 3 COO is small and can be ignored.

      [ OH ] = 0.075 × 10 3 mol 0.0705 L = 1.064 × 10 3 = 1.1 × 10 3 M ; pOH = 2.97 pH = 14.00 2.97 = 11.03

    1. [ OH ] = 2.25 × 10 3 mol 0.0850 L = 0.0265 M OH ; pOH = 1.577 ; pH = 14.00 1.577 = 12.423

    17.46

    1. Weak base problem: K b = 1.8 × 10 5 = [ NH 4 + ] [ OH ] [ NH 3 ]

      At equilibrium, [OH ] = x, [NH 3 ] = (0.030 – x); [NH 4 + ] = x

      1.8 × 10 5 = x 2 ( 0.050 x) x 2 0.050 ; x = [ OH ] = 9.487 × 10 4 = 9.5 × 10 4 M

      pH = 14.00 – 3.02 = 10.98


    (b–f) Calculate mol NH 3 and mol NH 4 + after the acid–base reaction takes place.

    0.050 M NH 3 × 0.0300 L = 1.5 × 10 –3 mol NH 3 present initially.

    NH 3 (aq)  + HCl(aq)  → NH 4 + (aq) + Cl (aq)
    (0.025 M × 0.0200 L) =
    (b) before rx 1.5 × 10 –3 mol 0.50 × 10 –3 mol 0 mol
    after rx 1.0 × 10 –3 mol 0 mol 5.0 × 10 –4 mol
    (0.025 M × 0.0590 L) =
    (c) before rx 1.5 × 10 –3 mol 1.475 × 10 –3 mol 0 mol
    after rx 2.5 × 10 –5 mol 0 mol 1.475 × 10 –3 mol
    (0.025 M × 0.0600 L) =
    (d) before rx 1.5 × 10 –3 mol 1.5 × 10 –3 mol 0 mol
    after rx 0 mol 0 mol 1.5 × 10 –3 mol
    (0.025 M × 0.0610 L) =
    (e) before rx 1.5 × 10 –3 mol 1.525 × 10 –3 mol 0 mol
    after rx 0 mol 2.5 × 10 –5 mol 1.5 × 10 –3 mol
    (0.025 M × 0.0650 L) =
    (f) before rx 1.5 × 10 –3 mol 1.625 × 10 –3 mol 0 mol
    after rx 0 mol 1.25 × 10 –4 mol 1.5 × 10 –3 mol
    1. Using the acid dissociation equilibrium for NH 4 + (so that we calculate [H + ] directly), NH 4 + (aq)  ⇌  H + (aq)   +  NH 3 (aq)

      K a = [ H + ] [NH 3 ] [ NH 4 + ] = K w K b for NH 3 = 1.0 × 10 14 1.8 × 10 5 = 5.56 × 10 10 = 5.6 × 10 10

      [ NH 3 ] = 1.0 × 10 3 mol 0.0500 L = 0.020 M ; [ NH 4 + ] = 5.0 × 10 4 mol 0.0500 L = 0.010 M

      [ H + ] = 5.56 × 10 10 [ NH 4 + ] [ NH 3 ] 5.56 × 10 10 ( 0.010 ) ( 0.020 ) = 2.78 × 10 10 ; pH = 9.56

      (We will assume [H + ] is small compared to [NH 3 ] and [NH 4 + ].)

    1. [ NH 3 ] = 2.5 × 10 5 mol 0.0890 L = 2.8 × 10 4 M ; [ NH 4 + ] = 1.475 × 10 3 mol 0.0890 L = 0.017 M

      [ H + ] = 5.56 × 10 10 (0 .017) ( 2.8 × 10 4 ) = 3.38 × 10 8 = 3.4 × 10 8 M ; pH = 7.47


    1. At the equivalence point, [H + ] = [NH 3 ] = x

      [ NH 4 + ] = 1.5 × 10 3 M 0.0900 L = 0.01667 = 0.017 M

      5.56 × 10 10 = x 2 0.01667 ; x = [ H + ] = 3.043 × 10 6 = 3.0 × 10 6 M ; pH = 5.52

    1. Past the equivalence point, [H + ] from the excess HCl determines the pH.

      [ H + ] = 2.5 × 10 5 mol 0.0910 L = 2.747 × 10 4 = 2.7 × 10 4 M ; pH = 3.56

    1. Past the equivalence point, [H + ] from the excess HCl determines the pH.

      [ H + ] = 1.25 × 10 4 mol 0.0950 L = 1.316 × 10 3 = 1.3 × 10 3 M ; pH = 2.88

    17.47 Analyze/Plan .  Calculate the pH at the equivalence point for the titration of several bases with 0.200 M HBr. The volume of 0.200 M HBr required in all cases equals the volume of base and the final volume = 2V base . The concentration of the salt produced at the

    equivalence point is 0.200 M × V base 2 V base = 0.100 M .

    In each case, identify the salt present at the equivalence point, determine its acid–base properties (Section 16.9), and solve the pH problem. Solve .

    1. NaOH is a strong base; the salt present at the equivalence point, NaBr, does not affect the pH of the solution. 0.100 M NaBr, pH = 7.00.
    2. HONH 2 is a weak base, so the salt present at the equivalence point is HONH 3 + Br–. This is the salt of a strong acid and a weak base, so it produces an acidic solution.
      image

      K a = [ H + ] [ HONH 2 ] [ HONH 3 + ] = K w K b = 1.0 × 10 14 1.1 × 10 8 = 9.09 × 10 7 = 9.1 × 10 7

      Assume x is small with respect to [salt].

      K a = x 2 /0.100; x = [H + ] = 3.02 × 10 –4 = 3.0 × 10 –4 M , pH = 3.52

    3. C 6 H 5 NH 2 is a weak base and C 6 H 5 NH 3 + Br is an acidic salt.

      0.100 M C 6 H 5 NH 3 + Br . Proceeding as in (b):

      K a = [H + ] [C 6 H 5 NH 2 ] [ C 6 H 5 NH 3 + ] = K w K b = 2.33 × 10 5 = 2.3 × 10 5

      [H + ] 2 = 0.100(2.33 × 10 –5 ); [H + ] = 1.52 × 10 –3 = 1.5 × 10 –3 M , pH = 2.82


    17.48   The volume of NaOH solution required in all cases is

    V base = V acid × M acid M base = ( 0.100 ) V acid ( 0.080 ) = 1.25 V acid

    The total volume at the equivalence point is V base + V acid = 2.25 V acid

    The concentration of the salt at the equivalence point is M acid V acid 2.25 V acid = 0.100 2.25 = 0.0444 M

    1. 0.0444 M NaBr, pH = 7.00
    2. 0.0444 M NaClO 2 ;  ClO 2 (aq) + H 2 O(l)  ⇌  HClO 2 (aq) + OH (aq)

      K b = [ HClO 2 ] [ OH ] [ ClO 2 ] = K w K a = 1.0 × 10 14 1.1 × 10 2 = 9.09 × 10 13 = 9.1 × 10 13

      [HClO 2 ] = [OH ]; [ClO 2 ] ≈ 0.0444 M

      [OH ] 2 ≈ 0.0444(9.09 × 10 –13 ); [OH ] = 2.01 × 10 –7 = 2.0 × 10 –7 M , pOH = 6.70; pH = 7.30

      Note that HClO 2 is a relatively strong acid (large K a value), so the pH at the equivalence point is not much greater than 7.0. Because [OH ] from the hydrolysis of ClO 2 is very small, the autoionization equilibrium should be considered for a more accurate value of the equivalence point pH.

      Let [H + ] = x, [OH ] = (2.0 × 10 –7 M + x); 1.0 × 10 –14 = (x)(2.0 × 10 –7 M + x)

      Solving the quadratic equation gives a pH of 7.38.

    3. C 6 H 5 COO (aq) + H 2 O(l)  ⇌  C 6 H 5 COOH(aq) + OH (aq)

      K b = [C 6 H 5 COO ] [OH ] [ C 6 H 5 COOH ] = K w K a = 1.0 × 10 14 6.3 × 10 5 = 1.59 × 10 10 = 1.6 × 10 10

      [OH ] 2 ≈ 0.0444(1.59 × 10 –8 ); [OH ] = 2.655 × 10 –6 = 2.7 × 10 –6 M , pH = 8.42

    Solubility Equilibria and Factors Affecting Solubility (Sections 17.4 and 17.5)

    17.49

    1. True.
    2. False. The solubility product of a slightly soluble salt is the square of the solubility if the salt contains one cation and one anion.
    3. False. The common-ion effect is in play for solubility equilibria as well as acid–base equilibria.
    4. True. The common-ion effect does not change the equilibrium constant.

    17.50

    1. MZ 2 has the larger numerical value for the solubility product constant, 4s 2 versus s 2 .
    2. The [M 2+ ] is the same in the two saturated solutions, because the molar solubilities are the same and there is one mole of [M 2+ ] in one mole of either salt.
    3. [M 2+ ] = 4 × 10 –4 M. After the addition, mol M 2+ increase, but so does the total volume. The [M 2+ ] remains the same.

    17.51 Analyze/Plan .  Follow the example in Sample Exercise 17.10. Solve .

    K sp = [Ag + ][I ]; K sp = [Sr 2+ ][SO 4 2– ]; K sp = [Fe 2+ ][OH ] 2 ; K sp = [Hg 2 2+ ][Br ] 2

    17.52

    1. False. Solubility is the amount (grams, moles) of solute that will dissolve in a certain volume of solution. Solubility-product constant is an equilibrium constant, the product of the molar concentrations of all the dissolved ions in solution.
    2. K sp = [Mn 2+ ][CO 3 2– ]; K sp = [Hg 2+ ][OH ] 2 ; K sp = [Cu 2+ ] 3 [PO 4 3– ] 2

    17.53 Analyze/Plan .  Follow the logic in Sample Exercise 17.11. Solve .

    1. CaF 2 (s)  ⇌  Ca 2+ (aq)  +  2 F (aq);    K sp = [Ca 2+ ][F ] 2

      The molar solubility is the moles of CaF 2 that dissolve per liter of solution. Each mole of CaF 2 produces 1 mol Ca 2+ (aq) and 2 mol F (aq).

      [Ca 2+ ] = 1.24 × 10 –3 M ; [F ] = 2 × 1.24 × 10 –3 M = 2.48 × 10 –3 M

      K sp = (1.24 × 10 –3 ) (2.48 × 10 –3 ) 2 = 7.63 × 10 –9

    2. SrF 2 (s)  ⇌  Sr 2+ (aq)  +  2 F (aq);    K sp = [Sr 2+ ][F ] 2

      Transform the gram solubility to molar solubility.

      1.1 × 10 2 g SrF 2 0.100 L × 1 mol SrF 2 125.6 g SrF 2 = 8.76 × 10 4 = 8.8 × 10 4 mol SrF 2 / L

      [Sr 2+ ] = 8.76 × 10 –4 M ; [F ] = 2(8.76 × 10 –4 M )

      K sp = (8.76 × 10 –4 ) (2(8.76 × 10 –4 )) 2 = 2.7 × 10 –9

    3. Ba(IO 3 ) 2 (s)  ⇌  Ba 2+ (aq)  +  2 IO 3 (aq);    K sp = [Ba 2+ ][IO 3 ] 2

      Because 1 mole of dissolved Ba(IO 3 ) 2 produces 1 mole of Ba 2+ , the molar solubility of Ba(IO 3 ) 2 = [Ba 2+ ]. Let x = [Ba 2+ ]; [IO 3 ] = 2x.

      K sp = 6.0 × 10 –10 = (x)(2x) 2 ; 4x 3 = 6.0 × 10 –10 ; x 3 = 1.5 × 10 –10 ; x = 5.3 × 10 –4 M

      The molar solubility of Ba(IO 3 ) 2 is 5.3 × 10 –4 mol/L.

    17.54

    1. PbBr 2 (s)  ⇌  Pb 2+ (aq) + 2 Br (aq)

      K sp = [Pb 2+ ][Br ] 2 ; [Pb 2+ ] = 1.0 × 10 –2 M , [Br ] = 2.0 × 10 –2 M

      K sp = (1.0 × 10 –2 M )(2.0 × 10 –2 M ) 2 = 4.0 × 10 –6

    2. AgIO 3 (s)  ⇌  Ag + (aq) + IO 3 (aq);    K sp = [Ag + ][IO 3 ]

      [ Ag + ] = [ IO 3 ] = 0.0490 g AgIO 3 1.00 L soln × 1 mol AgIO 3 282.8 g AglO 3 = 1.733 × 10 4 = 1.73 × 10 4 M

      K sp = (1.733 × 10 –4 M ) (1.733 × 10 –4 M ) = 3.00 × 10 –8

    3. Ca(OH) 2 (s)  ⇌  Ca 2+ (aq) + 2 OH (aq);    K sp = [Ca 2+ ][OH ] 2

      [Ca 2+ ] = x, [OH ] = 2x; K sp = 6.5 × 10 –6 = (x)(2x) 2

      6.5 × 10 –6 = 4x 3 ; x = [Ca 2+ ] = 0.01176 = 0.012 M ; [OH ] = 0.02351 = 0.024 M

      pH = 14 – pOH = 14 – 1.629 = 12.37


    17.55 Analyze/Plan .  Given gram solubility of a compound, calculate K sp . Write the dissociation equilibrium and K sp expression. Change gram solubility to molarity of the individual ions, taking the stoichiometry of the compound into account. Calculate K sp . Solve .

    CaC 2 O 4 (s)  ⇌  Ca 2+ (aq) + C 2 O 4 2– (aq);   K sp = [Ca 2+ ][C 2 O 4 2– ]

    [ Ca 2 + ] = [ C 2 O 4 2 ] = 0.0061 g CaC 2 O 4 1.00 L s oln × 1 mol CaC 2 O 4 128.1 g CaC 2 O 4 = 4.76 × 10 5 = 4.8 × 10 5 M

    K sp = (4.76 × 10 –5 M )(4.76 × 10 –5 M ) = 2.3 × 10 –9

    17.56   PbI 2 (s)  ⇌  Pb 2+ (aq) + 2 I (aq);    K sp = [Pb 2+ ][I ] 2

    [ Pb 2 + ] = 0.54 g PbI 2 1.00 L soln × 1 mol PbI 2 461.0 g PbI 2 = 1.17 × 10 3 = 1.2 × 10 3 M

    [I ] = 2[Pb 2+ ]; K sp = [Pb 2+ ](2[Pb 2+ ]) 2 = 4[Pb 2+ ] 3 = 4(1.17 × 10 –3 ) 3 = 6.4 × 10 –9

    17.57 Analyze/Plan .  Follow the logic in Sample Exercises 17.12 and 17.13. Solve .

    1. AgBr(s)  ⇌  Ag + (aq) + Br (aq);     K sp = [Ag + ][Br ] = 5.0 × 10 –13

      molar solubility = x = [Ag + ] = [Br ]; K sp = x 2

      x = (5.0 × 10 –13 ) 1/2 ; x = 7.1 × 10 –7 mol AgBr/L

    2. Molar solubility = x = [Br ]; [Ag + ] = 0.030 M + x

      K sp = (0.030 + x)(x) ≈ 0.030(x)

      5.0 × 10 –13 = 0.030(x); x = 1.7 × 10 –11 mol AgBr/L

    3. Molar solubility = x = [Ag + ]

      There are two sources of Br : NaBr(0.10 M ) and AgBr(x M )

      K sp = (x)(0.10 + x); assume x is small compared to 0.10 M .

      5.0 × 10 –13 = 0.10 (x); x ≈ 5.0 × 10 –12 mol AgBr/L

    17.58   LaF 3 (s)  ⇌  La 3+ (aq) + 3 F (aq);    K sp = [La 3+ ][F ] 3

    1. molar solubility = x = [La 3+ ]; [F ] = 3x

      K sp = 2 × 10 –19 = (x)(3x) 3 ; 2 × 10 –19 = 27 x 4 ; x = (7.41 × 10 –21 ) 1/4 , x = 9.28 × 10 –6 = 9 × 10 –6 M La 3+

      9.28 × 10 6 mol LaF 3 1 L × 195.9 g LaF 3 1 mol = 1.82 × 10 3 = 2 × 10 3 g LaF 3 / L

    2. molar solubility = x = [La 3+ ]

      There are two sources of F : KF (0.010 M ) and LaF 3 (3x M )

      K sp = (x)(0.010 + 3x) 3 ; assume x is small compared to 0.010 M .

      2 × 10 –19 = (0.010) 3 x; x = 2 × 10 –19 /1.0 × 10 –6 = 2 × 10 –13 M La 3+

      2 × 10 13 mol LaF 3 1 L × 195.9 g LaF 3 1 mol = 3.92 × 10 11 = 4 × 10 11 g LaF 3 /L


    1. molar solubility = x, [F ] = 3x, [La 3+ ] = 0.050 M + x

      K sp = (0.050 + x)(3x) 3 ; assume x is small compared to 0.050 M .

      2 × 10 –19 = (0.050)(27 x 3 ) = 1.35 x 3 ; x = (1.48 × 10 –19 ) 1/3 = 5.29 × 10 –7 = 5 × 10 –7 M

      5.29 × 10 7 mol LaF 3 1 L × 195.9 g LaF 3 1 mol = 1.04 × 10 4 = 1 × 10 4 g LaF 3 /L

    17.59 Analyze/Plan . Given a saturated solution of CaF 2 in contact with undissolved CaF 2 (s), consider the effect of adding CaCl 2 (s). The two salts have the Ca 2+ ion in common.

    Solve . As CaCl 2 is added, [Ca 2+ ] increases, K sp is exceeded, and additional CaF 2 precipitates until equilibrium is reestablished. At the new equilibrium position:

    1. Increase. The additional Ca 2+ from CaCl 2 decreases the solubility of CaF 2 .
    2. Increase. We have added CaCl 2 , which contains Ca 2+ .
    3. Decrease. After CaCl 2 is added, CaF 2 (s) precipitates, which decreases [F ].

    17.60   As KI is added, [I ] increases, K sp is exceeded, and additional PbI 2 precipitates until equilibrium is reestablished. At the new equilibrium position:

    1. Increase. The additional I from KI decreases the solubility of PbI 2 .
    2. Decrease. After KI is added, additional PbI 2 (s) precipitates.
    3. Increase. A small amount of the additional I precipitates as PbI 2 (s), but most of it increases the concentration of I ions in solution.

    17.61 Analyze/Plan .  We are asked to calculate the solubility of a slightly soluble hydroxide salt at various pH values. This is a common ion problem; pH tells us not only [H + ] but also [OH ], which is an ion common to the salt. Use pH to calculate [OH ] and then proceed as in Sample Exercise 17.13. Solve .

    Mn(OH) 2 (s)  ⇌  Mn 2+ (aq) + 2 OH (aq);    K sp = 1.6 × 10 –13

    Because [OH ] is set by the pH of the solution, the solubility of Mn(OH) 2 is just [Mn 2+ ].

    1. pH = 7.0, pOH = 14 – pH = 7.0, [OH ] = 10 –pOH = 1.0 × 10 –7 M

      K sp = 1.6 × 10 13 = [ Mn 2 + ] ( 1.0 × 10 7 ) 2 ; [ Mn 2 + ] = 1.6 × 10 13 1.0 × 10 14 = 16 M

      16 mol Mn(OH) 2 1 L × 88.95 g Mn(OH) 2 1 mol Mn(OH) 2 = 1423 = 1.4 × 10 3 g Mn(OH) 2 /L

      Check .  Note that the solubility of Mn(OH) 2 in pure water is 3.6 × 10 –5 M , and the pH of the resulting solution is 9.0. The relatively low pH of a solution buffered to pH 7.0 actually increases the solubility of Mn(OH) 2 .

    2. pH = 9.5, pOH = 4.5, [OH ] = 3.16 × 10 –5 = 3.2 × 10 –5 M

      K sp = 1.6 × 10 13 = [ Mn 2 + ] ( 3.16 × 10 5 ) 2 ; [ Mn 2 + ] = 1.6 × 10 13 1.0 × 10 9 = 1.6 × 10 4 M

      1.6 × 10 –4 M Mn(OH) 2 × 88.95 g/mol = 0.0142 = 0.014 g/L


    1. pH = 11.8, pOH = 2.2, [OH ] = 6.31 × 10 –3 = 6.3 × 10 –3 M

      K sp = 1.6 × 10 13 = [ Mn 2 + ] ( 6.31 × 10 3 ) 2 ; [ Mn 2 + ] = 1.6 × 10 13 3.98 × 10 5 = 4.0 × 10 9 M

      4.02 × 10 –9 M Mn(OH) 2 × 88.95 g/mol = 3.575 × 10 –7 = 3.6 × 10 –7 g/L

    17.62   Ni(OH) 2 (s)  ⇌  Ni 2+ (aq) + 2 OH (aq);    K sp = 6.0 × 10 –16

    Because the [OH ] is set by the pH of the solution, the solubility of Ni(OH) 2 is just [Ni 2+ ].

    1. pH = 8.0, pOH = 14 – pH = 6.0, [OH ] = 10 –pOH = 1 × 10 –6 M

      K sp = 6.0 × 10 16 = [ Ni 2 + ] ( 1.0 × 10 6 ) 2 ; [ Ni 2 + ] = 6.0 × 10 16 1.0 × 10 12 = 6.0 × 10 4 = 6 × 10 4 M

    2. pH = 10.0, pOH = 4.0, [OH ] = 1.0 × 10 –4 = 1 × 10 –4 M

      K sp = 6.0 × 10 16 = [ Ni 2 + ] [ 1.0 × 10 4 ] 2 ; [ Ni 2 + ] = 6.0 × 10 16 1.0 × 10 8 = 6.0 × 10 8 = 6 × 10 8 M

    3. pH = 12.0, pOH = 2.0, [OH ] = 1.0 × 10 –2 = 1 × 10 –2 M

      K sp = 6.0 × 10 16 = [ Ni 2 + ] [ 1.0 × 10 2 ] 2 ; [ Ni 2 + ] = 6.0 × 10 16 1.0 × 10 4 = 6.0 × 10 12 = 6 × 10 12 M

    17.63 Analyze/Plan .  Follow the logic in Sample Exercise 17.14. Solve .

    If the anion of the salt is the conjugate base of a weak acid, it will combine with H + , reducing the concentration of the free anion in solution, thereby causing more salt to dissolve. More soluble in acid: (a) ZnCO 3, (b) ZnS, (d) AgCN, (e) Ba 3 (PO 4 ) 2 .

    17.64   If the anion in the slightly soluble salt is the conjugate base of a strong acid, there will be no reaction.

    1. MnS(s) + 2 H + (aq) → H 2 S(aq) + Mn 2+ (aq)
    2. PbF 2 (s) + 2 H + (aq) → 2 HF(aq) + Pb 2+ (aq)
    3. AuCl 3 (s) + H + (aq) → no reaction
    4. Hg 2 C 2 O 4 (s) + 2 H + (aq) → H 2 C 2 O 4 (aq) + Hg 2 2+ (aq)
    5. CuBr(s) + H + (aq) → no reaction

    17.65 Analyze/Plan .  Follow the logic in Sample Exercise 17.15. Solve .

    The formation equilibrium is

    Ni 2+ (aq) + 6NH 3 (aq)  ⇌ Ni(NH 3 ) 6 2 + (aq) K f = [ Ni (NH 3 ) 6 2 + ] [ Ni 2 + ] [ NH 3 ] 6 = 1.2 × 10 9

    1 .25 g NiCl 2 × 1 mol NiCl 2 129 .62 g NiCl 2 × 1 0 .1000 L = 0 .096436 = 0 .0964 M NiCl 2


    Assume that nearly all the Ni 2+ is in the form Ni(NH 3 ) 6 2+ .

    [Ni(NH 3 ) 6 2+ ] = 0.0964 M ; [Ni 2+ ] = x; [NH 3 ] = 0.20 M

    1.2 × 10 9 = ( 0.0964 ) x(0 .20) 6 ; x = 1.26 × 10 6 = 1.3 × 10 6 M = [ Ni 2 + ]

    [Ni(NH 3 ) 6 2+ ] = 0.0964 M – 1.26 × 10 –6 M = 0.0964 M

    Note that our assumption that most of the Ni 2+ is present as Ni(NH 3 ) 6 2+ is true.

    17.66   NiC 2 O 4 (s)  ⇌  Ni 2+ (aq) + C 2 O 4 2– (aq);     K sp = [Ni 2+ ][C 2 O 4 2– ] = 4 × 10 –10

    When the salt has just dissolved, [C 2 O 4 2– ] will be 0.020 M . Thus, [Ni 2+ ] must be less than 4 × 10 –10 / 0.020 = 2 × 10 –8 M . To achieve this low [Ni 2+ ], we must complex the Ni 2+ ion with NH 3 : Ni 2+ (aq) + 6 NH 3 (aq)  ⇌  Ni(NH 3 ) 6 2+ (aq). Essentially all Ni(II) is in the form of the complex, so [Ni(NH 3 ) 6 2+ ] = 0.020. Find K f for Ni(NH 3 ) 6 2+ in Table 17.1.

    K f = [Ni(NH 3 ) 6 2 + ] [ Ni 2 + ] [ NH 3 ] 6 = ( 0.020 ) ( 2 × 10 8 ) [ NH 3 ] 6 = 1.2 × 10 9 ; [NH 3 ] 6 = 8.33 × 10 4 ; [ NH 3 ] = 0.307 = 0.3 M

    17.67 Analyze/Plan .  Calculate the solubility of AgI in pure water according to the method in Sample Exercise 17.12. Obtain K eq for the complexation reaction, making use of pertinent K sp and K f values from Appendix D.3 and Table 17.1. Write the dissociation equilibrium for Agl and the formation reaction for Ag(CN) 2 . Use algebra to manipulate these equations and their associated equilibrium constants to obtain the desired reaction and its equilibrium constant. Finally, use this K eq value to calculate the solubility of AgI in 0.100 M NaCN solution. Solve .

    1. AgI(s)  ⇌  Ag + (aq) + I (aq);     K sp = [Ag + ][I ] = 8.3 × 10 –17

      molar solubility = x = [Ag + ] = [I ]; K sp = x 2

      x = (8.3 × 10 –17 ) 1/2 ; x = 9.1 × 10 –9 mol AgI/L

    2. AgI(s) Ag + (aq) + I (aq) Ag + (aq) + 2 CN (aq) Ag(CN) 2 (aq) AgI(s) + 2 CN (aq) Ag(CN) 2 (aq) + I (aq)

      K = K sp × K f = [ Ag + ] [ I ] × [ Ag(CN) 2 ] [ Ag + ] [ CN ] 2 = ( 8.3 × 10 17 ) ( 1 × 10 21 ) = 8 × 10 4

    3. K is much greater than one for the reaction of AgI(s) with CN . This means that the reaction goes to completion. For a AgI(s) in 0.100 M NaCN solution, CN is the limiting reactant. Two moles of CN react with one mole of AgI, so the solublility of AgI in 0.100 M NaCN is (0.100/2) = 0.0500 M .

    17.68   According to Appendix D.3, K sp for Ag 2 S(s) is of the type

    Ag 2 S(s) + H 2 O(l) 2 Ag + (aq) + HS (aq) + OH (aq) HS (aq) + H + (aq) H 2 S(aq) 2[Ag + (aq) + 2 Cl (aq) AgCl 2 (aq)] Ag 2 S(s) + H 2 O(l) + H + (aq) + 4 Cl (aq) 2 AgCl 2 (aq) + H 2 S(aq) K sp
    1/K a1
    K f 2

    Add H + (aq) to each side to obtain the overall reacation

    Ag 2 S(s) + 2 H + (aq) + 4 Cl (aq) ⇌ 2 AgCl 2 (aq) + H 2 S(aq)

    K = K sp × K f 2 K a1 = ( 6 × 10 51 ) ( 1.1 × 10 5 ) 2 ( 9.5 × 10 8 ) = 7.64 × 10 34 = 8 × 10 34

    Precipitation and Separation of Ions (Section 17.6)

    17.69 Analyze/Plan .  Follow the logic in Sample Exercise 17.16. Precipitation conditions: will Q (see Chapter 15 ) exceed K sp for the compound? Solve .

    1. In base, Ca 2+ can form Ca(OH) 2 (s).

      Ca(OH) 2 (s)  ⇌  Ca 2+ (aq) + 2 OH (aq);    K sp = [Ca 2+ ][OH ] 2

      Q = [Ca 2+ ][OH ] 2 ; [Ca 2+ ] = 0.050 M ; pOH = 14 – 8.0 = 6.0; [OH ] = 1.0 × 10 –6 M

      Q = (0.050)(1.0 × 10 –6 ) 2 = 5.0 × 10 –14 ; K sp = 6.5 × 10 –6 (Appendix D.3)

      Q < K sp , no Ca(OH) 2 precipitates.

    2. Ag 2 SO 4 (s)  ⇌  2 Ag + (aq) + SO 4 2– (aq);     K sp = [Ag+] 2 [SO 4 2– ]

      [ Ag + ] = 0.050 M × 100 mL 110 mL = 4.545 × 10 2 = 4.5 × 10 2 M

      [ SO 4 2 ] = 0.050 M × 10 mL 110 mL = 4.545 × 10 3 = 4.5 × 10 3 M

      Q = (4.545 × 10 –2 ) 2 (4.545 × 10 –3 ) = 9.4 × 10 –6 ; K sp = 1.5 × 10 –5

      Q < K sp , no Ag 2 SO 4 precipitates.

    17.70

    1. Co(OH) 2 (s)  ⇌  Co 2+ (aq) + 2 OH (aq);  K sp = [Co 2+ ][OH ] 2 = 1.3 × 10 –15

      pH = 8.5; pOH = 14 – 8.5 = 5.5; [OH ] = 10 –5.5 = 3.16 × 10 –6 = 3 × 10 –6 M

      Q = (0.020)(3.16 × 10 –6 ) 2 = 2 × 10 –13 ; Q > K sp , Co(OH) 2 will precipitate.

    2. AgIO 3 (s)  ⇌  Ag + (aq) + IO 3 (aq);  K sp = [Ag+][IO 3 ] = 3.1 × 10 –8

      [ Ag + ] = 0.010 M Ag + × 0.020 L 0.030 L = 6.667 × 10 3 = 6.7 × 10 3 M

      [ IO 3 ] = 0.015 M IO 3 × 0.010 L 0.030 L = 5.000 × 10 3 = 5.0 × 10 3 M

      Q = (6.667 × 10 –3 )(5.00 × 10 –3 ) = 3.3 × 10 –5 ; Q > K sp , AgIO 3 will precipitate.


    17.71 Analyze/Plan .  We are asked to calculate pH necessary to precipitate Mn(OH) 2 (s) if the resulting Mn 2+ concentration is ≤ 1 μg/L.

    Mn(OH) 2 (s)  ⇌  Mn 2+ (aq) + 2 OH (aq); K sp = [Mn 2+ ][OH ] 2 = 1.6 × 10 –13

    At equilibrium, [Mn 2+ ][OH ] 2 = 1.6 × 10 –13 . Change concentration Mn 2+ (aq) to mol/L and solve for [OH ]. Solve .

    1 μ g Mn 2 + 1 L × 1 × 10 6 g 1 μ g × 1 mol Mn 2 + 54.94 g Mn 2 + = 1.82 × 10 8 = 2 × 10 8 M Mn 2 +

    1.6 × 10 –13 = (1.82 × 10 –8 )[OH ] 2 ; [OH ] 2 = 8.79 × 10 –6 ; [OH ] = 2.96 × 10 –3 = 3 × 10 –3 M

    pOH = 2.53; pH = 14 – 2.53 = 11.47 = 11.5

    17.72   PbI 2 (s) ⇌ Pb 2+ (aq) + 2 I (aq);  K sp = [Pb 2+ ][I ] 2 = 8.49 × 10 –9

    (This K sp value is taken from CRC Handbook of Chemistry and Physics , 74th edition.)

    [ Pb 2 + ] = 0.10 M × 0.2 mL 10.2 mL = 1.96 × 10 3 = 2 × 10 3 M ;

    [I ] = ( 8.49 × 10 9 1.96 × 10 3 M ) 1 / 2 = 2.08 × 10 3 = 2 × 10 3 M

    2.08 × 10 3 mol I 1 L × 126.90 g I 1 mol I × 0.0102 L = 2.69 × 10 3 g I = 3 × 10 3 g I

    17.73 Analyze/Plan .  We are asked which ion will precipitate first from a solution containing Pb 2+ (aq) and Ag + (aq) when I (aq) is added. Follow the logic in Sample Exercise 17.17. Calculate [I ] needed to initiate precipitation of each ion. The cation that requires lower [I ] will precipitate first. Solve .

    Ag + : K sp = [ Ag + ] [ I ] ; 8.3 × 10 17 = (2 .0 × 10 4 ) [ I ] ; [ I ] = 8.3 × 10 17 2.0 × 10 4 = 4.2 × 10 13 M I

    Pb 2 + : K sp = [ Pb 2 + ] [ I ] 2 ; 7.9 × 10 9 = (1 .5 × 10 3 ) [ I ] 2 ; [ I ] = ( 7.9 × 10 9 1.5 × 10 3 ) 1 / 2 = 2.3 × 10 3 M I

    AgI will precipitate first, at [I ] = 4.2 × 10 –13 M .

    17.74

    1. Precipitation will begin when Q = K sp .

      BaSO 4 : K sp = [Ba 2+ ][SO 4 2– ] = 1.1 × 10 –10

      1.1 × 10 –10 = (0.010)[SO 4 2– ]; [SO 4 2– ] = 1.1 × 10 –8 M

      SrSO 4 : K sp = [Sr 2+ ][SO 4 2– ] = 3.2 × 10 –7

      3.2 × 10 –7 = (0.010)[SO 4 2– ]; [SO 4 2– ] = 3.2 × 10 –5 M

      The [SO 4 2– ] necessary to begin precipitation is the smaller of the two values,

      1.1 × 10 –8 M SO 4 2– .

    2. Ba 2+ precipitates first, because it requires the smaller [SO 4 2– ].
    3. Sr 2+ will begin to precipitate when [SO 4 2– ] in solution (not bound in BaSO 4 ) reaches 3.2 × 10 –5 M .

    17.75 Analyze/Plan .  We are asked which ion will precipitate first when dilute Ag + (aq) is added to a solution containing 0.20 M CrO 4 2– , 0.10 M CO 3 2– , and 0.10 M Cl . The anions are present at different concentrations and their silver compounds have different stoichiometry, so we cannot directly compare K sp values. Follow the logic in Sample Exercise 17.17. Calculate [Ag + ] needed to initiate precipitation of each ion. The anion that requires lowest [Ag + ] will precipitate first, and so on. Solve .

    Ag 2 CrO 4 : K sp = [Ag + ] 2 [CrO 4 2– ] = 1.2 × 10 –12

    1.2 × 10 –12 = [Ag + ] 2 (0.20); [Ag + ] 2 = 6.0 × 10 –12 ; [Ag + ] = 2.4 × 10 –6 M

    Ag 2 CO 3 : K sp = [Ag + ] 2 [CO 3 2– ] = 8.1 × 10 –12

    8.1 × 10 –12 = [Ag + ] 2 (0.10); [Ag + ] 2 = 8.1 × 10 –11 ; [Ag + ] = 9.0 × 10 –6 M

    AgCl: K sp = [Ag + ][Cl ] = 1.8 × 10 –10

    1.8 × 10 –10 = [Ag + ](0.010); [Ag + ] = 1.8 × 10 –8

    AgCl requires the smallest [Ag + ] for precipitation and it will precipitate first. The other two will precipitate almost simultaneously.

    17.76   It is not appropriate to compare K sp values directly, because the stoichiometries of the the two precipitates are different.

    1. Precipitation will begin when Q = K sp .

      CaSO 4 : K sp = [Ca 2+ ][SO 4 2– ] = 2.4 × 10 –5

      2.4 × 10 –5 = (0.20)[SO 4 2– ]; [SO 4 2– ] = 1.2 × 10 –4 M

      Ag 2 SO 4 : K sp = [Ag + ] 2 [SO 4 2– ] = 1.5 × 10 –5

      1.5 × 10 –5 = (0.30) 2 [SO 4 2– ]; [SO 4 2– ] = 1.7 × 10 –4 M

      CaSO 4 requires the smaller [SO 4 2– ] for precipitation and it will precipitate first.

    2. The [SO 4 2– ] necessary to begin precipitation is the smaller of the two values,

      1.2 × 10 –4 M SO 4 2– .

      1.2 × 10 4 M = 1.0 M SO 4 2 × x L ( 0.010 + x L) ; x = ( 0.010 ) 1.2 × 10 4 = 1.2 × 10 6 L .

      We assume x is small compared to 0.010 L. The required volume is then 1.2 × 10 –6 L or 0.0012 mL or 1.2 μL. If one drop is approximately 0.2 mL, precipitation will begin as the first drop of 1.0 M Na 2 SO 4 solution is added.

    Qualitative Analysis for Metallic Elements (Section 17.7)

    17.77 Analyze/Plan .  Use Figure 17.23 and the description of the five qualitative analysis “groups” in Section 17.7 to analyze the given data. Ag + is in Group 1, Al 3+ is in Group 3, Mg 2+ is in Group 4, and Na + is in Group 5. Solve .

    The first two experiments eliminate Group 1 and 2 ions (Figure 17.23). The presence of a precipitate after the third experiment means that a Group 3 cation is present, in this case Al 3+ . The fact that no insoluble phosphates form in the filtrate from the third experiment rules out Group 4 ions. Ag + (Group 1) and Mg 2+ (Group 4) are definitely absent. Al 3+ (Group 3) is definitely present and Na + (Group 5) is possibly present.


    17.78   Initial solubility in water rules out CdS and HgO. Formation of a precipitate on addition of HCl indicates the presence of Pb(NO 3 ) 2 (formation of PbCl 2 ). Formation of a precipitate on addition of H 2 S at pH 1 probably indicates Cd(NO 3 ) 2 (formation of CdS). (This test can be misleading because enough Pb 2+ can remain in solution after filtering PbCl 2 to lead to visible precipitation of PbS.) Absence of a precipitate on addition of H 2 S at pH 8 indicates that ZnSO 4 is not present. The yellow flame test indicates presence of Na + . In summary, Pb(NO 3 ) 2 and Na 2 SO 4 are definitely present, Cd(NO 3 ) 2 is probably present, and CdS, HgO, and ZnSO 4 are definitely absent.

    17.79 Analyze/Plan .  We are asked to devise a procedure to separate various pairs of ions in aqueous solutions. In each case, refer to Figure 17.23 to find a set of conditions where the solubility of the two ions differs. Construct a procedure to generate these conditions. Solve .

    1. Cd 2+ is in Gp. 2, but Zn 2+ is not. Make the solution acidic using 0.2 M HCl; saturate with H 2 S. CdS will precipitate, but ZnS will not.
    2. Cr(OH) 3 is amphoteric, but Fe(OH) 3 is not. Add excess base; Fe(OH) 3 (s) precipitates, but Cr 3+ forms the soluble complex Cr(OH) 4 .
    3. Mg 2+ is a member of Gp. 4, but K + is not. Add (NH 4 ) 2 HPO 4 to a basic solution; Mg 2+ precipitates as MgNH 4 PO 4 , but K + remains in solution.
    4. Ag + is a member of Gp. 1, but Mn 2+ is not. Add 6 M HCl; precipitate Ag + as AgCl(s); Mn 2+ remains soluble.

    17.80

    1. Make the solution slightly acidic and saturate with H 2 S; CdS will precipitate, but Na + remains in solution.
    2. Make the solution acidic and saturate with H 2 S; CuS will precipitate, but Mg 2+ remains in solution.
    3. Add HCl; PbCl 2 precipitates. (It is best to carry out the reaction in an ice-water bath to reduce the solubility of PbCl 2 .)
    4. Add dilute HCl; AgCl precipitates, but Hg 2+ remains in solution.

    17.81

    1. Because phosphoric acid is a weak acid, the concentration of free PO 4 3– (aq) in an aqueous phosphate solution is low except in strongly basic media. In less basic media, the solubility product of the phosphates of interest is not exceeded.
    2. K sp for those cations in Group 3 is much larger. Thus, to exceed K sp , a higher

      [S 2– ] is required. This is achieved by making the solution more basic.

    3. They should all redissolve in strongly acidic solution; for example, in 12 M HCl (the chlorides of all Group 3 metals are soluble).

    17.82   The addition of (NH 4 ) 2 HPO 4 could result in precipitation of salts from metal ions of the other groups. The (NH 4 ) 2 HPO 4 will render the solution basic, so metal hydroxides as well as insoluble phosphates could form. It is essential to separate the metal ions of a group from other metal ions before carrying out the specific tests for that group.


    Additional Exercises

    17.83 Analyze/Plan . Follow the approach for deriving the Henderson–Hasselbach (H H) equation from the K a expression shown in Section 17.2. Begin with a general K b expression. Solve .

    B(aq) + H 2 O(l) ⇌ BH + (aq) + OH (aq); K b = [ BH + ] [ OH ] [ B ]

    pOH = –log[OH ]; rearrange K a to solve for [OH ]

    [ OH ] = K b [ B ] [ BH + ] ; take the –log of both sides

    –log[OH ] = –log K b + (–log[B] – (–log[BH + ] )

    pOH = pK b + log[BH + ] – log[B]

    pOH = pK b + log [ BH + ] [ B ]

    17.84   H 2 CO 3 (aq)  ⇌  H + (aq)  +  HCO 3 (aq) K a1 = 4.3 × 10 –7 pK a1 = 6.37
    HCO 3 (aq)  ⇌  H + (aq)  +  CO 3 2– (aq) K a2 = 5.6 × 10 –11 pK a2 = 10.25

    Use the two equilibrium constant expressions and the total carbonate concentration to solve for the three concentrations.

    [H + ] = 10 –5.60 = 2.512 × 10 –6 = 2.5 × 10 –6 M .

    K a1 = [ H + ] [ HCO 3 ] [ H 2 CO 3 ] , [ H 2 CO 3 ] = [ H + ] [ HCO 3 ] K a1 ; K a2 = [ H + ] [ CO 3 2 ] [ HCO 3 ] , [ CO 3 2 ] = K a2 [ HCO 3 ] [ H + ] [ H 2 CO 3 ] + [ HCO 3 ] + [ CO 3 2 ] = 1.0 × 10 5

    [ H + ] [ HCO 3 ] K a1 + [ HCO 3 ] + K a2 [ HCO 3 ] [ H + ] = 1.0 × 10 5

    Multiply by K a1 [H + ].

    [H + ] 2 [HCO 3 ] + K a1 [H + ] [HCO 3 ] + K a1 K a2 [HCO 3 ] = (1.0 × 10 –5 ) K a1 [H + ]

    [HCO 3 ]([H + ] 2 + K a1 [H + ] + K a1 K a2 ) = (1.0 × 10 –5 ) K a1 [H + ]

    [ HCO 3 ] = ( 1.0 × 10 5 ) K a1 [ H + ] [ H + ] 2 + K a1 [ H + ] + K a1 K a2 = ( 1.0 × 10 5 ) ( 4.3 × 10 7 ) ( 2.5 × 10 6 ) ( 2.5 × 10 6 ) 2 + ( 4.3 × 10 7 ) ( 2.5 × 10 6 ) + ( 4.3 × 10 7 ) ( 5.6 × 10 11 )

    [ HCO 3 ] = 1.468 × 10 –6 = 1.5 × 10 –6 M

    [ H 2 CO 3 ] = [ H + ] [ HCO 3 ] K a1 = 2.5 × 10 6 ( 1.5 × 10 6 ) 4.3 × 10 7 = 8.7 × 10 6 M

    [ CO 3 2 ] = K a2 [ HCO 3 ] [ H + ] = 5.6 × 10 11 ( 1.5 × 10 6 ) 2.5 × 10 6 = 3.4 × 10 11 M

    Check. First, pH of the raindrop (5.6) is less than pK a1 (6.37). We expect [H 2 CO 3 ] to be greater than [HCO 3 ], and it is. Second, the calculated total carbon species concentration is [H 2 CO 3 ] + [HCO 3 ] + [CO 3 2– ] = 8.7 × 10 –6 + 1.5 × 10 –6 + 3.4 × 10 –11 = 1.0 × 10 –5 M.

    The calculated results are self-consistent.


    17.85   The equilibrium of interest is

    HC 5 H 3 O 3 (aq) ⇌ H + (aq) + C 5 H 3 O 3 (aq); K a = 6.76 × 10 4 = [ H + ] [ C 5 H 3 O 3 ] [ HC 5 H 3 O 3 ]

    Begin by calculating [HC 5 H 3 O 3 ] and [C 5 H 3 O 3 ] for each case.

    1. 25 .0 g HC 5 H 3 O 3 0 .250 L soln × 1 mol HC 5 H 3 O 3 112 .1 g HC 5 H 3 O 3 = 0.8921 = 0.892 M HC 5 H 3 O 3

      30.0 g NaC 5 H 3 O 3 0.250 L soln × 1 mol NaC 5 H 3 O 3 134.1 g NaC 5 H 3 O 3 = 0.8949 = 0.895 M C 5 H 3 O 3

      [ H + ] = K a [ HC 5 H 3 O 3 ] [ C 5 H 3 O 3 ] = 6.76 × 10 4 (0 .8921 x) ( 0.8949 + x) 6.76 × 10 4 (0 .8921) ( 0.8949 )

      [H + ] = 6.74 × 10 –4 M , pH = 3.171

    2. For dilution, M 1 V 1 = M 2 V 2

      [ HC 5 H 3 O 3 ] = 0.250 M × 30.0 mL 125 mL = 0.0600 M

      [ C 5 H 3 O 3 ] = 0.220 M × 20.0 mL 125 mL = 0.0352 M

      [ H + ] 6.76 × 10 4 (0 .0600) 0.0352 = 1.15 × 10 3 M , pH = 2.938

      (yes, [H + ] is < 5% of 0.0352 M )

    3. 0.0850 M × 0.500 L = 0.0425 mol HC 5 H 3 O 3

      1.65 M × 0.0500 L = 0.0825 mol NaOH

      HC 5 H 3 O 3 (aq) + NaOH(aq) NaC 5 H 3 O 3 (aq) + H 2 O(l)
      initial 0.0425 mol 0.0825 mol
      reaction –0.0425 mol –0.0425 mol +0.0425 mol
      after 0 mol 0.0400 mol 0.0425 mol

      The strong base NaOH dominates the pH; the contribution of C 5 H 3 O 3 is negligible. This combination would be “after the equivalence point” of a titration. The total volume is 0.550 L.

      [ OH ] = 0.0400 mol 0.550 L = 0.0727 M ; pOH = 1.138 , pH = 12.862

    17.86 K a = [ H + ] [ In ] [ HIn ] ; at pH = 4.68 , [ HIn] = [ In ] ; [ H + ] = K a ; pH = pK a = 4.68

    17.87

    1. HA(aq) + B(aq) ⇌ HB + (aq) + A (aq) K eq = [ HB + ] [ A ] [ HA ] [ B ]
    2. Note that the solution is slightly basic because B is a stronger base than HA is an acid. (Or, equivalently, that A is a stronger base than HB + is an acid.) Thus, a little of the A is used up in reaction: A (aq) + H 2 O(l)  ⇌   HA(aq) + OH (aq). Because pH is not very far from neutral, it is reasonable to assume that the reaction in part (a) has gone far to the right, and that [A ] ≈ [HB + ] and [HA] ≈ [B]. Then,

    1. K a = [ A ] [H + ] [ HA ] = 8.0 × 10 5 ; when pH = 9.2 , [ H + ] = 6.31 × 10 10 = 6 × 10 10 M

      [ A ] [ HA ] = 8.0 × 10 5 / 6.31 × 10 10 = 1.268 × 10 5 = 1 × 10 5

      From the earlier assumptions, [ A ] [ HA ] = [ HB + ] [ B ] , so K eq [ A ] 2 [ HA ] 2 = 1.608 × 10 10 = 2 × 10 10

    1. K b for the reaction B(aq) + H 2 O(l) ⇌ BH + (aq) + OH (aq) can be calculated by noting that the equilibrium constant for the reaction in part (a) can be written as K = K a (HA) × K b (B) / K w . (You should prove this to yourself.) Then,

      K b (B) = K × K w K a ( HA) = ( 1.608 × 10 10 ) ( 1.0 × 10 14 ) 8.0 × 10 5 = 2.010 = 2

      K b (B) is larger than K a (HA), as it must be if the solution is basic.

    17.88

    1. K a = [ H + ] [ HCOO ] [ HCOOH ] ; [ H + ] = K a [HCOOH] [HCOO ]

      Buffer A: [HCOOH ] = [ HCOO ] = 1.00 mol 1.00 L = 1.00 M

      [ H + ] = 1.8 × 10 4 ( 1.00 M ) ( 1.00 M ) = 1.8 × 10 4 M , pH = 3.74

      Buffer B: [HCOOH ] = [ HCOO ] = 0.010 mol 1.00 L = 0.010 M

      [ H + ] = 1.8 × 10 4 ( 0.010 M ) ( 0.010 M ) = 1.8 × 10 4 M , pH = 3.74

      The pH values of the two buffers are equal because they both contain HCOOH and HCOONa and the [HCOOH] / [HCOO ] ratio is the same in both solutions.

    2. Buffer A has the greater capacity because it contains the greater absolute concentrations of HCOOH and HCOO .
    3. Buffer A: HCOO + HCl HCOOH  +   Cl
      1.00 mol 0.001 mol 1.00 mol
      0.999 mol 0 1.001 mol

      [ H + ] = 1.8 × 10 4 ( 1.001 ) ( 0.999 ) = 1.8 × 10 4 M , pH = 3.74

      (In a buffer calculation, volumes cancel and we can substitute moles directly into the K a expression.)

      Buffer B: HCOO + HCl HCOOH +   Cl
      0.010 mol 0.001 mol 0.010 mol
      0.009 mol 0 0.011 mol

      [ H + ] = 1.8 × 10 4 ( 0.011 ) ( 0.009 ) = 2.2 × 10 4 M , pH = 3.66


    1. Buffer A: 1.00 M HCl × 0.010 L = 0.010 mol H + added

      mol HCOOH = 1.00 + 0.010 = 1.01 mol

      mol HCOO = 1.00 – 0.010 = 0.99 mol

      [ H + ] = 1.8 × 10 4 ( 1.01 ) ( 0.99 ) = 1.8 × 10 4 M , pH = 3.74

      Buffer B: mol HCOOH = 0.010 + 0.010 = 0.020 mol = 0.020 M

      mol HCOO = 0.010 – 0.010 = 0.000 mol

      The solution is no longer a buffer; the only source of HCOO is the dissociation of HCOOH. Adding 10 mL of 1.00 M HCl exceeds the buffer capacity of this buffer.

      K a = [ H + ] [COO ] [ HCOOH ] = x 2 ( 0.020 x) M

      The extent of ionization is greater than 5%; from the quadratic formula,

      x = [H + ] = 1.8 × 10 –3 , pH = 2.74.

    17.89 0.15 mol CH 3 COOH 1 L soln × 0.750 L = 0.1125 = 0.11 mol CH 3 COOH

    0.1125 mol CH 3 COOH × 60.05 g CH 3 COOH 1 mol CH 3 COOH × 1 g gl acetic acid 0.99 g CH 3 COOH × 1.00 mL gl acetic acid 1.05 g gl acetic acid = 6.5 mL glacial acetic acid

    At pH 4.50, [H + ] = 10 –4.50 = 3.16 × 10 –5 = 3.2 × 10 –5 M ; this is small compared to

    0.15 M CH 3 COOH.

    K a = ( 3.16 × 10 5 ) [ CH 3 COO ] 0.15 = 1.8 × 10 5 ; [ CH 3 COO ] = 0.0854 = 0.085 M

    0 .0854 mol CH 3 COONa 1 L soln × 0 .750 L × 82 .03 g CH 3 COONa 1 mol CH 3 COONa = 5 .253 = 5 .25 g CH 3 COONa

    17.90

    1. For a monoprotic acid (one H + per mole of acid), at the equivalence point, moles OH added = moles H + originally present

      M B × V B = g acid/molar mass

      MM = g acid M B × V B = 0.2140 g 0.0950 M × 0.0300 L = 75.09 = 75.1 g/mol

    2. Addition of 15.0 mL of 0.0950 M NaOH is half way to the equivalence point of the titration. At this point, the solution in the flask is a buffer where [HA] = [A ] and pH of the solution equals pK a of HA. At 15.0 mL, pH = pK a = 6.50

      K a = 10 –6.50 = 3.16 × 10 –7 = 3.2 × 10 –7

    17.91

    1. For a monoprotic acid (one H + per mole of acid), at the equivalence point moles

      OH added = moles H + originally present

      M B × V B = g acid/molar mass

      MM = g acid M B × V B = 0.1687 g 0.1150 M × 0.0155 L = 94.642 = 94.6 g/mol


    1. initial mol HA = 0.1687 g 9 4.642 g/mol = 1.783 × 10 3 = 1.78 × 10 3 mol HA

      mol OH added to pH 2.85 = 0.1150 M × 0.00725 L = 8.338 × 10 –4 = 8.34 × 10 –4 mol OH

      HA(aq) + NaOH(aq) NaA(aq) + H 2 O
      before rx 1.783 × 10 –3 mol 0.834 × 10 –3 mol 0
      change –0.834 × 10 –3 mol –0.834 × 10 –3 mol 0.834 × 10 –3 mol
      after rx 0.949 × 10 –3 mol 0 0.834 × 10 –3 mol

      [ HA] = 9.49 × 10 4 mol 0.0325 L = 0.02919 = 0.0292 M

      [ A ] = 8.34 × 10 4 mol 0.0325 L = 0.02565 = 0.0257 M

      [ H + ] = 10 2.85 = 1.413 × 10 3 = 1.4 × 10 3

      The mixture after reaction (a buffer) can be described by the acid dissociation equilibrium.

      HA(aq) H + (aq) + A (aq)
      initial 0.0292 M 0 0.0257 M
      equil. (0.0292 – 1.4 × 10 –3 M ) 1.4 × 10 –3 M (0.0257 + 1.4 × 10 –3 ) M

      K a = [ H + ] [ A ] [ HA ] ( 1.413 × 10 3 ) ( 0.02707 ) ( 0.02778 ) = 1.4 × 10 3

      (Although we have carried extra figures through the calculation to avoid rounding errors, the data dictate an answer with 2 sig figs.)

    17.92   At the equivalence point of a titration, moles strong base added equals moles weak acid initially present. M B × V B = mol base added = mol acid initial.

    At the half-way point, the volume of base is one-half of the volume required to reach the equivalence point, and the moles base delivered equals one-half of the moles acid initially present. This means that one-half of the weak acid HA is converted to the conjugate base A . If exactly half of the acid reacts, mol HA = mol A and [HA] = [A ] at the half-way point.

    From Equation 17.9, pH = pK a + log [conj . base] [ conj . acid] = pK a + log [A ] [ HA ] .

    If [A ]/[HA] = 1, log(1) = 0 and pH = pK a of the weak acid being titrated.

    17.93   If 50.0 mL base is required to reach the equivalence point, addition of 25.0 mL of base is half-way to the equivalence point. At this point, [A ] = [HA] and [A ]/[HA] = 1. The pK a of the weak acid being titrated is equal to the pH of the solution, 3.62.


    17.94   Assume that H 3 PO 4 will react with NaOH in a stepwise fashion: (This is not unreasonable, because the three K a values for H 3 PO 4 are significantly different.)

    H 3 PO 4 (aq) + NaOH(aq)  → H 2 PO 4 (aq) + Na + (aq) + H 2 O(l)
    before 0.20 mol 0.30 mol 0 mol
    after 0 mol 0.10 mol 0.20 mol
    H 2 PO 4 (aq) + NaOH(aq)  → HPO 4 (aq) + Na + (aq) + H 2 O(l)
    before 0.20 mol 0.10 mol 0.25 mol
    after 0.10 mol 0 0.35 mol

    Thus, after all NaOH has reacted, the resulting 1.00 L solution is a buffer containing

    0.10 mol H 2 PO 4 and 0.35 mol HPO 4 2– . H 2 PO 4 (aq)  ⇌  H + (aq) + HPO 4 2– (aq)

    K a = 6.2 × 10 8 = [ HPO 4 2 ] [ H + ] [ H 2 PO 4 ] ; [ H + ] = 6.2 × 10 8 (0 .10 M ) 0.35 M = 1.77 × 10 8 = 1.8 × 10 8 M ; pH = 7.75

    17.95   The pH of a buffer system is centered around pK a for the conjugate acid component. For a diprotic acid, two conjugate acid/conjugate base pairs are possible.

    H 2 X(aq)  ⇌  H + (aq) + HX (aq); K a1 = 2 × 10 –2 ; pK a1 = 1.70

    HX (aq)  ⇌  H + (aq) + X 2– (aq); K a2 = 5.0 × 10 –7 ; pK a2 = 6.30

    Clearly, HX / X 2– is the more appropriate combination for preparing a buffer with pH = 6.50. The [H + ] in this buffer = 10 –6.50 = 3.16 × 10 –7 = 3.2 × 10 –7 M . Using the K a2 expression to calculate the [X 2– ] / [HX ] ratio:

    K a2 = [ H + ] [ X 2 ] [ HX ] ; K a2 [H + ] = [ X 2 ] [ HX ] = 5.0 × 10 7 3.16 × 10 7 = 1.58 = 1.6

    Because X 2– and HX are present in the same solution, the ratio of concentrations is also a ratio of moles.

    [ X 2 ] [ HX ] = ( mol X 2 /L soln mol HX /L soln ) = mol X 2 mol HX = 1.58 ; mol X 2 = ( 1.58 ) mol HX

    In the 1.0 L of 1.0 M H 2 X, there is 1.0 mol of material containing X 2– .

    Thus, mol HX + 1.58 (mol HX ) = 1.0 mol. 2.58 (mol HX ) = 1.0;

    mol HX = 1.0 / 2.58 = 0.39 mol HX ; mol X 2– = 1.0 – 0.39 = 0.61 mol X 2– .

    Thus, enough 1.0 M NaOH must be added to produce 0.39 mol HX and 0.61 mol X 2– .

    Considering the neutralization in a stepwise fashion (see discussion of titrations of polyprotic acids in Section 17.3).

    H 2 X(aq)  + NaOH(aq)  → HX (aq) + H 2 O(l)
    before 1.0 mol 1 mol 0
    after 0 0 1.0 mol

    HX (aq)  + NaOH(aq)  → X 2– (aq) + H 2 O(l)
    before 1.0 0.61
    change –0.61 –0.61 +0.61
    after 0.39 0 0.61

    Starting with 1.0 mol of H 2 X, 1.0 mol of NaOH is added to completely convert it to 1.0 mol of HX . Of that 1.0 mol of HX , 0.61 mol must be converted to 0.61 mol X 2– . The total moles of NaOH added is (1.00 + 0.61) = 1.61 mol NaOH.

    L NaOH = mol NaOH M NaOH = 1.61 mol 1 .0 M = 1.6 L of 1 .0 M NaOH

    17.96   CH 3 CH(OH)COO will be formed by reaction CH 3 CH(OH)COOH with NaOH.

    0.1000 M × 0.02500 L = 2.500 × 10 –3 mol CH 3 CH(OH)COOH; b = mol NaOH needed

    CH 3 CH(OH)COOH + NaOH CH 3 CH(OH)COO + H 2 O + Na +
    initial 2.500 × 10 –3 mol b mol
    rx –b mol –b mol +b mol
    after rx (2.500 × 10 –3 – b) mol 0 b mol

    K a = [ H + ] [ CH 3 CH(OH)COO ] [ CH 3 CH(OH)COOH ] ; K a = 1.4 × 10 4 ; [ H + ] = 10 pH = 10 3.75 = 1.778 × 10 4 = 1.8 × 10 4 M

    Because solution volume is the same for reaction CH 3 CH(OH)COOH and CH 3 CH(OH)COO , we can use moles in the equation for [H + ].

    K a = 1.4 × 10 4 = 1.778 × 10 4 (b) ( 2.500 × 10 3 b) ; 0.7874 ( 2.500 × 10 3 b) = b, 1 .969 × 10 3 = 1.7874 b,

    b = 1.10 × 10 –3 = 1.1 × 10 –3 mol OH

    (The precision of K a dictates that the result has 2 sig figs.)

    Substituting this result into the K a expression gives [H + ] = 1.8 × 10 –4 . This checks and confirms our result. Calculate volume NaOH required from M = mol/L.

    1.10 × 10 3 mol OH × 1 L 1 .000 mol × 1 μ L 1 × 10 6 L = 1.1 × 10 3 μ L (1 .1 mL)

    17.97

    1. PbCO 3 (s)  ⇌  Pb 2+ (aq) + CO 3 2− (aq)

      K sp = [Pb 2+ ][CO 3 2− ] = 7.4 × 10− 14 . molar solubility = s = [Pb 2+ ] = [CO 3 2− ]

      K sp = s 2 = 7.4 × 10− 14 . s = [Pb 2+ ] = 2.7203 × 10− 7 = 2.7 × 10− 7 M

    2. For very dilute aqueous solutions, assume the solution density is 1.0 g/mL.

      ppb = g solute 10 9 g solution = 1 × 10 6 g solute 1 × 10 3 g solution = μ g solute L solution

      2.7203 × 10 7 mol Pb 2 + L × 207.2 g Pb 2 + 1 mol Pb 2 + × 1 μ g 1 × 10 6 g = 56.365 μ g Pb 2 + L = 56 ppb

    3. The solubility of PbCO 3 increases as pH is lowered. When pH is lowered, [H + ] increases. The H + (aq) reacts with CO 3 2− (aq) to form HCO 3− and H 2 CO 3 (aq). This shifts the solubility equilibrium to the right and increases the solubility of PbCO 3 .

    1. A saturated solution of lead carbonate, with a lead concentration of 56 ppb, exceeds the EPA acceptable lead level of 15 ppb.

    17.98

    1. CdS: 8.0 × 10 –28 ; CuS: 6 × 10 –37 . CdS has greater molar solubility.
    2. PbCO 3 : 7.4 × 10 –14 ; BaCrO 4 : 2.1 × 10 –10 . BaCrO 4 has greater molar solubility.
    3. Because the stoichiometry of the two complexes is not the same, K sp values can not be compared directly; molar solubilities must be calculated from K sp values.

      Ni(OH) 2 : K sp = 6.0 × 10 –16 = [Ni 2+ ][OH ] 2 ; [Ni 2+ ] = x, [OH ] = 2x

      6.0 × 10 –16 = (x)(2x) 2 = 4x 3 ; x = 5.3 × 10 –6 M Ni 2+

      Note that [OH ] from the autoionization of water is less than 1% of [OH ] from Ni(OH) 2 and can be neglected.

      NiCO 3 : K sp = 1.3 × 10 –7 = [Ni 2+ ][CO 3 2– ]; [Ni 2+ ] = [CO 3 2– ] = x

      1.3 × 10 –7 = x 2 ; x = 3.6 × 10 –4 M Ni 2+

      NiCO 3 has greater molar solubility than Ni(OH) 2 , but the values are much closer than expected from inspection of K sp values alone.

    4. Again, molar solubilities must be calculated for comparison.

      Ag 2 SO 4 : K sp = 1.5 × 10 –5 = [Ag + ] 2 [SO 4 2– ]; [SO 4 2– ] = x, [Ag + ] = 2x

      1.5 × 10 –5 = (2x) 2 (x) = 4x 3 ; x = 1.6 × 10 –2 M SO 4 2–

      AgI:   K sp = 8.3 × 10 –17 = [Ag + ][I ]; [Ag + ] = [I ] = x

      8.3 × 10 –17 = x 2 ; x = 9.1 × 10 –9 M Ag +

      Ag 2 SO 4 has greater molar solubility than AgI.

    17.99

    1. K sp = 4.5 × 10 –9 = [Ca 2+ ][CO 3 2– ]; s = [Ca 2+ ] = [CO 3 2– ]

      s 2 = 4.5 × 10 –9 , s = 6.708 × 10 –5 = 6.7 × 10 –5

    2. CaCO 3 (s) Ca 2 + (aq) + CO 3 2‒ (aq) K sp CO 3 (aq) + H 2 O(l) HCO 3 (aq) + OH (aq) K b CaCO 3 (s) + H 2 O(l) Ca 2+ (aq) + HCO 3 (aq) + OH (aq) K

      K b = K w / K a for HCO 3

      K = K sp × K b = K sp × K w K a for HCO 3 = 4.5 × 10 9 × 1 × 10 14 5.6 × 10 11 = 8.036 × 10 13 = 8.0 × 10 13

    3. K = 8.036 × 10 –13 = [Ca 2+ ][HCO 3 ][OH ] = s 3 ; s = 9.297 × 10 –5 = 9.3 × 10 –5 M
    4. pH = 8.3, pOH = 14 – 8.3 = 5.7. [OH ] = 10 –5.7 = 1.995 × 10 –6 = 2 × 10 –6 M

      8.036 × 10 –13 = s 2 (1.995 × 10 –6 ), s = 6.346 × 10 –4 = 6 × 10 –4 M

    5. pH = 7.5, pOH = 14 – 7.5 = 6.5. [OH ] = 10 –6.5 = 3.162 × 10 –7 = 3 × 10 –7 M

      8.036 × 10 –13 = s 2 (3.162 × 10 –7 ), s = 1.549 × 10 –3 = 2 × 10 –3 M

      The drop in pH from 8.3 to 7.5 significantly increases (from 6.7 × 10 –5 M to 1.5 × 10 –3 M ) the molar solubility of CaCO 3 (s).


    17.100

    1. Hydroxyapatite: K sp = [Ca 2+ ] 5 [PO 4 3– ] 3 [OH ]

      Fluoroapatite: K sp = [Ca 2+ ] 5 [PO 4 3– ] 3 [F ]

    2. For each mole of apatite dissolved, one mole of OH or F is formed. Express

      molar solubility, s, in terms of [OH ] and [F ].

      Hydroxyapatite: [OH ] = s, [Ca 2+ ] = 5s, [PO 4 3– ] = 3s

      K sp = 6.8 × 10 –27 = (5s) 5 (3s) 3 (s) = 84,375 s 9

      s 9 = 8.059 × 10 –32 = 8.1 × 10 –32 .

      Use logs to find s. s = 3.509 × 10 –4 = 3.5 × 10 –4 M Ca 5 (PO 4 ) 3 OH.

      Fluoroapatite: [F ] = s, [Ca 2+ ] = 5s, [PO 4 3– ] = 3s

      K sp = 1.0 × 10 –60 = (5s) 5 (3s) 3 (s) = 84,375 s 9

      s 9 = 1.185 × 10 –65 = 1.2 × 10 –65 ; s = 6.109 × 10 –8 = 6.1 × 10 –8 M Ca 5 (PO 4 ) 3 F

    17.101

    1. K b for PO 4 3− = K w / K a3 = (1.0 × 10− 14 )/ (4.2 × 10− 13 ) = 0.02381 = 2.4 × 10− 2
    2. PO 4 3− (aq) + H 2 O(l)  ⇌  HPO 4 2− (aq) + OH (aq)

      [HPO 4 2− ] = [OH ] = x, [PO 4 3− ] = (1 × 10− 3 ) − x

      K b = 2.4 × 10 2 = [ HPO 4 2 ] [ OH ] [ PO 4 3 ] = x 2 ( 1 × 10 3 ) x

      Because K b is relatively large and [PO 4 3− ] is small, x may be significant relative to (1 × 10− 3 ) M . Use the quadratic formula to solve for x.

      x 2 + 2.4 × 10− 2 x − 2.4 × 10− 5 = 0; x = 9.615 × 10− 4 = 1 × 10− 3 M OH

      pOH = 3.017 = 3.0, pH = 10.983 = 11.0

    17.102 Analyze/Plan . Calculate the solubility of Mg(OH) 2 in 0.50 M NH 4 Cl. Find K sp for Mg(OH) 2 in Appendix D.3. NH 4 + is a weak acid, which will increase the solubility of Mg(OH) 2 . Combine the various interacting equilibria to obtain an overall reaction. Calculate K for this reaction and use it to calculate solubility (s) for Mg(OH) 2 in 0.50 M NH 4 Cl . Solve .

    Mg(OH) 2 (s) Mg 2 + (aq) + 2 OH (aq) 2 NH 4 + (aq) 2 NH 3 (aq) + 2 H + (aq) 2 H + (aq) + 2 OH (aq) 2 H 2 O(l) Mg(OH) 2 (s) + 2 NH 4 + (aq) + 2 H + (aq) + 2 OH (aq) Mg 2 + (aq) + 2 NH 3 (aq) + 2 OH ( aq ) + 2 H + ( aq ) + 2 H 2 O ( l )

    K sp
    K a
    1/K w

    Mg(OH) 2 (s) + 2 NH 4 + (aq) ⇌ Mg 2+ (aq)+ 2 NH 3 (aq) + 2 H 2 O(l)

    K = [ Mg 2 + ] [ NH 3 ] 2 [ NH 4 + ] 2 = K sp × K a 2 K w 2 ; K a for NH 4 + = K w K b for NH 3 ; K a K w = 1 K b

    K = K sp × K a 2 K w 2 = K sp K b 2 = 1.8 × 10 11 ( 1.8 × 10 5 ) 2 = 5.556 × 10 2 = 5.6 × 10 2


    Let [Mg 2+ ] = s, [NH 3 ] = 2s, [NH 4 + ] = 0.50 – 2s

    K = 5.6 × 10 2 = [ Mg 2 + ] [ NH 3 ] 2 [ NH 4 + ] 2 = s ( 2 s ) 2 ( 0.5 2 s ) 2 = 4 s 3 0.25 2 s + 4 s 2

    5.6 × 10 –2 (0.25 – 2s + 4s 2 ) = 4s 3 ; 4s 3 – 0.222s 2 + 0.111s – 1.39 × 10 –2 = 0

    Clearly, 2s is not small relative to 0.50. Solving the third-order equation, s = 0.1054 = 0.11 M . The solubility of Mg(OH) 2 in 0.50 M NH 4 Cl is 0.11 mol/L.

    Check . Substitute s = 0.1054 into the K expression.

    K = 4 ( 0.1054 ) 3 [ 0.50 2 ( 0.1054 ) ] 2 = 5.6 × 10 2 .

    The solubility and K value are consistent, to the precision of the K sp and K b values.

    17.103   K sp = [Ba 2+ ][MnO 4 ] 2 = 2.5 × 10 –10

    [ MnO 4 ] 2 = 2.5 × 10 10 / 2.0 × 10 8 = 0.0125 ; [ MnO 4 ] = 0.0125 = 0.11 M

    17.104   [Ca 2+ ][CO 3 2– ] = 4.5 × 10 –9 ; [Fe 2+ ][CO 3 2– ] = 2.1 × 10 –11

    Because [CO 3 2– ] is the same for both equilibria:

    [ CO 3 2 ] = 4.5 × 10 9 [ Ca 2 + ] = 2.1 × 10 11 [ Fe 2 + ] ; rearranging [Ca 2 + ] [Fe 2 + ] = 4.5 × 10 9 2.1 × 10 11 = 214 = 2.1 × 10 2

    17.105   PbSO 4 (s)  ⇌  Pb 2+ (aq) + SO 4 2– (aq);     K sp = 6.3 × 10 –7 = [Pb 2+ ] [SO 4 2– ]

    SrSO 4 (s)  ⇌   Sr 2+ (aq) + SO 4 2– (aq);      K sp = 3.2 × 10 –7 = [Sr 2+ ] [SO 4 2– ]

    Let x = [Pb 2+ ], y = [Sr 2+ ], x + y = [SO 4 2– ]

    x(x + y) y(x + y) = 6.3 × 10 7 3.2 × 10 7 ; x y = 1.9688 = 2.0 ; x = 1.969 y = 2.0 y

    y(1.969 y+y) = 3.2 × 10 –7 ; 2.969 y 2 = 3.2 × 10 –7 ; y = 3.283 × 10 –4 = 3.3 × 10 –4

    x = 1.969 y; x = 1.969(3.283 × 10 –4 ) = 6.464 × 10 –4 = 6.5 × 10 –4

    [Pb 2+ ] = 6.5 × 10 –4 M , [Sr 2+ ] = 3.3 × 10 –4 M , [SO 4 2– ] = (3.283 + 6.464) × 10 –4 = 9.7 × 10 –4 M

    17.106   MgC 2 O 4 (s)  ⇌  Mg 2+ (aq) + C 2 O 4 2– (aq)

    K sp = [Mg 2+ ][C 2 O 4 2– ] = 8.6 × 10 –5

    If [Mg 2+ ] is to be 3.0 × 10 –2 M , [C 2 O 4 2– ] = 8.6 × 10 –5 /3.0 × 10 –2 = 2.87 × 10 –3 = 2.9 × 10 –3 M

    The oxalate ion undergoes hydrolysis:

    C 2 O 4 2– (aq) + H 2 O(l)  ⇌  HC 2 O 4 (aq) + OH (aq)


    K b = [ HC 2 O 4 ] [ OH ] [ C 2 O 4 2 ] = 1.0 × 10 14 / 6.4 × 10 5 = 1.56 × 10 10 = 1.6 × 10 10

    [Mg 2+ ] = 3.0 × 10 –2 M , [C 2 O 4 2– ] = 2.87 × 10 –3 = 2.9 × 10 –3 M

    [HC 2 O 4 ] = (3.0 × 10 –2 – 2.87 × 10 –3 ) M = 2.71 × 10 –2 = 2.7 × 10 –2 M

    [ OH ] = 1.56 × 10 10 × [ C 2 O 4 2 ] [ HC 2 O 4 ] = 1.56 × 10 10 × ( 2.87 × 10 3 ) ( 2.71 × 10 2 ) = 1.652 × 10 11

    [OH ] = 1.7 × 10 –11 M ; pOH = 10.78, pH = 3.22

    17.107

    1. Express the molar solubility of Mg 3 (AsO 4 ) 2 as s. For each mole of Mg 3 (AsO 4 ) 2 that dissolves, 3 moles of Mg 2+ and 2 moles of AsO 4 3− are formed. In terms of s,

      [Mg 2+ ] = 3s and [AsO 4 3− ] = 2s.

      K sp = [Mg 2+ ] 3 [AsO 4 3− ] 2 = [3s] 3 [2s] 2 = 108s 5 = 2.1 × 10− 20 . s 5 = 1.9444 × 10− 22

      Use logs to find s. s = 4.547 × 10− 5 = 4.5 × 10− 5 M Mg 3 (AsO 4 ) 2 .

    2. A saturated solution of Mg 3 (AsO 4 ) 2 contains AsO 4 3– anion, the conjugate base of HAsO 4 2– . Base hydrolysis of AsO 4 3– changes the pH of the solution. Because H 3 AsO 4 is a polyprotic acid, base hydrolysis can occur in three steps. For H 3 AsO 4 , the pK a and corresponding pK b values are significantly different from each other and we can consider each step separately.

      AsO 4 3– (aq) + H 2 O(l)  ⇌  HAsO 4 2– (aq) + OH (aq)        pK b3 = 14 − pK a3

      HAsO 4 2– (aq) + H 2 O(l)  ⇌  H 2 AsO 4 (aq) + OH (aq)    pK b2 = 14 − pK a2

      H 2 AsO 4 (aq) + H 2 O(l)  ⇌  H 3 AsO 4 (aq) + OH (aq)      pK b1 = 14 − pK a1

      pK b3 = 14 − pK a3 = 14 – 11.50 = 2.5; K b3 = 3.1623 × 10 –3 = 3.2 × 10 –3

      [AsO 4 3– ] = 2s = 9.09 × 10 –5 M ; [OH ] = [HAsO 4 2– ] = x

      K b3 = [HAsO 4 2 ] [ OH ] [ AsO 4 3 ] = x 2 ( 9.09 × 10 5 x ) = 3.2 × 10 3

      Because K b3 is relatively large and [AsO 4 3– ] is small, we must use the quadratic equation to solve for x. x 2 + 3.16 × 10 –3 − 2.88 × 10 –7 = 0

      x = 8.847 × 10 –5 = 8.8 × 10 –5 M OH (aq); pOH = 4.05, pH = 9.95

      Now consider the second hydrolysis step.

      pK b2 = 14 − pK a2 = 14 – 6.98 = 7.02; K b2 = 9.550 × 10 –8 = 9.6 × 10 –8

      [H 2 AsO 4 ] = y; [OH ] = 8.8 × 10 –8 M + y; [HAsO 4 2– ] = 8.8 × 10 –5 M − y

      K b2 = [H 2 AsO 4 2 ] [ OH ] [ HAsO 4 2 ] = y ( 8.8 × 10 5 + y ) ( 8.8 × 10 5 y ) = 9.6 × 10 8

      With these values of K b2 and [HAsO 4 2– ], it is reasonable to assume y is small relative to 8.8 × 10 –5 .

      y = 9.6 × 10 –8 M ; [OH ] = 8.847 × 10 –5 + 9.550 × 10 –8 = 8.857 × 10 –5 = 8.9 × 10 –5 M .

      pOH = 4.05, pH = 9.95

      Because the second hydrolysis step did not produce a significant amount of OH (aq), and K b1 is even smaller than K b2 , we need not consider the third step. The pH of a saturated solution of Mg 3 (AsO 4 ) 2 is 9.95.

    17.108 Zn(OH) 2 (s) Zn 2 + (aq) + 2 OH (aq) K sp = 3.0 × 10 16 Zn 2 + (aq) + 4 OH (aq) Zn(OH) 4 (aq) K f = 4.6 × 10 17 Zn(OH) 2 (s) + 2 OH (aq) Zn(OH) 4 (aq) K = K sp × K f = 138 = 1.4 × 10 2

    K = 138 = 1.4 × 10 2 = [ Zn(OH) 4 2 ] [ OH ] 2

    If 0.015 mol Zn(OH) 2 dissolves, 0.015 mol Zn(OH) 4 2– should be present at equilibrium.

    [ OH ] 2 = ( 0.015 ) 138 ; [OH ] = 1.043 × 10 2 M ; [ OH ] 1.0 × 10 2 M or pH 12.02

    17.109

    1. Cd(OH) 2 (s)  ⇌  Cd 2+ (aq) + 2 OH (aq); K sp = 2.5 × 10 –14 = [Cd 2+ ][ OH ] 2 .

      [Cd 2+ ] = s; [OH ] = 2s; K sp = 2.5 × 10 –14 = 4s 3 .  s = 1.8 × 10 –5 M .

    2. Cd(OH) 2 (s) Cd 2 + (aq) + 2 OH (aq) K sp = 2.0 × 10 14 Cd 2 + (aq) + 4 Br (aq) CdBr 4 (aq) K f = 5 × 10 3 Cd(OH) 2 (s) + 4 Br (aq) CdBr 4 (aq) + 2 OH ( aq ) K

      K = K sp × K f = (2.5 × 10 –14 )( 5 × 10 3 ) = 1.25 × 10 –10 = 1 × 10 –10

      The desired molar solubility of Cd(OH) 2 is 1.0 × 10 –3 . Assume all soluble Cd 2+ is present as CdBr 4 2– . [CdBr 4 2– ] = 1.0 × 10 –3 ; [OH ] = 2(1.0 × 10 –3 ) = 2.0 × 10 –3 M .

      Let c = initial [NaBr] = initial [Br ]; [Br ] at equilibrium =

      c – 4(1.0 × 10 –3 ) = (c – 4.0 × 10 –3 ).

      K = 1.25 × 10 10 = [CdBr 4 2 ] [ OH ] 2 [ Br ] 4 = ( 1.0 × 10 3 ) ( 2.0 × 10 3 ) 2 ( c 4.0 × 10 3 ) 4

      Assume c is large relative to 4.0 × 10 –3 .

      (1.25 × 10 –10 )c 4 = 4.0 × 10 –9 ; c = (32) 1/4 = 2.378 = 2 M . The approximation is valid.

      4.0 × 10 –3 is about 0.2% of 2 M . Check this result in the equilibrium expression.

      K = ( 1.0 × 10 3 ) ( 2.0 × 10 3 ) 2 ( 2.378 4.0 × 10 3 ) 4 = 1.26 × 10 10 . Our calculations are consistent .

    Integrative Exercises

    17.110

    1. Complete ionic (CHO 2 = HCOO )

      H + (aq) + Cl (aq) + Na + (aq) + HCOO (aq) → HCOOH(aq) + Na + (aq) + Cl (aq)

      Na + and Cl are spectator ions.

      Net ionic: H + (aq) + HCOO (aq)  ⇌  HCOOH(aq)

    2. The net ionic equation in part (a) is the reverse of the dissociation of HCOOH.

      K = 1 K a = 1 1.8 × 10 4 = 5.55 × 10 3 = 5.6 × 10 3


    1. For Na + and Cl , this is just a dilution problem.

      M 1 V 1 = M 2 V 2 ; V 2 is 50.0 mL + 50.0 mL = 100.0 mL

      Cl : 0 .15 M × 50.0 mL 100.0 mL = 0.075 M ; Na + : 0.15 M × 50.0 mL 100.0 mL = 0.075 M

      H + and HCOO react to form HCOOH. Because K >> 1, the reaction essentially goes to completion.

      0.15 M × 0 .0500 mL = 7.5 × 10 3 mol H + 0.15 M × 0.0500 mL = 7.5 × 10 3 mol HCOO = 7.5 × 10 3 mol HCOOH

      Solve the weak acid problem to determine [H + ], [HCOO ], and [HCOOH] at equilibrium.

      K a = [ H + ] [ HCOO ] [ HCOOH ] ; [ H + ] = [ HCOO ] = x M ; [HCOOH] = ( 7.5 × 10 3 x) mol 0.100 L = ( 0.075 x) M

      1.8 × 10 4 = x 2 ( 0.075 x) x 2 0.075 ; x = 3.7 × 10 3 M H + and  HCOO

      [HCOOH] = (0.075 – 0.0037) = 0.071 M

      [ H + ] [ HCOOH ] × 100 = 3.7 × 10 3 0.075 × 100 = 4.9 % dissociation

      In summary:

      [Na + ] = [Cl ] = 0.075 M , [HCOOH] = 0.071 M , [H + ] = [HCOO ] = 0.0037 M

    17.111

    1. For a monoprotic acid (one H + per mole of acid), at the equivalence point

      moles OH added = moles H + originally present

      M B × V B = g acid/molar mass

      MM = g acid M B × V B = 0.1044 g 0.0500 M × 0.02210 L = 94.48 = 94.5 g/mol

    2. 11.05 mL is exactly half-way to the equivalence point (22.10 mL). When half of the unknown acid is neutralized, [HA] = [A ], [H + ] = K a and pH = pK a .

      K a = 10 –4.89 = 1.3 × 10 –5

    3. From Appendix D, Table D.1, acids with K a values close to 1.3 × 10 –5 are
      Name K a Formula Molar Mass
      propionic 1.3 × 10 –5 C 2 H 5 COOH 74.1
      butanoic 1.5 × 10 –5 C 3 H 7 COOH 88.1
      acetic 1.8 × 10 –5 CH 3 COOH 60.1
      hydroazoic 1.9 × 10 –5 HN 3 43.0

      Of these, butanoic has the closest match for K a and molar mass, but the agreement is not good.


    17.112 n = PV RT = 735 torr × 1 atm 760 torr × 7.5 L 295 K × mol-K 0 .08206 L-atm = 0.300 = 0.30 mol NH 3

    0.40 M × 0.50 L = 0.20 mol HCl

    HCl(aq)    + NH 3 (g)     → NH 4 + (aq)   + Cl (aq)
    before 0.20 mol 0.30 mol
    after 0 0.10 mol 0.20 mol 0.20 mol

    The solution will be a buffer because of the substantial concentrations of NH 3 and NH 4 + present. Use K a for NH 4 + to describe the equilibrium.

    NH 4 + (aq)  ⇌  NH 3 (aq)  +  H + (aq)

    equil. 0.20 – x 0.10 + x x

    K a = 1.0 × 10 14 1.8 × 10 5 = 5.56 × 10 10 = 5.6 × 10 10 ; K a = [ NH 3 ] [ H + ] [ NH 4 + ] ; [H + ] = K a [ NH 4 + ] [ NH 3 ]

    Because this expression contains a ratio of concentrations, volume will cancel and we can substitute moles directly. Assume x is small compared to 0.10 and 0.20.

    [ H + ] = 5.56 × 10 10 (0 .20) ( 0.10 ) = 1.111 × 10 9 = 1.1 × 10 9 M , pH = 8.95

    17.113 Calculate the initial M of aspirin in the stomach and solve the equilibrium problem to find equilibrium concentrations of C 8 H 7 O 2 COOH and C 8 H 7 O 2 COO . At pH = 2,

    [H + ] = 1 × 10 –2 .

    325 mg tablet × 2 tablets × 1 g 1000 mg × 1 mol C 8 H 7 O 2 COOH 180 .2 g C 8 H 7 O 2 COOH × 1 1 L = 3.61 × 10 3 = 4 × 10 3 M

    C 8 H 7 O 2 COOH(aq) C 8 H 7 O 2 COO (aq) + H + (aq)
    initial 3.61 × 10 –3 M 0 1 × 10 –2 M
    equil. (3.61 × 10 –3 – x) M x M (1 × 10 –2 + x) M

    K a = 3 × 10 5 = [ H + ] [ C 8 H 7 O 2 COO ] [ C 8 H 7 O 2 COOH ] = ( 0.01 + x) (x) ( 3.61 × 10 3 x) 0.01 x 3 .61 × 10 3

    x = [C 8 H 7 O 2 COO ] = 1.08 × 10 –5 = 1 × 10 –5 M

    % ionization = 1.08 × 10 5 M C 8 H 7 O 2 COO 3.61 × 10 3 M C 8 H 7 O 2 COOH × 100 = 0.3 %

    (% ionization is small, so the approximation was valid.)

    % aspirin molecules = 100.0% – 0.3% = 99.7% molecules

    17.114   According to Equation 13.4, S g = kP g

    S CO 2 = 3.1 × 10 2 mol L-atm × 1.10 atm = 0.0341 = 0.034 mol L = 0.034 M CO 2

    CO 2 (g) + H 2 O(l) → H 2 CO 3 (aq); 0.0341 M CO 2 = 0.0341 M H 2 CO 3


    Consider the stepwise dissociation of H 2 CO 3 (aq).

    H 2 CO 3 (aq) H + (aq) + HCO 3 (aq)
    initial 0.0341 M 0 0
    equil. (0.0341 – x) M x x

    K a1 = [ H + ] [ HCO 3 ] [ H 2 CO 3 ] = x 2 ( 0.0341 x) x 2 0.0341 4.3 × 10 7

    x 2 = 1.47 × 10 –8 ; x = 1.2 × 10 –4 M H + ; pH = 3.92

    K a2 = 5.6 × 10 –11 ; assume the second ionization does not contribute significantly to [H + ].

    17.115   Ca(OH) 2 (aq) + 2 HCl(aq) → CaCl 2 (aq) + 2 H 2 O

    mmol HCl = M × mL = 0.0983 M × 11.23 mL = 1.1039 = 1.10 mmol HCl

    mmol Ca(OH) 2 = mmol HCl/2 = 1.1039/2 = 0.55195 = 0.552 mmol Ca(OH) 2

    [ Ca 2 + ] = 0.55195 mmol 50.00 mL = 0.01104 = 0.0110 M

    [OH ] = 2[Ca 2+ ] = 0.02208 = 0.0221 M

    K sp = [Ca 2+ ][OH ] 2 = (0.01104)(0.02208) 2 = 5.38 × 10 –6

    The value in Appendix D.3 is 6.5 × 10 –6 , a difference of 17%. Because a change in temperature does change the value of an equilibrium constant, the solution may not have been kept at 25 o C . It is also possible that experimental errors led to the difference in K sp values.

    17.116 Π = M RT, M = Π RT = 21 torr 298 K × 1 atm 760 torr × mol-K 0.08206 L-atm = 1.13 × 10 3 = 1.1 × 10 3 M

    SrSO 4 (s)  ⇌  Sr 2+ (aq)  +   SO 4 2– (aq); K sp = [Sr 2+ ] [SO 4 2– ]

    The total particle concentration is 1.13 × 10 –3 M . Each mole of SrSO 4 that dissolves produces 2 mol of ions, so [Sr 2+ ] = [SO 4 2– ] = 1.13 × 10 –3 M /2 = 5.65 × 10 –4 = 5.7 × 10 –4 M .

    K sp = (5.65 × 10 –4 ) 2 = 3.2 × 10 –7

    17.117   For very dilute aqueous solutions, assume the solution density is 1 g/mL.

    ppb = g solute 10 9 g solution = 1 × 10 6 g solute 1 × 10 3 g solution = μ g solute L solution

    1. K sp = [Ag + ][Cl ] = 1.8 × 10 –10 ; [Ag + ] = (1.8 × 10 –10 ) 1/2 = 1.34 × 10 –5 = 1.3 × 10 –5 M

      1.34 × 10 5 mol Ag + L × 107.9 g Ag + 1 mol Ag + × 1 μ g 1 × 1 0 6 g = 1.4 × 10 3 μ g Ag + L = 1.4 × 10 3 ppb = 1.4 ppm

    2. K sp = [Ag + ][Br ] = 5.0 × 10 –13 ; [Ag + ] = (5.0 × 10 –13 ) 1/2 = 7.07 × 10 –7 = 7.1 × 10 –7 M

      7.07 × 10 7 mol Ag + L × 107.9 g Ag + 1 mol Ag + × 1 μ g 1 × 1 0 6 g = 76 ppb


    1. K sp = [Ag + ][I ] = 8.3 × 10 –17 ; [Ag + ] = (8.3 × 10 –17 ) 1/2 = 9.11 × 10 –9 = 9.1 × 10 –9 M

      9.11 × 10 9 mol Ag + L × 107.9 g Ag + 1 mol Ag + × 1 μ g 1 × 1 0 6 g = 0.98 ppb

      AgBr(s) would maintain [Ag + ] in the correct range.

    17.118   To determine precipitation conditions, we must know K sp for CaF 2 (s) and calculate Q under the specified conditions. K sp = 3.9 × 10 –11 = [Ca 2+ ][F ] 2

    [Ca 2+ ] and [F ]. The term 1 ppb means 1 part per billion or 1 g solute per billion g solution. Assume that the density of this very dilute solution is the density of water.

    1 ppb = 1 g solute 1 × 10 9 g solution × 1 g solution 1 mL solution × 1 × 10 3 mL 1 L = 1 × 10 6 g solute 1 L solution

    1 × 10 6 g solute 1 L solution × 1 μ g 1 × 10 6 g = 1 μ g/1 L

    8 ppb Ca 2 + × 1 μ g 1 L = 8 μ g Ca 2 + 1 L = 8 × 10 6 g Ca 2 + 1 L × 1 mol Ca 2 + 40 g = 2 × 10 7 M Ca 2 +

    1 ppb F × 1 μ g 1 L = 1 μ g F 1 L = 1 × 10 6 g F 1 L × 1 mol F 19 .0 g = 5 × 10 8 M F

    Q = [Ca 2+ ][F ] 2 = (2 × 10 –7 )(5 × 10 –8 ) 2 = 5 × 10 –22

    5 × 10 –22 < 3.9 × 10 –11 , Q < K sp , no CaF 2 will precipitate

    17.119

    1. CH 3 CH(OH)COOH(aq) + HCO 3 (aq)  →  H 2 CO 3 (aq) + CH 3 CH(OH)COO (aq)

      H 2 CO 3 (aq)  ⇌  H 2 O(l) + CO 2 (g)

    2. 2.16 g NaHCO 3 mL × 236 .6 mL 48 tsp × 1 mol NaHCO 3 84 .01 g NaHCO 3 × 0.5 tsp = 0.06337 = 0.0634 mol NaHCO 3

      mol NaHCO 3 = mol HCO 3 ;

      at neutralization, mol HCO 3 = mol CH 3 CH(OH)COOH

      0.06337 mol CH 3 CH(OH)COOH 1 cup milk × 1 cup 236 .6 mL × 1000 mL L = 0.2678 = 0.268 M CH 3 CH(OH)COOH

    3. 5/9( o F – 32) = o C; 5/9(350 – 32) = 176.67 = 177 o C; 177 o C + 273 = 450 K

      mol CO 2 = mol HCO 3

      V = nRT P = 0.06337 mol × 450 K 1 atm × 0.08206 L-atm mol-K = 2.34 L CO 2

    17.120   Statements (b) and (d) follow from this observation.

    When HF behaves like this, it is acting like a base, rather than an acid. This agrees with statements (b) and (d) and renders (a) incorrect. As for (c), if HF were thermodynamically unstable, it would decompose rather than gain H + .


     

    18 Chemistry of the Environment

    Visualizing Concepts

    18.1 Analyze. Given that one mole of an ideal gas at 1 atm and 298 K occupies 22.4 L, is the volume of one mole of ideal gas in the middle of the stratosphere greater or less than 22.4 L?

    Plan .  Consider the relationship between pressure, temperature, and volume of an ideal gas. Use Figure 18.1 to estimate the pressure and temperature in the middle of the stratosphere, and compare the two sets of temperature and pressure.

    Solve .  According to the ideal-gas law, PV = nRT, so V = nRT/P. Because n and R are constant for this exercise, V is proportional to T/P.

    1. Greater. The stratosphere ranges from 10 to 50 km, so the middle is at approximately 30 km. At this altitude, T ≈ 230 K, P ≈ 40 torr (from Figure 18.1). Because we are comparing T/P ratios, either atm or torr can be used as pressure units; we will use torr.

      At sea level: T/P = 298 K/760 torr = 0.39

      At 30 km: T/P = 230 K/40 torr = 5.75

      The proportionality constant (T/P) is much greater at 30 km than sea level, so the volume of 1 mol of an ideal gas is greater at this altitude. The decrease in temperature at 30 km is more than offset by the substantial decrease in pressure.

    2. No. Volume is proportional to T/P, not simply T. The relative volumes of one mole of an ideal gas at 50 km and 85 km depend on the temperature and pressure at the two altitudes. From Figure 18.1,

      50 km: T ≈ 270 K, P ≈ 20 torr, T/P = 270 K/20 torr = 13.5

      85 km: T ≈ 190 K, P < 0.01 torr, T/P = 190 K/0.01 torr = 19,000

      Again, the slightly lower temperature at 85 km is more than offset by a much lower pressure. One mole of an ideal gas will occupy a much larger volume at

      85 km than 50 km.

    3. The thermosphere, stratopause, and low-altitude troposphere. Gases behave most ideally at high temperature and low pressure. Pressure is minimum and temperature is high in the thermosphere. The stratopause (the boundary between the stratosphere and mesosphere) and the troposphere at low altitude are other regions with temperature maxima and relatively low pressures.

    18.2   Molecules in the upper atmosphere tend to have multiple bonds because they have sufficiently high bond dissociation enthalpies (Table 8.3) to survive the incoming high- energy radiation from the Sun. According to Table 8.3, for the same two bonded atoms, multiple bonds have higher bond dissociation enthalpies than single bonds. Molecules with single bonds are likely to undergo photodissociation in the presence of the high- energy, short-wavelength solar radiation present in the upper atmosphere.


    18.3

    1. A = troposphere, 0–10 km; B = stratosphere, 12–50 km; C = mesosphere, 50–85 km.
    2. Ozone is a pollutant in the troposphere and filters UV radiation in the stratosphere.
    3. Infrared radiation from Earth is most strongly reflected back in the troposphere.
    4. Assuming the “boundary” between the stratosphere and mesosphere is at 50 km, only region C in the diagram is involved in an aurora borealis.
    5. The concentration of water vapor is greatest near Earth’s surface in region A and decreases with altitude. Water’s single bonds are susceptible to photodissociation in regions B and C, so its concentration is likely to be very low in these regions. The relative concentration of CO 2 , with strong double bonds, increases in regions B and C, because it is less susceptible to photodissociation.

    18.4 Analyze .  Given granite, marble, bronze, and other solid materials, what observations and measurements indicate whether the material is appropriate for an outdoor sculpture? If the material changes (erodes) over time, what chemical processes are responsible?

    Plan. An appropriate material resists chemical and physical changes when exposed to environmental conditions. An inappropriate material undergoes chemical reactions with substances in the troposphere, degrading the structural strength of the material and the sculpture. Solve .

    1. The appearance and mass of the material upon environmental exposure are indicators of both chemical and physical changes. If the appearance and mass of the material are unchanged after a period of time, the material is well suited for the sculpture because it is inert to chemical and physical changes. Changes in the color or texture of the material’s surface indicate that a chemical reaction has occurred, because a different substance with different properties has formed. A decrease in mass indicates that some of the material has been lost, by either chemical reaction or physical change. An increase in mass indicates corrosion. If the mass of the material is unchanged, it is probably inert to chemical and physical environmental changes and suitable for sculpture.
    2. The two main chemical processes that lead to erosion are reaction with acid rain and corrosion or air oxidation, which is encouraged by acid conditions (see Section 20.8).

      Acid rain is primarily H 2 SO 3 and/or H 2 SO 4 , which reacts directly with carbonate minerals such as marble and limestone. Acidic conditions created by acid rain encourage corrosion of metals such as iron, steel, and bronze. Corrosion produces metal oxides, which may or may not cling to the surface of the material. If the oxides are washed away, the material will lose mass after corrosion. Physical erosion due to the effects of wind and rain on soft materials such as sandstone also causes mass to decrease.


    18.5   The Sun.

    18.6 Analyze/Plan .  Given salinity, calculate salt concentration in ppm. A salinity of 35 denotes that there are 35 g of dry salt per kg of seawater.

    ppm = g salt/1 × 10 6 g seawater. Solve .

    35 g salt 1 kg seawater × 1 kg seawater 1000 g seawater × 1 × 10 6 = 3.5 × 10 4 ppm salt

    35,000 ppm total salt – 500 ppm salt remain = 34,500 ppm salt must be removed

    34 , 500 ppm salt removed 35 , 000 ppm salt total × 100 = 98 .57 = 99% salt removed

    Of the total dissolved salts in seawater, 99% must be removed in order for the water to be considered freshwater.

    18.7   CO 2 (g) dissolves in seawater to form H 2 CO 3 (aq). The basic pH of the ocean encourages ionization of H 2 CO 3 (aq) to form HCO 3 (aq) and CO 3 2– (aq). Under the correct conditions, carbon is removed from the ocean as CaCO 3 (s) (sea shells, coral, chalk cliffs). As carbon is removed, more CO 2 (g) dissolves to maintain the balance of complex and interacting acid–base and precipitation equilibria.

    18.8   The first stage at the Carlsbad plant is a physical filtration using anthracite coal (carbon), sand, and gravel as the porous medium. This filtration will remove particles larger than the smallest filter pore size from the remaining solution. It will not remove dissolved salts; ions are much smaller than the pore size of any part of the filter.

    18.9 Plan. Follow the yellow arrows on the diagram to find potential routes for environmental contamination at a fracking well site. Solve .

    Above ground, the two main avenues for escape of contaminants are leakage and evaporation. Methane, hydrogen sulfide, and other petroleum compounds that are gases at atmospheric conditions can leak from the well head. Heavier petroleum compounds (those containing from three to six carbon atoms) are also released by fracking, mix with the resulting aqueous solution, and end up in the wastewater ponds. Volatile organic compounds produced by fracking and from the fracking liquid can evaporate from the ponds. Wastewwater ponds, if unlined, can seep into nearby water sources or over flow due to rainfall.

    Below ground, petroleum gases and fracking liquid can migrate into groundwater, both deep and shallow aquifers. (Fracking mobilizes petroleum and gases, which enables these underground “leaks.”)

    The extent of underground leaking depends on the integrity of the well plumbing and casings, as well as the real permeability of the “impermeable” layers. Although there are several avenues for contamination, they can be minimized by careful attention to geology, engineering, and oversight.

    18.10  Some of the missing CO 2 is absorbed by “land plants” (vegetation other than trees). These plants are directly or indirectly used as food by land animals, or they complete an uninterrupted life cycle. Either way, natural decomposition occurs and the remains are incorporated into the soil. (Fossil fuels were formed millions of years ago by natural decomposition of buried dead organisms.) Soil is the largest land-based carbon reservoir. The amount of carbon-storing capacity of soil is affected by erosion, soil fertility, and other complex factors. For more details, search the Internet for ”carbon budget.”


    Earth’s Atmosphere (Section 18.1)

    18.11

    1. The temperature profile of the atmosphere (Figure 18.1) is the basis of its division into regions. The center of each peak or trough in the temperature profile corresponds to a new region.
    2. Troposphere, 0–12 km; stratosphere, 12–50 km; mesosphere, 50–85 km; thermosphere, 85–110 km.

    18.12

    1. Boundaries between regions of the atmosphere are at maxima and minima (peaks and valleys) in the atmospheric temperature profile. For example, in the troposphere, temperature decreases with altitude, whereas in the stratosphere, it increases with altitude. The temperature minimum is the tropopause boundary.
    2. From Figure 18.1, atmospheric pressure in the troposphere ranges from 760 torr to 200 torr, whereas pressure in the stratosphere ranges from 200 torr to 20 torr. Gas density (g/L) is directly proportional to pressure. The much lower density of the stratosphere means it has the smaller mass, despite having a larger volume than the troposphere.

    18.13 Analyze/Plan .  Given O 3 concentration in ppm, calculate partial pressure. Use the definition of ppm to get mol fraction O 3 . For gases, mole fraction = pressure fraction. Use the ideal-gas law to find mol O 3 /L air and Avogadro’s number to get molecules.

    P O 3 = X O 3 × P atm ; 0.441 ppm O 3 = 0.441 mol O 3 1 × 10 6 mol air = 4.41 × 10 7 = X O 3 Solve .

    1. P O 3 = X O 3 × P atm = 4.41 × 10 7 (0 .67 atm) = 2 .955 × 10 7 = 3.0 × 10 7 atm
    2. n = PV RT = 2.955 × 10 7 atm × 1 .0 L 298 K × mol-K 0.08206 L-atm = 1.208 × 10 8 = 1.2 × 10 8 mol O 3

      1.208 × 10 8 mol O 3 × 6 .022 × 10 23 molecules mol = 7.277 × 10 15 = 7.3 × 10 15 O 3 molecules

    18.14 P Ar = X Ar × P atm ; P Ar = 0.00934 ( 1.05 bar) = 0.009807 = 9.81 × 10 3 bar

    P Ar = 0 .009807 bar × 10 5 Pa bar × 760 torr 101,325 Pa = 7.3559 = 7.36 torr

    P CO 2 = X CO 2 × P atm ; P CO 2 = 0.000400 ( 1.05 bar) = 0.0004200 = 4.20 × 10 4 bar

    P CO 2 = 0 .0004200 bar × 10 5 Pa bar × 760 torr 101,325 Pa = 0.3150 = 0.315 torr

    18.15 Analyze/Plan .  Given CO concentration in ppm, calculate number of CO molecules in 1.0 L air at given conditions. ppm CO X O 3 atm CO mol CO molecules CO . Use the ideal-gas law to change atm CO to mol CO and Avogadro’s number to get molecules. Solve .


    3.5 ppm CO = 3.5 mol CO 1 × 10 6 mol air = 3.5 × 10 6 = X CO

    P CO = X CO × P atm = 3.5 × 10 6 × 759 torr × 1 atm 760 torr = 3.495 × 10 6 = 3.5 × 10 6 atm

    n CO = P CO V RT = 3.495 × 10 6 atm × 1 .0 L 295 K × mol-K 0.08206 L-atm = 1.444 × 10 7 = 1.4 × 10 7 mol CO

    1.444 × 10 7 mol CO × 6 .022 × 10 23 molecules mol = 8.695 × 10 16 = 8.7 × 10 16 CO molecules

    18.16

    1. ppm Ne = mol Ne/1 × 10 6 mol air; X Ne = 1.818 × 10 –5 mol Ne/mol air

      1.818 × 10 5 mol Ne 1 mol air = x mol Ne 1 × 10 6 mol air ; x = 18.18 ppm Ne

    2. P Ne = X Ne × P atm = 1.818 × 10 5 × 730 torr × 1 atm 760 torr = 1.7462 × 10 5 = 1.75 × 10 5 atm

      T = 296 K

      n Ne V = P Ne RT = 1.7462 × 10 5 atm 296 K × mol-K 0 .08206 L-atm = 7.1892 × 10 7 = 7.19 × 10 7 mol/L

      7.1892 × 10 7 mol Ne L × 6.022 × 10 23 atoms mol = 4.3293 × 10 17 = 4.33 × 10 17 Ne atoms/L

    18.17 Analyze/Plan .  Given bond dissociation energy in kJ/mol, calculate the wavelength of a single photon that will rupture a C–Br bond. kJ/mol → J/molecule. λ = hc/E.

    (λ = hc/E describes the energy/wavelength relationship of a single photon.) Solve .

    1. 276 × 10 3 J 1 mol × 1 mol 6 .022 × 10 23 molecules = 4.583 × 10 19 = 4.58 × 10 19 J/molecule

      λ = hc E = ( 6.626 × 10 34 J-sec) (3 .00 × 10 8 m/sec) 4.583 × 10 19 J = 4.337 × 10 7 m = 434 nm

    2. This 434 nm wavelength is visible electromagnetic radiation.

    18.18 339 × 10 3 J 1 mol × 1 mol 6 .022 × 10 23 molecules = 5.6294 × 10 19 = 5.63 × 10 19 J/molecule

    λ = hc E = ( 6.626 × 10 34 J-sec) (3 .00 × 10 8 m/sec) 5.6294 × 10 19 J = 3.53 × 10 7 m = 353 nm

    293 × 10 3 J 1 mol × 1 mol 6 .022 × 10 23 molecules = 4.8655 × 10 19 = 4.87 × 10 19 J/molecule

    λ = ( 6.626 × 10 34 J-sec) (3 .00 × 10 8 m/sec) 4.8655 × 10 19 J = 4.09 × 10 7 m = 409 nm

    Photons of wavelengths longer than 409 nm cannot cause rupture of the C–Cl bond in either CF 3 Cl or CCl 4 . Photons with wavelengths between 409 and 353 nm can cause C–Cl bond rupture in CCl 4 , but not in CF 3 Cl.


    18.19

    1. Photodissociation is cleavage of the O O bond such that two neutral O atoms are produced: O 2 (g) → 2O(g).

      Photoionization is absorption of a photon with sufficient energy to eject an electron from an O 2 molecule: O 2 (g) + hν → O 2 + + e .

    2. Photoionization of O 2 requires 1205 kJ/mol. Photodissociation requires only 495 kJ/mol. At lower elevations, solar radiation with wavelengths corresponding to 1205 kJ/mol or shorter has already been absorbed, whereas the longer wavelength radiation has passed through relatively well. Below 90 km, the increased concentration of O 2 and the availability of longer wavelength radiation cause the photodissociation process to dominate.

    18.20   Photodissociation of N 2 is relatively unimportant compared to photodissociation of O 2 for two reasons. The bond dissociation energy of N 2 , 941 kJ/mol, is much higher than that of O 2 , 495 kJ/mol. Photons with a wavelength short enough to photodissociate N 2 are not as abundant as the ultraviolet photons that lead to photodissociation of O 2 . Also, N 2 does not absorb these photons as readily as O 2 so even if a short-wavelength photon is available, it may not be absorbed by an N 2 molecule.

    18.21

    1. A wavelength of 145 nm is in the ultraviolet portion of the electromagnetic spectrum. (See Figure 6.4.)
    2. Analyze/Plan . E = hc/λ. 145 nm = 1.45 x 10 –7 m. Change J/photon to kJ/mol. Compare to the bond energy of O 2 , 495 kJ/mol. Solve .

      E = hc λ = ( 6.626 × 10 34 J-sec) (3 .00 × 10 8 m/sec) 1.45 × 10 7 m = 1.37 × 10 18 J/photon

      1.37 × 10 18 J photon × 1 kJ 1000 J × 6 .022 × 10 23 photons 1 mol = 825.55 = 826 kJ/mol

      The 145-nm photon has more than enough energy to photodissociate O 2 .

      According to Table 18.3, the photoionization energy of O 2 is 1205 kJ/mol. The 145-nm photon does not have enough energy to photoionize O 2 .

    18.22

    1. Energy is inversely related to wavelengths, E = hc/λ. Therefore, photons from UV-C, with the shortest wavelengths, have the highest energy and are most harmful to living tissue.
    2. Atmospheric N 2 , O 2 , and atomic oxygen absorb wavelengths shorter than 240 nm, a portion of UV-C. In the absence of ozone, wavelengths in UV-A and UV-B are not absorbed.
    3. No, even when appropriate concentrations of ozone are present in the stratosphere, not all UV light is absorbed before reaching Earth’s surface. Nearly all of UV-C is absorbed by N 2 , O 2 , and atomic oxygen, and the remainder is absorbed by stratospheric ozone. Some, but not all of UV-B is absorbed by ozone. The UV-A region is not filtered by ozone, N 2 , O 2 , or atomic oxygen.

    Human Activities and Earth’s Atmosphere (Section 18.2)

    18.23   The oxidation state of oxygen in O 3 , O 2 , and O is zero (0). Reactions in which oxygen changes only from one of these species to another do not involve changes in oxidation state. Examples: O(g) + O(g) → O 2 (g); 2 O 3 (g) → 3 O 2 (g).

    Ozone depletion reactions that involve a halogen oxide such as ClO do involve a change in oxidation state for oxygen. In ClO, the oxidation state of oxygen is either +1 or +2, but it is not zero. A reaction involving ClO and one of the oxygen species with a zero oxidation state does involve a change in the oxidation state of oxygen atoms.

    18.24   Reactions (a) and (b).

    It is those star* reactions, Equations 18.3–18.5. For example, the reaction of O 2 (g) + O(g) is exothermic and produces a high-energy O 3 *(g) molecule with 105 kJ of energy to disperse. This energy is transferred through collisions, primarily to N 2 (g) and O 2 (g) molecules. The overall kinetic energy (translational, vibrational, and rotational energy) of these molecules (M*) increases and the temperature of the stratosphere is kept relatively high. (Recall that the temperature of a gas is directly proportional to its average kinetic energy.)

    18.25

    1. A chlorofluorocarbon is a compound that contains chlorine, fluorine, and carbon. A hydrofluorocarbon contains hydrogen, fluorine, and carbon; it contains hydrogen in place of chlorine.
    2. CFCs are harmful because they undergo photodissociation to produce Cl atoms that catalyze the destructions of ozone. HFCs are potentially less harmful to the ozone layer because they contain no C–Cl bonds. Their relatively stronger C–F bonds require more energy to undergo photodissociation, energy that is unlikely to be available in the stratosphere. (HFCs are still a powerful greenhouse gas. Montreal Protocal members have recently agreed to limit HFCs as well as CFCs.)

    18.26   32 e , 16 e pr

    image

    CFC–11, CFCl 3 , contains C–Cl bonds that can be cleaved by UV light in the stratosphere to produce Cl atoms. It is chlorine in atomic form that catalyzes the destruction of stratospheric ozone. CFC–11 is chemically inert and resists decomposition in the troposphere, so that it eventually reaches the stratosphere in molecular form.

    18.27

    1. Analyze/Plan .  Given bond enthalpies in kJ/mol, calculate the maximum wavelength of a single photon that will rupture a C–F and a C–Cl bond, respectively. kJ/mol → J/molecule. λ = hc/E. (λ = hc/E describes the energy/wavelength relationship of a single photon.) Solve .

      485 × 10 3 J 1 mol × 1 mol 6 .022 × 10 23 C F bonds = 8.054 × 10 19 = 8.05 × 10 19 J/C F bond

      λ = hc E = ( 6.626 × 10 34 J-sec) (3 .00 × 10 8 m/sec) 8.054 × 10 19 J = 2.47 × 10 7 m = 247 nm


      328 × 10 3 J 1 mol × 1 mol 6 .022 × 10 23 C Cl bonds = 5.447 × 10 19 = 5.45 × 10 19 J/C Cl bond

      λ = hc E = ( 6.626 × 10 34 J-sec) (3 .00 × 10 8 m/sec) 5.447 × 10 19 J = 3.65 × 10 7 m = 365 nm

    1. The maximum wavelength that can dissociate a C–F bond is 247 nm; shorter wavelengths are also effective. Since most wavelengths shorter than 240 nm are absorbed in the upper atmosphere, few effective photons will reach the lower atmosphere. We don’t expect the photodissociation of C–F bonds to be significant in the lower atmosphere. (The dissociation of C–Cl bonds will be significant.)

    18.28

    1. The products of the reaction are ClO(g) and O 2 (g).
    2. The average bond dissociation enthalpy of C–Cl is 328 kJ/mol, while that of C–Br is 276 kJ/mol (Table 8.3). Since a C–Cl bond has a higher dissociation enthalpy, the maximum wavelength required to dissociate it will be shorter and more energetic than the wavelength required to dissociate a C–Br bond. A photon capable of dissociating a C–Cl bond will also have energy sufficient to dissociate a C–Br bond.
    3. Yes, we expect the substance CFBr 3 to accelerate depletion of the ozone layer. Photons capable of dissociating C–Cl bonds are also capable of dissociating C–Br bonds. And, longer effective wavelengths are also available in the lower atmosphere. Bromine atoms will react with ozone to form BrO(g) and O 2 (g). Ozone will be depleted.

    18.29 Analyze/Plan . Write and balance equations for the reaction of NO(g) and NO 2 (g) with water. First assume a simple acid–base reaction with a single product, HNO 3 (aq); these equations can’t be balanced. Reaction of NO(g) and NO 2 (g) to produce HNO 3 (aq) [or HNO 2 (aq)] are redox reactions. An Internet search reveals that the reactions are as shown below. Solve .

    2 NO 2 (g) + H 2 O(l)  ⇌  HNO 2 (aq) + HNO 3 (aq)

    2 NO(g) + O 2 (aq) + H 2 O(l)  ⇌  HNO 2 (aq) + HNO 3 (aq) or

    4 NO 2 (g) + O 2 (aq) + 2 H 2 O(l)  ⇌  4 HNO 3 (aq)

    4 NO(g) + 3 O 2 (aq) + 2 H 2 O(l)  ⇌  4 HNO 3 (aq)

    18.30   Rainwater is naturally acidic because of the presence of CO 2 (g) in the atmosphere. All oxides of nonmetals produce acidic solutions when dissolved in water. Even in the absence of polluting gases such as SO 2 , SO 3 , NO, and NO 2 , CO 2 causes rainwater to be acidic. The important equilibria are:

    CO 2 (g) + H 2 O(l)  ⇌  H 2 CO 3 (aq)  ⇌  H + (aq) + HCO 3 (aq)

    18.31

    1. Acid rain is primarily H 2 SO 4 (aq).

      H 2 SO 4 (aq) + CaCO 3 (s) → CaSO 4 (s) + H 2 O(l) + CO 2 (g)

    2. The CaSO 4 (s) would be much less reactive with acidic solution, because it would require a strongly acidic solution to shift the relevant equilibrium to the right.

      CaSO 4 (s) + 2 H + (aq)  ⇌  Ca 2+ (aq) + 2 HSO 4 (aq)

      Note, however, that CaSO 4 (s) is brittle and easily dislodged; it provides none of the structural strength of limestone.


    18.32

    1. Fe(s) + O 2 (g) + 4 H 3 O + (aq) → Fe 2+ (aq) + 6 H 2 O(l)
    2. No. Silver is a ”noble” metal. It is relatively resistant to oxidation, and much more resistant than iron. In Table 4.5, The Activity Series of Metals in Aqueous Solution, Ag is much, much lower than Fe and it is below hydrogen, whereas Fe is above hydrogen. This means that Fe is susceptible to oxidation by acid, but Ag is not.

    18.33 Analyze/Plan .  Given wavelength of a photon, place it in the electromagnetic spectrum, calculate its energy in kJ/mol, and compare it to an average bond dissociation energy. Use Figure 6.4; E(J/photon) = hc/λ. J/photon → kJ/mol. Solve .

    1. Ultraviolet (Figure 6.4).
    2. E photon = hc/ λ = 6.626 × 10 34 J-s × 3 .00 × 10 8 m/s 335 × 10 9 m = 5.934 × 10 19 = 5.93 × 10 19 J/photon

      5.934 × 10 19 J 1 photon × 6.022 × 10 23 photons 1 mol × 1 kJ 1000 J = 357 kJ/mol

    3. The average C–H bond energy from Table 8.3 is 413 kJ/mol. The energy calculated in part (b), 357 kJ/mol, is the energy required to break 1 mol of C–H bonds in formaldehyde, CH 2 O. The C–H bond energy in CH 2 O must be less than the “average” C–H bond energy.
    4. image

    18.34

    1. Visible (Figure 6.4).
    2. E photon = hc/ λ = 6.626 × 10 34 J-s × 3 .00 × 10 8 m/s 420 × 10 9 m = 4.733 × 10 19 = 4.73 × 10 19 J/photon

      4.733 × 10 19 J 1 photon × 6.022 × 10 23 photons 1 mol × 1 kJ 1000 J = 285 kJ/mol

    3. image

    18.35

    1. Four sources transfer energy to the atmosphere, surface radiation, evapotranspiration, incoming solar radiation, and convective heating. Surface radiation makes the largest contribution. The total energy absorbed by the atmosphere is [350 + 78 + 67 + 24] = 519 W/m 2 .
    2. Of the 519 W/m 2 transferred to the atmosphere, 324 W/m 2 are radiated back to the surface. The percentage is (324/519) × 100 = 62.4%.

    18.36   There is a much larger temperature variation from day to night on Mars than on Earth. This indicates that the greenhouse gases in Mars’ atmosphere do not absorb infrared radiation as effectively as those on Earth. Even though atmospheric pressure at the surface of Mars is only 6 × 10 –3 atm, 96% of that is due to CO 2 . The partial pressure of CO 2 in Mars’ atmosphere is then 0.0058 atm (0.006 atm to 1 sig fig). The partial pressure of CO 2 in Earth’s atmosphere is only 0.0004 atm, because the amount of CO 2 in Earth’s atmosphere is so much smaller. If CO 2 were the main greenhouse gas contributing to daily temperature variation, Mars, with more CO 2 , would have a smaller variation than Earth, not larger. The presence of potent greenhouse gas H 2 O(g) in Earth’s atmosphere is the main reason for our small overnight temperature variation. Thus, the composition of the Mars atmosphere, the absence of water vapor, plays the largest role in the wide daily temperature variation.

    Earth’s Water (Section 18.3)

    18.37 Analyze/Plan .  Given salinity and density, calculate molarity. A salinity of 5.6 denotes that there are 5.6 g of dry salt per kg of water. 1.03 g/mL = 1.03 kg/L Solve .

    5.6 g NaCl 1 kg soln × 1.03 kg soln 1 L soln × 1 mol NaCl 58.44 g NaCl × 1 mol Na + 1 mol NaCl = 0.0987 = 0.099 M Na +

    18.38   If the phosphorous is present as H 2 PO 4 , there is a 1:1 ratio between the molarity of phosphorus and molarity of phosphate. Thus, we can calculate the molarity based on the given mass of P.

    0.07 g P 1 × 10 6 g H 2 O × 1 mol P 31 g P × 1 mol PO 4 3 1 mol P × 1 × 10 3 g H 2 O 1 L H 2 O = 2.26 × 10 6 = 2 × 10 6 M PO 4 3

    18.39 Analyze/Plan . Given the power of sunlight per square meter striking Earth’s surface, the enthalpy of evaporation of water, and specific heat capacity of water, calculate the amount of energy delivered by the Sun over a 12-hour day. Use this amount of energy to calculate: (a) how many grams of water can be evaporated and (b) the temperature of a 10.0 cm by 1 square meter volume of water after 12 hours in the sunlight, assuming no evaporation. Calculate the mass of this volume of water using density at 25 o C. Solve .

    1. 168 W m 2 × 1 J/s 1 W = 168 J m 2 -s

      168 J m 2 -s × 1.00 m 2 × 12 h × 60 min 1 h × 60 s 1 min × 1 k J 1000 J = 7257.6 = 7.26 × 10 3 k J

      7257.6 kJ × 1 mol H 2 O 40.67 kJ × 18.02 g H 2 O 1 mol H 2 O = 3215.7 = 3.22 × 10 3 g H 2 O

    2. 1.00 m 2 × 10.0 cm × ( 100 ) 2 cm 2 1 m 2 × 0.99707 g 1 cm 3 = 99 , 707 = 9.97 × 10 4 gH 2 O

      7257.6 kJ × 1000 J 1 kJ × 1 g- o C 4.184 J × 1 99 , 707 g H 2 O = 17.397 = 17.4 ° C

      The final temperature is 26 o C + 17.4 o C = 43.4 o C.


    18.40

    1. 168 W m 2 × 1 J/s 1 W = 168 J m 2 -s

      168 J m 2 -s × 1.00 m 2 × 12 h × 60 min 1 h × 60 s 1 min × 1 k J 1000 J = 7257.6 = 7.26 × 10 3 k J

      7257.6 kJ × 1 mol H 2 O 6.01 kJ × 18.02 g H 2 O 1 mol H 2 O = 21 , 761 = 2.18 × 10 4 g H 2 O ( ice )

    2. 1.00 m 2 × 1.00 cm × ( 100 ) 2 cm 2 1 m 2 × 0.99987 g 1 cm 3 = 9998.7 = 1.00 × 10 4 g H 2 O ( ice at 0 ° C )

      7257.6 kJ × 1000 J 1 kJ × 1 g- o C 2.032 J × 1 9998.7 g H 2 O = 357.21 = 357.2 o C

      Assuming no phase changes, the final temperature is –5 o C + 357 o C = 352 o C. Clearly the ice melts. This agrees with the result from part (a), which shows that sunlight striking 1.00 square meter of ice for 12 hours provides enough energy to melt 2.18 × 10 4 g ice, twice the mass in the first centimeter of a square meter of ice.

    18.41 Analyze/Plan .  g Mg(OH) 2 → mol Mg(OH) 2 → mol ratio → mol CaO → g CaO. Solve .

    1000 lb Mg(OH) 2 × 453.6 g lb × 1 mol Mg(OH) 2 58.33 g Mg(OH) 2 × 1 mol CaO 1 mol Mg(OH) 2 × 56.08 g CaO 1 mol CaO = 4.361 × 10 5 g CaO

    18.42   0.05 ppb Au = 0.05 g Au/1 × 10 9 g seawater

    $ 1 , 000 , 000 × 1 troy oz Au $1300 × 31.1035 g troy o z = 2.3926 × 10 4 g = 2.39 × 10 4 g Au needed

    2.3926 × 10 4 g Au × 1 × 10 9 g seawater 0 .05 g Au × 1 mL seawater 1 .03 g seawater × 1 L 1000 mL = 4.6458 × 10 11 = 5 × 10 11 L seawater

    5 × 10 11 L seawater is needed if the process is 100% efficient; because it is only 50% efficient, twice as much seawater is needed.

    4.6458 × 10 11 × 2 = 9.2916 × 10 12 = 9 × 10 12 L seawater

    Note that the 1 sig fig in 0.05 ppb Au limits the precision of the calculation.

    18.43 Analyze/Plan . Use molar concentrations of the six major ions in seawater from Table 18.5. Calculate charge in coulombs by multiplying molarity × integer charge × Faraday’s constant (coulombs/mol). Sum the coulombic charges of the anions, the cations, and compare. Solve .

    0.55 mol Cl L seawater × 1 mol charge mol Cl × 9.64853365 C mol = 5.30669351 = 5 .3 C

    0.028 mol SO 4 2 L seawater × 2 mol charge mol SO 4 2 × 9.64853365 C mol = 0.54031788 = 0.54 C

    0.47 mol Na + L seawater × 1 mol charge mol K + × 9.64853365 C mol = 4.53481082 = 4.5 C


    0.054 mol Mg 2 + L seawater × 2 mol charge mol Mg 2 + × 9.64853365 C mol = 1.04204163 = 1.0 C

    0.010 mol Ca 2 + L seawater × 2 mol charge mol Mg 2 + × 9.64853365 C mol = 0.19297067 = 0.19 C

    0.010 mol K + L seawater × 1 mol charge mol K + × 9.64853365 C mol = 0.0964853365 = 0.096 C

    anion charge: [5.30669351 + 0.54031788] = 5.84701139 = 5.8 C

    cation charge: [4.53481082 + 1.04204164 + 0.19297067 + 0.0964853365] = 5.86630846 = 5.9 C

    The two numbers vary in the third significant figure. This is not surprising, because the molarities of the various ions are given to two significant figures.

    18.44

    1. We need to replace the 18 billion gal of water per day used for irrigation. One billion is 1 × 10 9 .

      18 × 10 9 gal d × 365 d yr × 3.7854 L gal × 1 dm 3 L × 1 m 3 ( 10 ) 3 dm 3 × 1 km 3 ( 1000 ) 3 m 3 × 1 6 × 10 5 km 2 = 4.145 × 10 5 = 4 × 10 5 km/yr

      4.145 × 10 5 km yr × 1000 m km × 100 cm m × 1 in 2.54 cm = 1.632 = 2 in/yr is needed

      However, only 2% of rainfall actually recharges to aquifer, so (1.632/0.02) = 81.59 = 80 in/year annual rainfall is required to replace water removed for irrigation. (Data limits the calculated result to 1 sig fig.)

    2. The process of dissolving accounts for the presence of arsenic in well water. If minerals in the rock feeding or holding the aquifer are somewhat soluble, ions can dissolve in the water. If arsenic, usually in the form of arsenic oxide anions, is present in the somewhat soluble minerals, it can leach into the aquifer and find its way into wells.

    Human Activities and Water Quality (Section 18.4)

    18.45 Analyze/Plan .  Given temperature and the concentration difference between the two solutions, (Δ M = 0.22 – 0.01 = 0.21 M ), calculate the minimum pressure for reverse osmosis. Use the relationship Π = M RT from Section 13.5. This is the pressure required to halt osmosis from the more dilute (0.01 M ) to the more concentrated (0.22 M ) solution. Slightly more pressure will initiate reverse osmosis. Solve .

    Π = Δ M RT = 0.21 mol L × 0.08206 L-atm mol-K × 298 K = 5.135 = 5.1 atm

    The minimum pressure required to initiate reverse osmosis is greater than 5.1 atm.

    18.46   Calculate the total ion concentration of seawater by summing the molarities given in Table 18.5. Then use Π = Δ M RT to calculate pressure.

    M total = 0.55 + 0.47 + 0.028 + 0.054 + 0.010 + 0.010 + 2.3 × 10 –3 + 8.3 × 10 –4 + 4.3 × 10 –4 + 9.1 × 10 –5 + 7.0 × 10 –5 = 1.1257 = 1.13 M


    Π = ( 1.1257 0.02 ) mol L × 0.08206 L × atm mol-K × 297 K = 26.948 = 26.9 atm

    Check .  The largest numbers in the molarity sum have 2 decimal places, so M total has 2 decimal places and 3 sig figs. Δ M also has 2 decimal places and 3 sig figs so the calculated pressure has 3 sig figs. Units are correct.

    18.47 Analyze/Plan .  Under aerobic conditions, excess oxygen is present and decomposition leads to oxidized products, the element in its maximum oxidation state combined with oxygen. Under anaerobic conditions, little or no oxygen is present so decomposition leads to reduced products, the element in its minimum oxidation state combined with hydrogen. Solve .

    1. CO 2 , HCO 3 , H 2 O, SO 4 2– , NO 3 .
    2. CH 4 (g), H 2 S(g), NH 3 (g).

    18.48

    1. Decomposition of organic matter by aerobic bacteria depletes dissolved O 2 . A low dissolved oxygen concentration indicates the presence of organic pollutants.
    2. In general, gas solubility decreases with increasing temperature. According to Figure 13.16, the solubility of O 2 (g) at 20 o C is approximately 1.4 m M , and at 30 o C is 1.2 m M . This is a 14.3% decrease in solubility over a typical atmospheric temperature range.

    Colder natural water has a greater maximum possible O 2 (g) solubility. A general increase in global average temperature accompanied by an increase in water temperature decreases water quality by decreasing the amount of dissolved oxygen.

    18.49 Analyze/Plan .  Given the balanced equation, calculate the amount of one reactant required to react exactly with a certain amount of the other reactants. Solve the stoichiometry problem. g C 18 H 29 SO 3 → mol → mol ratio → mol O 2 → g O 2 . Solve .

    10.0 g C 18 H 29 SO 3 × 1 mol C 18 H 29 SO 3 325 g C 18 H 29 SO 3 × 51 mol O 2 2 mol C 18 H 29 SO 3 × 32.0 g O 2 1 mol O 2 = 25.1 g O 2

    Notice that the mass of O 2 required is 2.5 times greater than the mass of biodegradable material.

    18.50   Water at 9 ppm O 2 is 50% depleted when the concentration drops by 4.5 ppm.

    1 , 200 , 000 persons × 59 g O 2 1 person × 1 × 10 6 g H 2 O 4.5 g O 2 × 1 L H 2 O 1 × 10 3 g H 2 O = 1.57 × 10 10 = 2 × 10 10 L H 2 O

    18.51 Analyze/Plan . The reaction is metathesis. Solve .

    Mg 2+ (aq) + Ca(OH) 2 (s) → Mg(OH) 2 (s) + Ca 2+ (aq)

    [The excess Ca 2+ (aq) is removed as CaCO 3 by naturally occurring bicarbonate or added Na 2 CO 3 .]

    18.52 Analyze/Plan .  Given [Ca 2+ ] and [Mg 2+ ], calculate mol Ca(OH) 2 and Na 2 CO 3 needed to remove the cations. Consider the chemical equations and reaction stoichiometry for each ion. Solve .


    Ca(OH) 2 is added to remove Mg 2+ as Mg(OH) 2 (s); Na 2 CO 3 removes the original and added Ca 2+ .

    Mg 2+ (aq) + Ca(OH) 2 (aq) → Mg(OH) 2 (s) + Ca 2+ (aq)

    Ca 2+ (aq) + Na 2 CO 3 (aq) → CaCO 3 (s) + Na 2+ (aq)

    One mol Ca(OH) 2 is needed for each mol of Mg 2+ (aq) present.

    1.200 × 10 3 L H 2 O × 7.0 × 10 4 mol Mg 2 + L × 1 mol Ca(OH) 2 1 mol Mg(OH) 2 × = 0.84 mol Ca(OH) 2

    Total mol Ca 2+ = mol Ca 2+ originally present + mol Ca 2+ from added Ca(OH) 2

    1.200 × 10 3 L H 2 O × 5 .0 × 10 4 mol Ca 2 + L = 0.60 mol Ca 2 + (aq) original

    0.84 mol Ca 2+ added + 0.60 mol Ca 2+ (aq) original = 1.44 mol Ca 2+ total = 1.44 mol Na 2 CO 3

    Softening requires 0.84 mol Ca(OH) 2 and 1.44 mol Na 2 CO 3 .

    18.53

    1. Trihalomethanes are a class of molecules with one central carbon atom bound to one hydrogen and three halogen atoms. They are produced by the reaction of dissolved chlorine with organic matter naturally present in water, and are by-products of water disinfection via chlorination.
    2. image

    18.54

    1. The most likely origin of bromate ion, BrO 3 , in municipal water supplies is oxidation of dissolved bromide ion, Br . Bromide can react with ozone in a two-step process to form bromate. The ozone might be produced photochemically, or be part of the water disinfection process.
    2. BrO 3 is an oxidizing agent. Hyponitrite ion, NO–, has one less O atom than nitrite ion, NO 2– .

      BrO 3 (aq) + 2 NO (aq) → BrO (aq) + 2 NO 2 (g) or

      BrO 3 (aq) + NO (aq) → BrO (aq) + NO 3 (g)

    Green Chemistry (Section 18.5)

    18.55   The fewer steps in a process, the less waste (solvents as well as unusable by-products) is generated. It is probably true that a process with fewer steps requires less energy at the site of the process, and it is certainly true that the less waste the process generates, the less energy is required to clean or dispose of the waste.

    18.56   Catalysts increase the rate of a reaction by lowering activation energy, E a . For an uncatalyzed reaction that requires extreme temperatures and pressures to generate product at a viable rate, finding a suitable catalyst reduces the required temperature and/or pressure, which reduces the amount of energy used to run the process. A catalyst can also increase rate of production, which would reduce the net time and thus energy required to generate a certain amount of product.


    18.57

    1. image
      • Prevention (1). The alternative process eliminates production of 3-chlorobenzoic acid by-product, chlorine-containing waste that must be treated.
      • Atom economy (2) . Most of the starting atoms are in the final product.
      • Less hazardous chemical synthesis (3) and Inherently safer for accident prevention (12) . The starting material of the alternative process, if it is less concentrated than 30% by mass, is not shock sensitive, and the by-product is nontoxic water. The low molar mass of water means that a small amount of ”waste” is generated.
      • Catalysis (9) and Design for energy efficiency (6) . The alternative process is catalyzed, which could mean that the process will be more energy efficient than the Baeyer–Villiger reaction (see Solution 18.58).
      • Raw materials should be renewable (7) . The catalyst can be recovered from the reaction mixture and reused. We don’t have information about solvents or other auxiliary substances.

    18.58   scCO 2 achieves maximum conversion much faster than CH 2 Cl 2 solvent. This reduces processing time, temperature, and energy requirements. It also results in fewer unwanted by-products to be separated and processed. Although use of scCO 2 can increase the amount of a greenhouse gas released to the environment, it eliminates use of CH 2 Cl 2 , which is implicated in stratospheric ozone depletion. Use of scCO 2 rather than CH 2 Cl 2 is a good green trade-off. (If the CO 2 used to form scCO 2 can be captured from some other industrial process, the net release of CO 2 is the same, and the use of CH 2 Cl 2 is avoided.)

    (In either solvent, the reaction is catalyzed, which usually leads to decreased processing temperatures and times, and greater energy efficiency.)

    18.59

    1. Water as a solvent is much “greener” than benzene, which is a known carcinogen. Water fits criteria: (5) safer solvent, (7) renewable feedstock, and (12) inherently safer for accident prevention.
    2. Reaction temperature of 500 K rather than 1000 K is “greener,” according to criteria (6) design for energy efficiency and (12) inherently safer chemistry for accident prevention. Also, low temperature is less likely to produce undesirable by-products that have to be separated and treated as waste, which fits criterion (1).
    3. Sodium chloride as a by-product rather than chloroform (CHCl 3 ) is “greener,” according to criteria: (1) prevention, (3) less hazardous chemical systems, and (12) inherently safer.

    18.60

    1. The catalyzed reaction that can be run close to room temperature and for a shorter time is definitely greener, according to criteria (6) design for energy efficiency and (9) catalysis.

    1. The reagent obtained from corn husks is greener, by criteria (7) use of renewable feedstocks.
    1. Neither process is totally “ungreen,” because recycling of unavoidable by-products is always desirable. However, by criterion (2) atom economy, the process that produces no by-products is greener.

    Additional Exercises

    18.61

    1. Acid rain is rain with an elevated [H + ] and thus a low pH. The additional H + is produced by the dissolution of sulfur and nitrogen oxides such as SO 3 (g) and NO 2 (g) in rain droplets to form sulfuric and nitric acid, H 2 SO 4 (aq) and HNO 3 (aq).

      High levels of CO 2 , mostly the product of burning fossil fuels, can also increase the pH of rain. The effect is not as great as for dissolved sulfur and nitrogen oxides, because H 2 CO 3 is a much weaker acid than sulfuric or nitric acid.

    2. A g reenhouse gas absorbs infrared or “heat” radiation emitted from Earth’s surface and serves to maintain a relatively constant temperature on the surface. These include H 2 O(g), CH 4 , and CO 2 . A significant buildup of greenhouse gases in the atmosphere could cause a corresponding increase in the average surface temperature and stimulate global climate.
    3. Photochemical smog is an unpleasant collection of atmospheric pollutants initiated by photochemical dissociation of NO 2 to form NO and O atoms. The major components are NO(g), NO 2 (g), CO(g), and unburned hydrocarbons, all produced by automobile engines, and O 3 (g), ozone.
    4. Ozone depletion is the reduction of O 3 concentration in the stratosphere, most notably over Antarctica. It is caused by reactions between O 3 and Cl atoms originating from CFCs, as well as other chlorine-containing organic compounds. Depletion of the ozone layer allows damaging ultraviolet radiation disruptive to the ecosystem to reach Earth's surface.

    18.62   MM avg at the surface = 83.8(0.17) + 16.0(0.38) + 32.0(0.45) = 34.73 = 35 g/mol

    Next, calculate the percentage composition at 200 km. The fractions can be “normalized” by saying that the 0.45 fraction of O 2 is converted into two 0.45 fractions of O atoms, then dividing by the total fractions, 0.17 + 0.38 + 0.45 + 0.45 = 1.45:

    MM avg = 83.8 ( 0.17 ) + 16.0 ( 0.38 ) + 16.0 ( 0.90 ) 1.45 = 23.95 = 24 g/mol

    18.63   Stratospheric ozone is formed and destroyed in a cycle of chemical reactions. The decomposition of O 3 to O 2 and O produces oxygen atoms, an essential ingredient for the production of ozone. Although single O 3 molecules exist for only a few seconds, new O 3 molecules are constantly reformed. This cyclic process ensures a finite concentration of O 3 in the stratosphere available to absorb ultraviolet radiation. (This explanation assumes that the cycle is not disrupted by outside agents such as CFCs.)


    18.64

    2[Cl(g) + O 3 (g) ClO(g) + O 2 (g)] 2[ClO(g) + h ν O(g) + Cl(g)] O(g) + O(g) O 2 (g) 2 Cl(g) + 2 O 3 ( g ) + 2 ClO(g) + 2 O(g) 2 ClO(g) + 3 O 2 (g) + 2 Cl(g) 2 O 3 (g) Cl 3 O 2 (g) [ 18.7 ] [18 .9] [ 18.10 ]

    Note that Cl(g) fits the definition of a catalyst in this reaction.

    18.65   CFCs, primarily CFCl 3 and CF 2 Cl 2 , are chemically inert and water insoluble. These properties make them valuable as propellants, refrigerants, and foaming agents because they are virtually unreactive in the troposphere (lower atmosphere) and do not initiate or propagate undesirable reactions. Further, they are water insoluble and not removed from the atmosphere by rain; they do not end up in the fresh water supply.

    These properties render CFCs a long-term problem in the stratosphere . Because CFCs are inert and water insoluble, they are not removed from the troposphere by reaction or dissolution and have very long lifetimes. Virtually the entire mass of released CFCs eventually diffuses into the stratosphere where conditions are right for photo-dissociation and the production of Cl atoms. Cl atoms catalyze the destruction of ozone, O 3 .

    18.66

    1. The production of Cl atoms in the stratosphere is the result of the photodissociation of a C–Cl bond in the CFC molecule.

      CF 2 Cl 2 ( g ) h ν CF 2 Cl ( g ) + Cl(g)

      According to Table 8.3, the bond dissociation energy of a C–Br bond is

      276 kJ/mol, whereas the value for a C–Cl bond is 328 kJ/mol. Photodissociation of CBrF 3 to form Br atoms requires less energy than the production of Cl atoms and should occur readily in the stratosphere.

    2. CBrF 3 (g) h ν CF 3 (g) + Br(g)

      Br(g) + O 3 (g) → BrO(g) + O 2 (g)

      Also, under certain conditions

      BrO(g) + BrO(g) → Br 2 O 2 (g)

      Br 2 O 2 (g) + hν → O 2 (g) + 2Br(g)

    18.67

    1. A CFC has C–Cl bonds and C–F bonds. In an HFC, the C–Cl bonds are replaced by C–H bonds.
    2. The longer a halogen-containing molecule exists in the stratosphere, the greater the likelihood that it will encounter light with energy sufficient to dissociate a carbon–halogen bond. Free halogen atoms catalyze the destruction of ozone.
    3. The bond dissociation enthalpy of a C–F bond is 485 kJ/mol, much more than for a C–Cl bond, 328 kJ/mol (Table 8.3). Although HFCs have long lifetimes in the stratosphere, it is infrequent that light with energy sufficient to dissociate a C–F bond will reach an HFC molecule. F atoms are much less likely than Cl atoms to be produced by photodissociation in the stratosphere.

    1. The main disadvantage of HFCs as replacements for CFCs is that they are potent greenhouse gases. Although HFCs are far less threatening to stratospheric ozone, they may contribute to global climate change.

    18.68   From Section 18.2:

    N 2 (g) + O 2 (g) ⇌ 2 NO(g)  ΔH = +180.8 kJ [18.11]

    2 NO(g) + O 2 (g) ⇌ 2 NO 2 (g) ΔH = – 113.1 kJ [18.12]

    In an endothermic reaction, heat is a reactant. As the temperature of the reaction increases, the addition of heat favors formation of products and the value of K increases. The reverse is true for exothermic reactions; as temperature increases, the value of K decreases. Thus, K for reaction [18.11], which is endothermic, increases with increasing temperature and K for reaction [18.12], which is exothermic, decreases with increasing temperature.

    18.69

    1. CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g)
    2. 2 CH 4 (g) + 3 O 2 (g) → 2 CO(g) + 4 H 2 O(g)
    3. vol CH 4 → vol O 2 → volume air ( X O 2 = 0.20948 )

      Equal volumes of gases at the same temperature and pressure contain equal numbers of moles (Avogadro’s law). If 2 moles of O 2 are required for 1 mole of CH 4 , 2.0 L of pure O 2 are needed to burn 1.0 L of CH 4 .

      vol O 2 = X O 2 × vol air = vol O 2 X O 2 = 2.0 L 0.20948 = 9.5 L air

    18.70

    1. 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g)

      SO 2 (g) + O 3 (g) → SO 3 (g) + O 2 (g)

      SO 2 (g) + H 2 O(l)→ H 2 SO 3 (l, aerosol)

      SO 3 (g) + H 2 O(l)→ H 2 SO 4 (l, aerosol)

    2. The finely dispersed liquid droplets in the aerosol reflect sunlight into space. Less warming solar radiation reaches Earth’s surface.
    3. In the stratosphere, aerosol particles act as a heterogeneous catalyst for ozone destruction by halogens. That is, they provide a platform to attract and orient the reactants in ozone depletion processes. The depletion reactions occur at a greater rate than in the absence of the aerosol catalyst.

    18.71

    1. According to Section 13.3, the solubility of gases in water decreases with increasing temperature. From the graph, this is also true for CO 2 (g). Comparing the graph in this exercise with Figure 13.16, the general shape of the solubility versus temperature curve for CO 2 (g) is similar to that for other gases. [Although the solubility units are different on the two graphs, it seems that CO 2 (g) is significantly more soluble than the gases in Figure 13.16.]
    2. If the solubility of CO 2 (g) in the ocean decreased because of climate change, more CO 2 (g) would be released into the atmosphere, perpetuating a cycle of increasing temperature and concomitant release of CO 2 (g) from the ocean.

    18.72   Most of the 390 watts/m 2 radiated from Earth’s surface is in the infrared region of the spectrum. Tropospheric gases, particularly H 2 O(g), CH 4 (g), and CO 2 (g), absorb much of this radiation and prevent it from escaping into space (Figures 18.12 and 18.13). The energy absorbed by these so-called greenhouse gases warms the atmosphere close to Earth’s surface and makes the planet livable.

    18.73   Given 168 watts/m 2 at 10% efficiency, find the land area needed to produce 12,000 megawatts. 13,200 megawatts = 13,200 × 10 6 = 1.32 × 10 10 watts.

    168 watts/m 2 (0.10) = 16.8 watts/m 2 solar energy possible with current technology.

    1.32 × 10 10 watts × 1 m 2 16.8 watts = 7.857 × 10 8 = 7.9 × 10 8 m 2

    The land area of New York City is 830 km 2 , which is 830 × 10 6 m 2 . The area needed for solar energy harvesting to provide peak power would then be 7.857 × 10 8 m 2 830 × 10 6 m 2 = 0.95 times the land area of New York City.

    18.74

    1. NO(g) + hν → N(g) + O(g)
    2. NO(g) + hν → NO + (g) + e
    3. NO(g) + O 3 (g) → NO 2 (g) + O 2 (g)
    4. 3 NO 2 (g) + H 2 O(l) → 2 HNO 3 (aq) + NO(g)

    18.75

    1. CO 3 2– is a relatively strong Brønsted–Lowry base and produces OH in aqueous solution according to the hydrolysis reaction:

      CO 3 2– (aq) + H 2 O(l)  ⇌  HCO 3 (aq) + OH (aq),    K b = 1.8 × 10 –4

      If [OH (aq)] is sufficient for the reaction quotient, Q, to exceed K sp for Mg(OH) 2 , the solid will precipitate.

    2. 125 mg Mg 2 + 1 kg soln × 1 g Mg 2 + 1000 mg Mg 2 + × 1.00 kg soln 1.00 L soln × 1 mol Mg 2 + 24.305 g Mg 2 + = 5.143 × 10 3 = 5.14 × 10 3 M Mg 2 +

      4.0 g Na 2 CO 3 1.0 L soln × 1 mol CO 3 2 106.0 g Na 2 CO 3 = 0.03774 = 0.038 M CO 3 2

      K b = 1.8 × 10 4 = [ HCO 3 ] [ OH ] [ CO 3 2 ] x 2 0.03774 ; x = [ OH ] = 2.606 × 10 3 = 2.6 × 10 3 M

      (This represents 6.9% hydrolysis, but the result will not be significantly different using the quadratic formula.)

      Q = [Mg 2+ ][OH–] 2 = (5.143 × 10 –3 )(2.606 × 10 –3 ) 2 = 3.5 × 10 –8

      K sp for Mg(OH) 2 = 1.6 × 10 –12 ; Q > K sp , so Mg(OH) 2 will precipitate.

    18.76

    1. 15 ppb = 15 g Pb in 1 × 10 9 g solution. For very dilute solutions, assume the density of the solution is 1.0 g/mL. 1.0 × 10 9 g solution. = 1.0 × 10 9 mL solution.

      15 g Pb 1.0 × 10 9 g solution × 1 mol Pb 106.42 g Pb × 1000 mL 1 L = 1.4 × 10 7 M


    1. Change μg/dL to g/1.0 × 10 9 mL.

      1.6 μ g 1.0 × 10 9 g solution × 1 × 10 6 g μ g × 1 dL 100 mL × 1 × 10 9 mL = 16 ppb Pb

    18.77 Plan .  Calculate the volume of air above Los Angeles and the volume of pure O 3 that would be present at the 84 ppb level. For gases at the same temperature and pressure, volume fractions equal mole fractions. Solve.

    V air = 4000 mi 2 × ( 1.6093 ) 2 km 2 mi 2 × ( 1000 ) 2 m 2 1 km 2 × 100 m × 1 L 1 × 10 3 m 3 = 1.036 × 10 15 = 1.0 × 10 15 L air

    84 ppb O 3 = 84 mol O 3 1 × 10 9 mol air = 8.4 × 10 8 = X O 3

    V (pure O 3 ) = 8.4 × 10 –8 (1.036 × 10 15 L air) = 8.702 × 10 7 = 8.7 × 10 7 L O 3

    Values for P and T are required to calculate mol O 3 from volume O 3 , using the ideal-gas law. Because these are not specified in the exercise, we will make a reasonable assumption for a sunny April day in Los Angeles. The city is near sea level and temperatures are moderate throughout the year, so P = 1 atm and T = 25 °C (78 °F) are reasonable values. PV = nRT, n = PV/RT.

    n = 1.000 atm × 8 .702 × 10 7 L 298 K × mol-K 0 .08206 L-atm = 3.558 × 10 6 = 3.6 × 10 6 mol O 3

    Check .  Using known conditions to make reasonable estimates and assumptions is a valuable skill for problem solving. Knowing when assumptions are required is an important step in the learning process.

    Integrative Exercises

    18.78

    1. 0.016 ppm NO 2 = 0.016 mol NO 2 1 × 10 6 mol air = 1.6 × 10 8 = X NO 2

      P NO 2 = X NO 2 × P atm = 1.6 × 10 8 ( 755 torr ) = 1.208 × 10 5 = 1.2 × 10 5 torr

    2. n = PV RT ; molecules = n × 6 .022 × 10 23 molecules mol = PV RT × 6.022 × 10 23 molecules mol V = 15 ft × 14 ft × 8 ft × 12 3 in 3 ft 3 × 2.54 3 cm 3 in 3 × 1 L 1000 cm 3 = 4.757 × 10 4 = 5 × 10 4 L

      1.208 × 10 5 torr × 1 atm 760 torr × 4.757 × 10 4 L 293 K × mol-K 0.08206 L-atm × 6.022 × 10 23 molecules mol = 1.894 × 10 19 = 2 × 10 19 molecules

    18.79

    1. 8 , 376 , 726 tons coal × 83 ton C 100 ton coal × 44.01 ton CO 2 12.01 ton C = 2.5 × 10 7 ton CO 2

      8 , 376 , 726 tons coal × 2 .5 ton S 100 ton coal × 64.07 ton SO 2 32.07 ton S = 4.2 × 10 5 ton SO 2


    1. CaO(s) + SO 2 (g) → CaSO 3 (s)

      4.18 × 10 5 ton SO 2 × 55 ton SO 2 removed 100 ton SO 2 produced × 120.15 ton CaSO 3 64.07 ton SO 2 = 4.3 × 10 5 ton CaSO 3

    18.80 Coarse sand is removed by coarse sand filtration. Finely divided particles and some bacteria are removed by precipitation with aluminum hydroxide. Remaining harmful bacteria are removed by ozonation. Trihalomethanes are removed by either aeration or activated carbon filtration; use of activated carbon might be preferred because it does not involve release of TCMs into the atmosphere. Dissolved organic substances are oxidized (and rendered less harmful, but not removed) by both aeration and ozonation. Dissolved nitrates and phosphates are not removed by any of these processes, but are rendered less harmful by adequate aeration.

    18.81   Calculate the molar concentration of impurity that would have an absorbance of 0.0001. This is the minimum concentration of the impurity detectable by absorption spectroscopy.

    A = εbc; A = absorbance, ε = extinction coefficient, b = path length, c = molarity. The common path length is 1 cm.

    c = A ε b = 0.0001 × M × cm 3.45 × 10 3 × 1 1 cm = 2.8986 × 10 8 = 3 × 10 8 M

    Because we do not have the identity of the impurity, we cannot calculate the corresponding concentration in ppb. We can calculate a maximum molar mass for the impurity, such that a 3 × 10 –8 M solution is 50 ppb. A concentration of 50 ppb corresponds to 50 g impurity per 10 9 L solution.

    5 0 g impurity 10 9 g solution × 1000 g solution L solution × 1 L solution 2 .8986 × 10 8 mol impurity = 1725 g impurity/mol

    In this calculation, molar mass is directly proportional to ppm concentration. This means that a 50 ppm solution or any impurity with a molar mass less than or equal to 1725 g/mol will be observable by absorption spectroscopy. Concentrations less than 50 ppm are probably observable, because 1725 is a large molar mass. (The calculated molar mass is more correctly represented with one sig fig as 2 × 10 3 g/mol.)

    18.82

    1. image
    2. Δ H = 2D(O H) D(O H) = D(O H) = 463 kJ/mol

      463 kJ mol H 2 O × 1 mol H 2 O 6.022 × 10 23 molecules × 1000 J kJ = 7.688 × 10 19 = 7.69 × 10 19 J/H 2 O molecule

      λ = hc Δ E = 6.626 × 10 34 J-sec × 2 .998 × 10 8 m/s 7.688 × 10 19 J = 2.58 × 10 7 m = 258 nm

      This wavelength is in the UV region of the spectrum, close to the visible.


    1. OH(g) + O 3 (g) HO 2 (g) + O 2 (g) HO 2 (g) + O(g) OH(g) + O 2 (g) OH(g) + O 3 (g) + HO 2 (g) + O(g) HO 2 (g) + 2 O 2 (g) + OH(g) O 3 (g) + O(g) 2 O 2 ( g )

      OH(g) is the catalyst in this overall reaction, another pathway for the destruction of ozone.

    18.83   According to Equation 14.12, ln([A] t / [A] o ) = –kt. [A] t = 0.10 [A] o .

    ln(0.10 [A] o / [A] o ) = ln(0.10) = –(2 × 10 –6 s –1 ) t

    t = –ln(0.10) / 2 × 10 –6 s –1 = 1.151 × 10 6 s

    1.151 × 10 6 s × 1 min 60 s × 1 h 60 min × 1 day 24 h = 13.3 days (1 × 10 days)

    The value of the rate constant limits the result to 1 sig fig. This implies that there is minimum uncertainty of ±1 in the tens place of our answer. Realistically, the remediation could take anywhere from 1 to 20 days.

    18.84

    1. ClO(g) + O 3 (g) → ClO 2 (g) + O 2 (g)

      Δ H i = Δ H f ° ClO 2 (g) + Δ H f ° O 2 (g) Δ H f ° ClO(g) Δ H f ° O 3 (g)

      ΔH i = 102 + 0 – 101 – (142.3) = –141 kJ

    2. ClO 2 (g) + O(g) → ClO(g) + O 2 (g)

      Δ H ii = Δ H f ° ClO(g) + Δ H f ° O 2 (g) Δ H f ° ClO 2 (g) Δ H f ° O(g)

      ΔH ii = 101 + 0 – 102 – (247.5) = –249 kJ

    (overall) ClO(g) + O 3 (g) + ClO 2 (g) + O(g) → ClO 2 (g) + O 2 (g) + ClO(g) + O 2 (g)

    O 3 (g) + O(g) → 2O 2 (g)

    ΔH = ΔH i + ΔH ii = –141 kJ + (–249) kJ = –390 kJ

    Because the enthalpies of both (i) and (ii) are distinctly exothermic, it is possible that the ClO – ClO 2 pair could be a catalyst for the destruction of ozone.

    18.85

    1. Assume the density of water at 20 °C is the same as at 25 °C.

      1.00 gal × 4 qt 1 gal × 1 L 1 .057 qt × 1000 mL 1 L × 0.99707 g H 2 O 1 mL = 3773 = 3.77 × 10 3 g H 2 O

      The H 2 O(l) must be heated from 20 °C to 100 °C and then vaporized at 100 °C.

      3.773 × 10 3 g H 2 O × 4 .184 J g ° C × 80 ° C × 1 kJ 1000 J = 1263 = 1.3 × 10 3 kJ

      3.773 × 10 3 g H 2 O × 1 mol H 2 O 18 .02 g H 2 O × 40 .67 kJ mol H 2 O = 8516 = 8.52 × 10 3 kJ

      energy = 1263 kJ + 8516 kJ = 9779 = 9.8 × 10 3 kJ/gal H 2 O


    1. According to Solution 5.18, 1 kwh = 3.6 × 10 6 J.

      9779 kJ gal H 2 O × 1000 J kJ × 1 kwh 3 .6 × 10 6 J × $ 0.085 kwh = $ 0.23 /gal

    1. $ 0.23 $ 1.26 × 100 = 18 % of the total cost is energy

    18.86

    1. A rate constant of M –1 s –1 is indicative of a reaction that is second order overall. For the reaction given, the rate law is probably rate = k[O][O 3 ]. (Although rate = k[O] 2 or k[O 3 ] 2 are possibilities, it is difficult to envision a mechanism consistent with either one that would result in two molecules of O 2 being produced.)
    2. Yes. Most atmospheric processes are initiated by collision. One could imagine an activated complex of four O atoms collapsing to form two O 2 molecules. Also, the rate constant is large, which is less likely for a multistep process. The reaction is analogous to the destruction of O 3 by Cl atoms (Equation 18.7), which is also second order with a large rate constant.
    3. Δ H o = 2 Δ H f o O 2 (g) Δ H f o O(g) Δ H f o O 3 (g)

      Δ H o = 0 247.5 kJ 142.3 kJ = 389.8 kJ

      The reaction is exothermic, so energy is released; the reaction would raise the temperature of the stratosphere.

    18.87

    1. Holding one reactant concentration constant and changing the other, evaluate the effect this has on the initial rate. Use these observations to write the rate law.

      Compare Experiments 1 and 3. [O 3 ] is constant, [H] doubles, initial rate doubles. The reaction is first order in [H].

      Compare Experiments 2 and 1. [H] is constant, [O 3 ] doubles, initial rate doubles. The reaction is first order in [O 3 ].

      rate = k[O 3 ][H]

    2. Calculate a value for the rate constant for each experiment and average them to obtain a single representative value.

      rate = k[O 3 ][H]; k = rate/[O 3 ][H]

      k 1 = 1.88 × 10 14 M / s ( 5.17 × 10 33 M ) ( 3.22 × 10 26 M ) = 1 . 1293 × 10 44 = 1 . 13 × 10 44

      k 2 = 9.44 × 10 15 M / s ( 2.59 × 10 33 M ) ( 3.25 × 10 26 M ) = 1 . 1215 × 10 44 = 1 . 12 × 10 44

      k 3 = 3.77 × 10 14 M / s ( 5.19 × 10 33 M ) ( 6.46 × 10 26 M ) = 1 . 1245 × 10 44 = 1 . 12 × 10 44

      k avg = (1.1293 × 10 44 + 1.1215 × 10 44 + 1.1245 × 10 44 )/3 = 1.1251 × 10 44 = 1.13 × 10 44 M –1 s –1


    18.88   rate = k[CF 3 CH 2 F][OH]. k = 1.6 × 10 8 M –1 s –1 at 4 °C.

    [CF 3 CH 2 F] = 6.3 × 10 8 molecules/cm 3 , [OH] = 8.1 × 10 5 molecules/cm 3

    Change molecules/cm 3 to mol/L ( M ) and substitute into the rate law.

    6.3 × 10 8 molecules cm 3 × 1 mol 6.022 × 10 23 molecules × 1000 cm 3 1 L = 1 .0462 × 10 –12 = 1 .0 × 10 –12 M CF 3 CH 2 F

    8.1 × 10 5 molecules cm 3 × 1 mol 6.022 × 10 23 molecules × 1000 cm 3 1 L = 1.3451 × 10 15 = 1.3 × 10 15 M OH

    rate = 1.6 × 10 8 M -s × 1.0462 × 10 12 M × 1.3451 × 10 15 M = 2.2515 × 10 19 = 2.3 × 10 19 M / s

    18.89

    1. According to Table 18.1, the mole fraction of CO 2 in air is 0.000375.

      P CO 2 = X CO 2 × P atm = 0.000400 (1 .00 atm) = 4 .00 × 10 4 atm C CO 2 = kP CO 2 = 3.1 × 10 2 M / atm × 4 .00 × 10 4 atm = 1 .24 × 10 5 = 1.2 × 10 5 M

    2. H 2 CO 3 is a weak acid, so the [H + ] is regulated by the equilibria:

      H 2 CO 3 (aq)   ⇌   H + (aq) + HCO 3 (aq)     K a1 = 4.3 × 10 –7

      HCO 3 (aq)   ⇌   H + (aq) + CO 3 2– (aq)     K a2 = 5.6 × 10 –11

      Because the value of K a2 is small compared to K a1 , we will assume that most of the H + (aq) is produced by the first dissociation.

      K a1 = 4.3 × 10 7 = [ H + ] [ HCO 3 ] [ H 2 CO 3 ] ; [ H + ] = [ HCO 3 ] = x, [ H 2 CO 3 ] = 1.24 × 10 5 x

      Because K a1 and [H 2 CO 3 ] have similar values, we cannot assume x is small compared to 1.2 × 10 –5 .

      4.3 × 10 7 = x 2 ( 1.24 × 10 5 x) ; 5.332 × 10 12 4.3 × 10 7 x = x 2

      0 = x 2 + 4.3 × 10 –7 x – 5.332 × 10 –12

      x = 4.3 × 10 7 ± ( 4.3 × 10 7 ) 2 4 ( 1 ) ( 5.332 × 10 12 ) 2 ( 1 )

      x = 4.3 × 10 7 ± 1.85 × 10 13 + 2.133 × 10 11 2 = 4.3 × 10 7 ± 4.638 × 10 6 2

      The negative result is meaningless; x = 2.104 × 10 –6 = 2.1 × 10 –6 M H + ; pH = 5.68

      Because this [H + ] is quite small, the [H + ] from the autoionization of water might be significant. Calculation shows that for [H + ] = 2.1 × 10 –6 M from H 2 CO 3 , [H + ] from H 2 O = 5.2 × 10 –9 M , which we can ignore.


    18.90

    1. Al(OH) 3 (s)  ⇌  Al 3+ (aq) + 3 OH (aq) K sp = 1.3 × 10 –33 = [Al 3+ ] [OH ] 3

      This is a precipitation conditions problem. At what [OH ] (we can get pH from [OH ]) will Q = 1.3 × 10 –33 , the requirement for the onset of precipitation?

      Q = 1.3 × 10 –33 = [Al 3+ ][OH ] 3 . Find the molar concentration of Al 2 (SO 4 ) 3 and thus [Al 3+ ].

      5.0 lb Al 2 (SO 4 ) 3 2000 gal H 2 O × 453.6 g 1 lb × 1 mol Al 2 (SO 4 ) 3 342.2 g Al 2 ( SO 4 ) 3 × 1 gal 4 qt × 1 qt 0 .946 L = 8.758 × 10 4 M Al 2 ( SO 4 ) 3 = 1.752 × 10 3 = 1.8 × 10 3 M Al 3 +

      Q = 1.3 × 10 –33 = (1.752 × 10 –3 )[OH ] 3 ; [OH ] 3 = 7.42 × 10 –31

      [OH ] = 9.054 × 10 –11 = 9.1 × 10 –11 M ; pOH = 10.04; pH = 14 – 10.04 = 3.96

    2. CaO(s) + H 2 O(l) → Ca 2+ (aq) + 2 OH (aq); [OH ] = 9.054 × 10 –11 mol/L

      mol OH = 9.054 × 10 11 mol 1 L × 2000 gal × 4 qt 1 gal × 0.946 L 1 qt = 6.852 × 10 7 = 6.9 × 10 7 mol OH

      6.852 × 10 7 mol OH × 1 mol CaO 2 mol OH × 56.1 g CaO 1 mol CaO × 1 lb 453 .6 g = 4.2 × 10 8 lb CaO

      This is a very small amount of CaO, about 20 μg.

    18.91

    1. Process (i) is greener, because it does not involve the toxic reactant phosgene (COCl 2 ) and the by-product is water, not HCl.
    2. Reaction (i): C in CO 2 is linear with sp hybridization; C in R–N=C=O is linear with sp hybridization; C in the urethane monomer is trigonal planar with sp 2 hybridization. Reaction (ii): C in COCl 2 is trigonal planar with sp 2 hybridization; C in R–N=C=O is linear with sp hybridization; C in the urethane monomer is trigonal planar with sp 2 hybridization.
    3. Traditionally, industrial processes are conducted at higher temperatures to speed up reactions and encourage formation of product. However, this is not a green solution, because it requires additional energy. Using Le Châtelier’s principle, we could either “push” or “pull” the reaction toward products. The “push” requires that we increase the amount of reactants, again not a green approach. The greenest way to promote formation of the isocyanate is to “pull” the reaction forward by removing by-product from the reaction mixture. In reaction (i), remove water; in reaction (ii), remove HCl.

    18.92

    1. The various forms of carbonate in water are related by the following equilibria:

      H 2 CO 3 ( aq ) H + ( aq ) + HCO 3 ( aq ) K a1 = 4 .3×10 –7 HCO 3 ( aq ) H + ( aq ) + CO 3 2– ( aq ) K a2 = 5 .6×10 –11

      K a1 = 4.3 × 10 7 = [ H + ] [ HCO 3 ] [ H 2 CO 3 ] ; K a2 = 5.6 × 10 11 = [ H + ] [ CO 3 2 ] [ HCO 3 ]

      [H + ] = 10 –pH = 10 –5.6 = 2.5119 × 10 –6 = 3 × 10 –6 M

      Also, [H 2 CO 3 ] + [HCO 3 ] + [CO 3 2– ] = 1.0 × 10 –5 M


      We now have 3 equations in 3 unknowns, so we can solve explicitly for one. Solve for [HCO 3 ] (because it appears in both K a expressions) and then substitute to find [H 2 CO 3 ] and [CO 3 2– ].

      1.0 × 10 5 = [ H + ] [ HCO 3 ] K a1 + [ HCO 3 ] + K a2 [ HCO 3 ] [ H + ]

      1.0 × 10 5 = 2.5119 × 10 6 [ HCO 3 ] 4.3 × 10 7 + [ HCO 3 ] + 5.6 × 10 11 [ HCO 3 ] 2.5119 × 10 6

      1.0 × 10 5 = 5.8416 [ HCO 3 ] + [ HCO 3 ] + 2.2294 × 10 5 [ HCO 3 ]

      [ HCO 3 ] = 1.0 × 10 5 6.8416 = 1.4616 × 10 6 = 1.5 × 10 6 M

      Note that [CO 3 2– ] is very small compared to [H 2 CO 3 ] and [HCO 3 ].

      [ H 2 CO 3 ] = ( 2.5119 × 10 6 ) ( 1.4616 × 10 6 ) 4.3 × 10 7 = 8.5383 × 10 6 = 8.5 × 10 6 M

      [ CO 3 2 ] = ( 5.6 × 10 11 ) ( 1.4616 × 10 6 ) 2.5119 × 10 6 = 3.2586 × 10 11 = 3.3 × 10 11 M

      Check . 1.5 × 10 –6 M + 8.5 × 10 –6 M + 3.3 × 10 –11 M = 1.0 × 10 –5 M

    1. To test for sulfur-containing species, we must first remove the various forms of carbonate. One method is to exploit the solubility differences between carbonate and sulfate salts. Most sulfates are soluble, whereas most carbonates are not. However, K sp values for carbonates are relatively large, and [CO 3 2– ] in the raindrop is very small. Precipitating insoluble carbonates will shift the acid dissociation equilibria to the right, but precipitation may not be the best method for effectively removing carbonates.

      A different method involves removing carbonates as CO 2 (g). Heating the rainwater will decrease the solubility of CO 2 (g), which will bubble off as a gas. Slightly acidifying the solution will encourage this process, by shifting the acid dissociation equilibria toward H 2 CO 3 and CO 2 (g).

      After removal of carbonates, sulfates are precipitated with Ba 2+ (aq). The amount of precipitate is small, but it does cause turbidity in the solution. Turbidity is detected by instrumental methods that measure light scattering by colloids.


     

    19 Chemical Thermodynamics

    Visualizing Concepts

    19.1

    1. image
    2. ΔS is positive, because the disorder of the system increases. Each gas has greater motional freedom as it expands into the second bulb, and there are many more possible arrangements for the mixed gases.

      By definition, ideal gases experience no attractive or repulsive intermolecular interactions, so ΔH for the mixing of ideal gases is zero, assuming heat exchange only between the two bulbs.

    3. The process is spontaneous and, therefore, irreversible. It is inconceivable that the gases would reseparate.
    4. The entropy change of the surroundings is related to ΔH for the system. Because we are mixing ideal gases and ΔH = 0, ΔH surr is also zero, assuming heat exchange only between the two bulbs.

    19.2

    1. Based on experience, the process is spontaneous. We know that 1,1-difluoro-ethane is a gas at atmospheric pressure, so the pressure inside the can must be much greater than atmospheric in order for the substance to be liquefied. When the nozzle is pressed and the system is open to the lower pressure of the atmosphere, the liquid vaporizes spontaneously. The 1,1-difluoroethane gas escapes the nozzle without external assistance.
    2. We expect q sys to be positive. We know that ΔH is positive for the vaporization of a gas. Because the change does not occur at constant pressure, q sys and ΔH are not equal, but the sign of q sys is still positive.
    3. ΔS is definitely positive for this process. Because the process is spontaneous and ΔH is positive, ΔS must be positive and large so that ΔG is negative. It is also true that the system, the 1,1-difluoroethane molecules, occupy a larger volume and have greater motional freedom after vaporization.
    4. The operation of the keyboard cleaner definitely depends more on entropy change than heat flow.

    19.3

    1. The process depicted is a change of state from a solid to a gas. ΔS is positive because of the greater motional freedom of the particles. ΔH is positive because both melting and boiling are endothermic processes.

    1. The sign of ΔS surr is negative, and the magnitude is less than or equal to ΔS sys . If the process is spontaneous, the second law states that ΔS univ ≥ 0. Because ΔS sys is positive, ΔS surr must be negative. If the change occurs via a reversible pathway, ΔS univ = 0 and ΔS surr = –ΔS sys . If the pathway is irreversible, the magnitude of ΔS sys is greater than the magnitude of ΔS surr , but the sign of ΔS surr is still negative.

    19.4  Both ΔH and ΔS for this reaction are positive.

    The reaction involves breaking five blue–blue and twenty blue–red bonds and then forming twenty blue–red bonds. The net change is breaking five blue–blue bonds. Enthalpies for bond breaking (Section 8.8) are always positive.

    In the depicted reaction, both reactants and products are in the gas phase (they are far apart and randomly placed). There are twice as many molecules (or moles) of gas in the products, so ΔS is positive for this reaction.

    19.5 Analyze/Plan .  Consider the physical changes that occur when a substance is heated. How do these changes affect the entropy of the substance? Solve .

    1. Both 1 and 2 represent changes in entropy at constant temperature; these are phase changes. Because 1 happens at a lower temperature, it represents melting (fusion), and 2 represents vaporization.
    2. The substance changes from solid to liquid in 1, from liquid to gas in 2. The larger volume and greater motional freedom of the gas phase causes ΔS for vaporization to (always) be larger than ΔS for fusion.
    3. For a perfect crystal at T = 0 K, the value of S is zero. This is the third law of thermodynamics.

    19.6

    1. We expect the enthalpy of combustion of the two isomers to be very similar. The molecular formulas of the two molecules are the same, so the balanced chemical equations for the two combustion reactions are identical. In the calculation of combustion enthalpy from standard enthalpies of formation of products and reactants, the only difference will be in the standard enthalpies of formation of the two isomers.
    2. We expect n -pentane to have the higher standard molar entropy. The rod-shaped n -pentane has more possible vibrational and rotational motions than the almost-spherical neopentane. That is, n -pentane has greater motional energy, which results in a higher standard molar entropy than that of neopentane.

    19.7

    1. At 300 K, ΔH = TΔS. Because ΔG = ΔH – TΔS, ΔG = 0 at this point. When ΔG = 0, the system is at equilibrium.
    2. The reaction is spontaneous when ΔG is negative. This condition is met when TΔS > ΔH. From the diagram, TΔS > ΔH when T > 300 K. The reaction is spontaneous at temperatures above 300 K.

    19.8

    1. At equilibrium, ΔG = 0. On the diagram, ΔG = 0 at 250 K. The system is at equilibrium at 250 K.
    2. A reaction is spontaneous when ΔG is negative. The reaction is spontaneous at temperatures greater than 250 K.

    1. ΔG = ΔH – TΔS, in the form of y = b + mx. ΔH is the y intercept of the graph (where T = 0) and is positive.
    2. The slope of the graph is –ΔS. The slope is negative, so ΔS is positive. [Also, ΔG decreases as T increases, so the TΔS term must become more negative and ΔS is positive.]

    19.9

    1. Analyze .  The boxes depict three different mixtures of reactants and products for the reaction A 2 + B 2 ⇌ 2AB.

      Plan . K c = 1 = [ AB ] 2 [ A ] [ B ] . Calculate Q for each box, using number of molecules as a measure of concentration. If Q = 1, the system is at equilibrium . Solve .

      Box 1: K = ( 3 ) 2 ( 3 ) ( 3 ) = 1

      Box 2: Q = ( 1 ) 2 ( 4 ) ( 4 ) = 1 16 = 0.0625 = 0.06

      Box 3: Q = ( 7 ) 2 ( 1 ) ( 1 ) = 49 1 = 49

      Box 1 is at equilibrium.

    2. Box 2.
    3. Qualitatively, Box 3 is farthest from equilibrium, so it has the largest magnitude of ΔG (driving force to reach equilibrium), then Box 2, and then Box 1, where ΔG = 0.

      Box 1 < Box 2 < Box 3

      Quantitatively, ΔG = ΔG° – RTlnQ. For Box 1, ΔG = 0 and K = 1, so ΔG° = 0.

      Box 2: ΔG = 0 – RTln(0.0625) = 2.77 RT

      Box 3: ΔG = 0 – RTln(49) = –3.89 RT

      Quantitative treatment confirms the order for magnitude of ΔG as

      Box 1 < Box 2 < Box 3.

    19.10

    1. True. When ΔG = 0, the reaction is at equilibrium.
    2. False. At equilibrium, there is a mixture of reactants and products.
    3. False. There are fewer moles of gas in the products than the reactants, so ΔS is negative.
    4. False. At the left and right extremes of the graph, reactants and products are gases at 1 atm pressure; they are in their standard states. The quantity “x” is the difference in free energy between reactants and products in their standard states, ΔG°.
    5. True. ΔG is a measure of the driving force for a reaction to reach equilibrium.

    Spontaneous Processes (Section 19.1)

    19.11 Analyze/Plan .  Follow the logic in Sample Exercise 19.1. Solve .

    1. Spontaneous; at ambient temperature, ripening happens without intervention.
    2. Spontaneous; sugar is soluble in water, and even more soluble in hot coffee.

    1. Spontaneous; N 2 molecules are stable relative to isolated N atoms.
    2. Spontaneous; under certain atmospheric conditions, lightning occurs.
    3. Nonspontaneous; CO 2 and H 2 O are in contact continuously at atmospheric conditions in nature and do not form CH 4 and O 2 .

    19.12

    1. Nonspontaneous; at 1 atm, ice does not melt spontaneously at temperatures below its normal melting point.
    2. Nonspontaneous; a mixture cannot be separated without outside intervention.
    3. Spontaneous.
    4. Spontaneous. The reaction is spontaneous but slow unless encouraged by a catalyst or spark.
    5. Spontaneous; the very polar HCl molecules readily dissolve in water to form concentrated HCl(aq).

    19.13

    1. True, assuming the conditions are the same for the forward and reverse reactions.
    2. False. A spontaneous process occurs without outside intervention. This definition says nothing about how quickly the process occurs. Spontaneity is a thermodynamic property, while rate is a kinetic property.
    3. False. All spontaneous processes are real processes and real processes are irreversible.
    4. True.
    5. False. The maximum amount of work can be accomplished by a reversible process.

    19.14

    1. Yes. While most spontaneous processes are exothermic, some, such the melting of ice at room temperature, are endothermic.
    2. Yes. Melting is an example of a process that is spontaneous at one temperature, the melting point, but nonspontaneous at other temperatures. Other phase changes are also examples of this behavior.
    3. No. While the forward and reverse processes can be induced, the processes cannot be reversed with an infinitesimally small change in some property of the system.
    4. Yes. The reversible pathway can accomplish the maximum amount of work on its surroundings because no entropy is lost to the universe. ΔS universe = 0.

    19.15 Analyze/Plan .  Define the system and surroundings. Use the appropriate definition to answer the specific questions. Solve .

    1. Water is the system. Heat must be added to the system to evaporate the water. The process is endothermic.
    2. At 1 atm, the reaction is spontaneous at temperatures above 100 °C.
    3. At 1 atm, the reaction is nonspontaneous at temperatures below 100 °C.
    4. The two phases are in equilibrium at 100 °C.

    19.16

    1. Exothermic. If melting requires heat and is endothermic, freezing must be exothermic.
    2. At 1 atm (indicated by the term normal freezing point), the freezing of n -octane is spontaneous at temperatures below –57 °C.
    3. At 1 atm, the freezing of n -octane is nonspontaneous at temperatures above –57 °C.
    4. At 1 atm and –57 °C, the normal freezing point of n -octane, the solid and liquid phases are in equilibrium. That is, at the freezing point, n -octane molecules escape to the liquid phase at the same rate as liquid n -octane solidifies, assuming no heat is exchanged between n -octane and the surroundings.

    19.17 Analyze/Plan . Consider the definitions of the terms reversible , isothermal , and state function to answer the questions. Solve .

    1. No. Temperature is a state function, so a change in temperature does not depend on pathway. (As we are talking about an ideal gas, a reversible pathway may be possible for this change in state. An irreversible pathway is always possible.)
    2. No. An isothermal process occurs at constant temperature.
    3. No. ΔE is a state function. ΔE = q + w; q and w are not state functions. Their values do depend on path, but their sum, ΔE, does not.

    19.18

    1. Yes, because ΔE is a state function. (1 → 2) = –ΔE (2 → 1)
    2. No. We can say nothing about the values of q and w because we have no information about the paths.
    3. The magnitudes of the work are equal, but the signs are opposite. If the changes of state are reversible, the two paths are the same and w (1 → 2) = –w (2 → 1). This is the maximum realizable work from this system.

    19.19 Analyze/Plan .  Define the system and surroundings. Use the appropriate definition to answer the specific questions. Solve .

    1. An ice cube can melt reversibly at the conditions of temperature and pressure where the solid and liquid are in equilibrium. At 1 atm external pressure, the normal melting point of water is 0 °C.
    2. No. We know that melting is endothermic, so ΔH for melting ice is a positive, nonzero value. Also, since ΔH is a state function, the nonzero value is independent of path. Whether the ice cube melts reversibly or irreversibly, ΔH for the process is not zero.

    19.20

    1. The detonation of an explosive is definitely not reversible. The farflung debris from the explosion (system and surroundings) cannot be perfectly reconstructed, even with a large input of energy.
    2. The quantity q is related to ΔH. As the detonation is highly exothermic, q is large and negative.

      If only P–V work is done and P is constant, ΔH = q. Although these conditions probably do not apply to a detonation, we can still predict the sign of q, based on ΔH, if not its exact magnitude.


    1. The sign (and magnitude) of w depend on the path of the process, the exact details of how the detonation is carried out. It seems clear, however, that work will be done by the system on the surroundings in almost all circumstances (buildings collapse, earth and air are moved), so the sign of w is probably negative.

    Entropy and the Second Law of Thermodynamics (Section 19.2)

    19.21

    1. True.
    2. False. For a reversible process, the entropy change of the universe is zero.
    3. True.
    4. False. For a reversible process, the entropy change to the system need not be zero, but it must be matched by the entropy change to the surroundings.

    19.22

    1. True.
    2. False. For an irreversible process, the net entropy change of the system and surroundings must be positive.
    3. False. For a spontaneous (and thus irreversible) process, the net entropy change of the system and surroundings must be positive. The entropy change of the system could be negative if the entropy change of the surroundings was large and positive.
    4. True. For an isothermal process, ΔS = q rev /T.

    19.23

    1. Br 2 (l) → Br 2 (g), entropy increases, more mol gas in products, greater motional freedom.
    2. Δ S = Δ H T = 29.6 kJ mol Br 2 (l) × 1.00 mol Br 2 ( l) × 1 ( 273.15 + 58.8 ) K × 1000 J 1 kJ = 89.2 J/K

    19.24

    1. Ga(l) → Ga(s), ΔS is negative, less motional freedom
    2. Δ H = 6 0. 0 g Ga × 1 mol Ga 69 .723 g Ga × 5.59 kJ mol Ga = 4.81046 = 4.81 kJ

      Δ S = Δ H T = 4.81046 kJ × 1000 J 1 kJ × 1 ( 273.15 + 29.8 ) K = 15.9 J/K

    19.25

    1. False. The second law states that entropy is conserved for a reversible process.
    2. True. In a reversible process, ΔS sys + ΔS surr = 0. If ΔS sys is positive, ΔS surr must be negative.
    3. False. Because ΔS univ must be positive for a spontaneous process, ΔS surr must be greater than –4.2 J/K. Entropy is not conserved for a spontaneous process.

    19.26

    1. Not necessarily. The only thing we know for sure is that the entropy of the universe increases for a spontaneous process.
    2. ΔS surr is positive and greater than the magnitude of the decrease in ΔS sys .
    3. ΔS sys = 78 J/K.

    19.27 Analyze .  Consider ΔS for the isothermal expansion of 0.200 mol of an ideal gas at 27 °C and an initial volume of 10.0 L.

    1. Whenever an ideal gas expands isothermally, we expect an increase in entropy, or positive ΔS, owing to the greater volume available for motion of the particles.
    2. Plan .  Use the relationship ΔS sys = nR ln(V 2 /V 1 ), Equation 19.3.

      Solve . ΔS sys = 0.200 (8.314 J/mol-K)(ln [18.5 L/10.0 L]) = 1.02 J/K.

      Check .  We expect ΔS to be positive when the motional freedom of a gas increases, and our calculation agrees with this prediction.

    3. No. The temperature at which the expansion occurs is not needed to calculate the entropy change, as long as the process is isothermal.

    19.28

    1. According to Boyle’s law, pressure and volume are inversely proportional at constant amount and temperature. If the pressure of an ideal gas increases, volume decreases. We expect a decrease in entropy, or negative ΔS, for the isothermal compression of an ideal gas, owing to the smaller volume available for motion of the particles.
    2. According to Boyle’s law, P 1 V 1 = P 2 V 2 at constant n and T.

      0.750 atm × V 1 = 1.20 atm × V 2 ; V 2 /V 1 = 0.750 atm/1.20 atm = 0.62500 = 0.625

      ΔS sys = nR ln (V 2 /V 1 ) = 0.600 mol (8.314 J/mol-K)(ln 0.625) = –2.34 J/K

      Check .  An increase in pressure results in a decrease in volume at constant T, so we expect ΔS to be negative, and it is.

    3. No. The temperature at which the compression (increase in pressure, decrease in volume) occurs is not needed to calculate the entropy change, as long as the process is isothermal.

    The Molecular Interpretation of Entropy and the Third Law of Thermodynamics (Section 19.3)

    19.29

    1. Yes, the expansion is spontaneous.
    2. The ideal gas is the system, and everything else, including the vessel containing the vacuum, is the surroundings. There is literally nothing inside the vessel containing the vacuum, no gas molecules and no physical barriers. As the ideal gas expands into the vacuum, there is nothing for it to “push back,” so no work is done. Mathematically, w = –P ext ΔV. Because the gas expands into a vacuum, P ext = 0 and w = 0.
    3. Entropy. The “driving force” for the expansion of the gas is the increase in entropy associated with greater volume, more motional freedom, and more possible positions for the gas particles.

    19.30

    1. A thermodynamic state is a set of conditions, usually temperature and pressure, that defines the properties of a bulk material. A microstate is a single possibility for all the positions and kinetic energies of all the molecules in a sample; it is a snapshot of positions and speeds at a particular instant.
    2. According to Equation 19.5 (Boltzmann law), the more possible microstates for a macroscopic state, the greater the entropy of the state. If S decreases going from A to B, then A has more microstates than B. Or, if ΔS is negative, the number of microstates decreases.
    3. According to part (b), if the number of microstates available to a system decreases, ΔS sys is negative. For a spontaneous process, ΔS univ is positive, so ΔS surr is positive (and the magnitude is greater than that of ΔS sys ).

    19.31

    1. The higher the temperature, the broader the distribution of molecular speeds and kinetic energies available to the particles. At higher temperature, the wider range of accessible kinetic energies leads to more microstates for the system.
    2. A decrease in volume reduces the number of possible positions for the particles and leads to fewer microstates for the system.
    3. Going from liquid to gas, particles have greater translational motion, which increases the number of positions available to the particles and the number of microstates for the system.

    19.32

    1. ΔH vap for H 2 O at 25 °C = 44.02 kJ/mol; at 100 °C = 40.67 kJ/mol

      Δ S = q rev T = 44.02 kJ mol × 1000 J kJ × 1 298 K = 148 J/mol-K Δ S = q rev T = 40.67 kJ mol × 1000 J kJ × 1 373 K = 109 J/mol-K

    2. At both temperatures, the liquid → gas phase transition is accompanied by an increase in entropy, as expected. That the magnitude of the increase is greater at the lower temperature requires some explanation.

      In the liquid state, there are significant hydrogen bonding interactions between H 2 O molecules. This reduces the number of possible molecular positions and the number of microstates. Liquid water at 100° has sufficient kinetic energy to have broken many hydrogen bonds, so the number of microstates for H 2 O(l) at 100° is greater than the number of microstates for H 2 O(l) at 25 °C. The difference in the number of microstates upon vaporization at 100 °C is smaller, and the magnitude of ΔS is smaller.

    19.33 Analyze/Plan .  Consider the conditions that lead to an increase in entropy: more mol gas in products than reactants, increase in volume of sample and, therefore, number of possible arrangements, more motional freedom of molecules, and so on. Solve .

    1. More gaseous particles means more possible arrangements and greater disorder; ΔS is positive.

    1. S sys increases when a banana ripens. Starches are polysaccharides, large molecules that break down into sugars, smaller monosaccharides and disaccharides, when ripening occurs. Formation of the sugars increases the number of molecules and the entropy.

      S sys clearly increases in 19.11 (b), where there is an increase in volume and possible arrangements for the sample.

      In 19.11 (c), the system goes from two moles of gaseous reactants to one mole of gaseous products, and S sys decreases.

      In 19.11 (d), the entropy of the universe clearly increases, but the definition of the system in a lightning strike is more problematic.

      In 19.11 (e), the specified state is room temperature and 1 atm pressure. This means that H 2 O is present as a liquid; there is then 1 mol of gaseous reactants (CO 2 ) and 3 mol of gaseous products (CH 4 and 2 O 2 ), so S sys increases. (The reaction is not spontaneous because of the very large positive ΔH sys for the reaction as written.)

    19.34

    1. Solids are much more ordered than gases, so ΔS is negative.
    2. The entropy of the system increases in Exercise 19.12 (a) and (e). There is more motional freedom for the system in both cases. In (b), (c), and (d), there is less motional freedom after the change and the entropy of the system decreases.

    19.35 Analyze/Plan .  Consider the conditions that lead to an increase in entropy: more mol gas in products than reactants, increase in volume of sample and, therefore, number of possible arrangements, more motional freedom of molecules, and so on. Solve .

    1. S increases; translational motion is greater in the liquid than the solid.
    2. S decreases; volume and translational motion decrease going from the gas to the liquid.
    3. S increases; volume and translational motion are greater in the gas than the solid.

    19.36

    1. When temperature increases, the range of accessible molecular speeds and kinetic energies increases. This produces more microstates and an increase in entropy.
    2. When the volume of a gas increases (even at constant T), there are more possible positions for the particles, more microstates, and greater entropy.
    3. When equal volumes of two miscible liquids are mixed, the volume of the sample and, therefore, the number of possible arrangements increases. This produces more microstates and an increase in entropy.

    19.37

    1. False. The entropy of a pure crystalline substance at absolute zero is zero.
    2. True.
    3. False. Monoatomic gases have no rotational or vibrational states.
    4. True.

    19.38

    1. True. (From the Boltzmann relationship, S = k lnW.)
    2. False. The translational, rotational, and vibrational degrees of freedom of CO 2 (or any molecule) are determined by its structure. As long as heating does not change the basic molecular structure, it does not change the degrees of freedom. The additional kinetic energy due to heating is distributed as more translational, vibrational, and rotational motion. That is, number of microstates [see Solution 19.30 (a)] increases, but the degrees of freedom do not.
      image
    3. False. At a given temperature, CO 2 (g) has more microstates and, thus, greater entropy than Ar(g). Because CO 2 (g) is a triatomic molecule, it has multiple rotational and vibrational microstates not available to monatomic Ar(g).

    19.39 Analyze/Plan .  Consider the factors that lead to higher entropy: more mol gas in products than reactants, increase in volume of sample and, therefore, number of possible arrangements, more motional freedom of molecules, and so on. Solve .

    1. Ar(g) (gases have higher entropy due primarily to much larger volume)
    2. He(g) at 1.5 atm (larger volume and more motional freedom)
    3. 1 mol of Ne(g) in 15.0 L (larger volume provides more motional freedom)
    4. CO 2 (g) (more motional freedom)

    19.40

    1. One mole of O 3 (g) at 300 °C, 0.01 atm (O 3 (g) is a more complex molecule and has more vibrational degrees of freedom.)
    2. 1 mol H 2 O(g) at 100 °C, 1 atm [larger volume occupied by H 2 O(g)]
    3. 0.5 mol CH 4 (g) at 298 K, 20-L volume (more complex molecule, more rotational and vibrational degrees of freedom)
    4. 100 g of Na 2 SO 4 (aq) at 30 °C (more motional freedom in aqueous solution)

    19.41 Analyze/Plan .  Consider the markers of an increase in entropy for a chemical reaction: liquids or solutions formed from solids, gases formed from either solids or liquids, increase in mol gas during reaction. Solve .

    1. ΔS negative (moles of gas decrease)
    2. ΔS positive (gas produced, increased disorder)
    3. ΔS negative (moles of gas decrease)
    4. ΔS is small and probably positive [moles of gas same in reactants and products, H 2 O(g) is more structurally complex than H 2 (g)]

    19.42

    1. Au(l) → Au(s); negative ΔS, less motional freedom in the solid
    2. Cl 2 (g) → 2 Cl(g); positive ΔS, moles of gas increase
    3. CO(g) + 2 H 2 (g) → CH 3 OH(l); negative ΔS, moles of gas decrease
    4. 3 Ca(NO 3 ) 2 (aq) + 2 (NH 4 ) 3 PO 4 (aq) → Ca 3 (PO 4 ) 2 (s) + 6 NH 4 NO 3 (aq); ΔS is negative, less motional freedom, fewer moles of ions in aqueous solution

    Entropy Changes in Chemical Reactions (Section 19.4)

    19.43

    1. image
    2. Boiling water, at 100 °C, has a much larger entropy change than melting ice at 0 °C. Before and after melting, H 2 O molecules are touching. And there is actually a small decrease in volume going from solid to liquid water. Boiling drastically increases the distance between molecules and the volume of the sample. The increase in available molecular positions is much greater for boiling than melting, so the entropy change is also greater.

    19.44  Melting = –126.5 °C; boiling = 97.4 °C

    image

    19.45 Analyze/Plan .  Given two molecules in the same state, predict which will have the higher molar entropy. In general, for molecules in the same state, the more atoms in the molecule, the more degrees of freedom, the greater the number of microstates, and the higher the standard entropy, S°.

    1. C 2 H 6 (g) has more degrees of freedom and larger S°.
    2. CO 2 (g) has more degrees of freedom and larger S°.

    19.46  Propylene will have a higher S° at 25 °C. At this temperature, both are gases, so there are no lattice effects. Because they have the same molecular formula, only the details of their structures are different. In propylene, there is free rotation around the C–C single bond, whereas in cyclopropane the 3-membered ring severely limits rotation. The greater motional freedom of the propylene molecule leads to a higher absolute entropy.

    19.47 Analyze/Plan .  Consider the conditions that lead to an increase in entropy: more mol gas in products than reactants, increase in volume of sample and, therefore, number of possible arrangements, more motional freedom of molecules, and so on. Solve .

    1. Sc(g) will have the higher standard entropy at 25 °C. In general, the gas phase of a substance has a larger S° than the solid phase because of the greater volume and motional freedom of the molecules. Sc(s), 34.6 J/mol-K; Sc(g), 174.7 J/mol-K.
    2. NH 3 (g) will have the higher standard entropy at 25 °C. Molecules in the gas phase have more motional freedom than molecules in solution. NH 3 (g), 192.5 J/mol-K; NH 3 (aq), 111.3 J/mol-K.
    3. O 3 (g) will have the higher standard entropy at 25 °C. The triatomic molecule will have more vibrational degrees of freedom than the diatomic molecule. O 2 (g), 205.0 J/K; O 3 (g), 237.6 J/K.
    4. C(graphite) will have the higher standard entropy at 25 °C. Diamond is a network covalent solid with each C atom tetrahedrally bound to four other C atoms. Graphite consists of sheets of fused planar 6-membered rings with each C atom bound in a trigonal planar arrangement to three other C atoms. The internal entropy in graphite is greater because there is translational freedom among the planar sheets of C atoms while there is very little vibrational freedom within the network covalent diamond lattice. C(diamond), 2.43 J/mol-K; C(graphite) 5.69 J/mol-K.

    19.48

    1. C 6 H 6 (g) will have the higher standard entropy at 25 °C. Molecules in the gas phase have larger volume and greater motional freedom than those in the liquid state. C 6 H 6 (l), 172.8 J/K; C 6 H 6 (g), 269.2 J/K.
    2. CO 2 (g) will have the higher standard entropy at 25 °C. The more complex CO 2 molecule has more vibrational degrees of freedom and a slightly higher entropy. CO(g), 197.9 J/mol-K; CO 2 (g), 213.6 J/mol-K.
    3. Two moles of NO 2 (g) will have the higher standard entropy at 25 °C. More particles have a greater number of arrangements or microstates. 1 mol N 2 O 4 (g), 304.3 J/K; 2 mol NO 2 (g), 2(240.45) = 480.90 J/K.
    4. HCl(g) will have a greater standard entropy at 25 °C. The greater motional freedom of HCl molecules in the gas phase outweighs the greater number of particles in HCl(aq). [HCl(aq) is ionized into H + (aq) and Cl (aq).] HCl(g), 186.69 J/K; HCl(aq), 56.5 J/K.

    19.49  For elements with similar structures, the heavier the atoms, the lower the vibrational frequencies at a given temperature. This means that more vibrations can be accessed at a particular temperature resulting in a greater absolute entropy for the heavier elements.

    19.50

    1. C(diamond), S° = 2.43 J/mol-K; C(graphite), S° = 5.69 J/mol-K. Diamond is a network covalent solid with each C atom tetrahedrally bound to four other C atoms. Graphite consists of sheets of fused planar 6-membered rings with each C atom bound in a trigonal planar arrangement to three other C atoms. The internal entropy in graphite is greater because there is translational freedom among the planar sheets of C atoms whereas there is very little translational or vibrational freedom within the covalent-network diamond lattice.
    2. S° for buckminsterfullerene will be ≥ 10 J/mol-K. S° for graphite is twice S° for diamond, and S° for the fullerene should be higher than that of graphite. The 60-atom “bucky” balls have more flexibility than graphite sheets. Also, the balls have translational freedom in three dimensions, whereas graphite sheets have it in only two directions. Because of the ball structure, there is more empty space in the fullerene lattice than in graphite or diamond; essentially, 60 C-atoms in fullerene occupy a larger volume than 60 C-atoms in graphite or diamond. Thus, the fullerene has additional “molecular” complexity, more degrees of translational freedom, and occupies a larger volume, all features that point to a higher absolute entropy.

    19.51 Analyze/Plan .  Follow the logic in Sample Exercise 19.5. Solve .

    1. ΔS° = S° C 2 H 6 (g) – S° C 2 H 4 (g) – S° H 2 (g)

      = 229.5 – 219.4 – 130.58 = –120.5 J/K

      ΔS° is negative because there are fewer moles of gas in the products.

    2. ΔS° = 2S° NO 2 (g) – ΔS° N 2 O 4 (g) = 2(240.45) – 304.3 = +176.6 J/K

      ΔS° is positive because there are more moles of gas in the products.

    3. ΔS° = ΔS° BeO(s) + ΔS° H 2 O(g) – ΔS° Be(OH) 2 (s)

      = 13.77 + 188.83 – 50.21 = +152.39 J/K

      ΔS° is positive because the product contains more total particles and more moles of gas.

    4. ΔS° = 2S° CO 2 (g) + 4S° H 2 O(g) – 2S° CH 3 OH(g) – 3S° O 2 (g)

      = 2(213.6) + 4(188.83) – 2(237.6) – 3(205.0) = +92.3 J/K

      ΔS° is positive because the product contains more total particles and more moles of gas.

    19.52

    1. ΔS° = S° NH 4 NO 3 (s) – S° HNO 3 (g) – S° NH 3 (g)

      = 151 – 266.4 – 192.5 = –307.9 = –308 J/K

      ΔS° is large and negative because all reactants are gases (2 moles) and the product is a solid.


    1. ΔS° = 4S° Fe(s) + 3S° O 2 (g) – 2S° Fe 2 O 3 (s)

      = 4(27.15) + 3(205.0) – 2(89.96) = 543.68 = 543.7 J/K

      ΔS° is large and positive because the reaction produces 3 moles of gas and the reactant is a solid.

    1. ΔS° = S° CaCl 2 (s) + S° CO 2 (g) + S° H 2 O(l) – S° CaCO 3 (s) – 2S° HCl(g)

      = 104.6 + 213.6 + 69.91 – 92.88 –2(186.69) = –78.15 J/K

      ΔS° is small and negative because the products contain one fewer mole of gas, but one more mole of liquid. Note the very small standard entropy for H 2 O(l), owing to its strength of hydrogen bonding. If the products included one mole of a different liquid, the magnitude of the entropy change would be even smaller.

    1. ΔS° = S° C 6 H 6 (l) + 6S° H 2 (g) – 3S° C 2 H 6 (g)

      = 172.8 + 6(130.58) – 3(229.5) = 267.78 = 267.8 J/K

      ΔS° is positive because there are more moles of gas in the products.

    Gibbs Free Energy (Sections 19.5 and 19.6)

    19.53

    1. Yes. ΔG = ΔH – TΔS
    2. No. If ΔG is positive, the process is nonspontaneous.
    3. No. There is no relationship between ΔG and rate of reaction. A spontaneous reaction, one with a –ΔG, may occur at a very slow rate. For example, 2 H 2 (g) + O 2 (g) → 2 H 2 O(g), ΔG = –457 kJ is very slow if not initiated by a spark.

    19.54

    1. No. ΔG = ΔG° + RT lnQ. The relative magnitudes of ΔG and ΔG° depend on the value of Q.
    2. For a process that occurs at constant temperature and pressure, the system is at equilibrium when ΔG = 0.
    3. No. Activation energy is related to the rate constant, k. The sign and magnitude of ΔG give no information about rate.

    19.55 Analyze/Plan .  Consider the definitions of ΔH°, ΔS°, and ΔG°, along with sign conventions. ΔG° = ΔH° – TΔS°. Solve .

    1. ΔH° is negative; the reaction is exothermic.
    2. ΔS° is negative; the reaction leads to decrease in disorder (increase in order) of the system.
    3. ΔG° = ΔH° – TΔS° = –35.4 kJ – 298 K (–0.0855 kJ/K) = –9.921 = –9.9 kJ
    4. At 298 K, ΔG° is negative. If all reactants and products are present in their standard states, the reaction is spontaneous (in the forward direction) at this temperature.

    19.56

    1. ΔH° is positive; the reaction is endothermic.
    2. ΔS° is positive; the reaction leads to an increase in disorder.
    3. ΔG° = ΔH° – TΔS° = 23.7 kJ – 298 K (0.0524 kJ/K) = 8.0848 = 8.08 kJ

    1. At 298 K, ΔG° is positive. If all reactants and products are present in their standard states, the reaction is spontaneous in the reverse direction at this temperature; it is nonspontaneous in the forward direction.

    19.57 Analyze/Plan .  Follow the logic in Sample Exercises 19.6 and 19.7. Calculate ΔH° according to Equation 5.31, ΔS° by Equation 19.8, and ΔG° by Equation 19.14. Then, use ΔH° and ΔS° to calculate ΔG° using Equation 19.12, ΔG° = ΔH° – TΔS°. Solve .

    1. ΔH° = 2(–268.61) – [0 + 0] = –537.22 kJ

      ΔS° = 2(173.51) – [130.58 + 202.7] = 13.74 = 13.7 J/K

      ΔG° = 2(–270.70) – [0 + 0] = –541.40 kJ

      ΔG° = –537.22 kJ – 298(0.01374) kJ = –541.31 kJ

    2. ΔH° = –106.7 – [0 + 2(0)] = –106.7 kJ

      ΔS° = 309.4 – [5.69 + 2(222.96)] = –142.21 = –142.2 J/K

      ΔG° = –64.0 – [0 + 2(0)] = –64.0 kJ

      ΔG° = –106.7 kJ – 298(–0.14221) kJ = –64.3 kJ

    3. ΔH° = 2(–542.2) – [2(–288.07) + 0] = –508.26 = –508.3 kJ

      ΔS° = 2(325) – [2(311.7) + 205.0] = –178.4 = –178 J/K

      ΔG° = 2(–502.5) – [2(–269.6) + 0] = –465.8 kJ

      ΔG° = –508.26 kJ – 298(–0.1784) kJ = –455.097 = –455.1 kJ

      (The discrepancy in ΔG° values is because of experimental uncertainties in the tabulated thermodynamic data.)

    4. ΔH° = –84.68 + 2(–241.82) – [2(–201.2) + 0] = –165.92 = –165.9 kJ

      ΔS° = 229.5 + 2(188.83) – [2(237.6) + 130.58] = 1.38 = 1.4 J/K

      ΔG° = –32.89 + 2(–228.57) – [2(–161.9) + 0] = –166.23 = –166.2 kJ

      ΔG° = –165.92 kJ – 298(0.00138) kJ = –166.33 = –166.3 kJ

    19.58

    1. ΔH° =2( –1139.7) – 4(0) + 3(0) = –2279.4 kJ

      ΔS° = 2(81.2) – 4(23.6) – 3(205.0) = –547.0 J/K

      ΔG° = 2(–1058.1) – 4(0) – 3(0) = –2116.2 kJ

      ΔG° = –2279.4 kJ –298 K(–0.5470 kJ/K) = –2116.4 kJ

    2. ΔH° = –553.5 – 393.5 – (–1216.3) = 269.3 kJ

      ΔS° = 70.42 + 213.6 – 112.1 = 171.92 = 171.9 J/K

      ΔG° = –525.1 – 394.4 – (–1137.6) = 218.1 kJ

      ΔG° = 269.3 kJ – 298 K (0.1719 kJ/K) = 218.1 kJ


    1. Assume the reactant is P(g), not P(s).

      ΔH° = 2(–1594.4) + 5(0) – 2(316.4) – 10( –268.61) = –1135.5 kJ

      ΔS° = 2(300.8) + 5(130.58) – 2(163.2) –10(173.51) = –807.0 J/K

      ΔG° = 2(–1520.7) + 5(0) – 2(280.0) –10(–270.70) = –894.4 kJ

      ΔG° = –1135.5 kJ – 298 K(–0.8070 kJ/K) = –895.014 = –895.0 kJ

      (The small discrepancy in ΔG° values is because of experimental uncertainties in tabulated thermodynamic data.)

    1. ΔH° = –284.5 – (0) –(0) = –284.5 kJ

      ΔS° = 122.5 – 64.67 – 205.0 = –147.2 J/K

      ΔG° = –240.6 – (0) – (0) = –240.6 kJ

      ΔG° = –284.5 kJ – 298 K (–0.1472 kJ/K) = –240.634 = –240.6 kJ

    19.59 Analyze/Plan .  Follow the logic in Sample Exercise 19.7. Solve .

    1. ΔG° = 2ΔG° SO 3 (g) – [2ΔG° SO 2 (g) + ΔG° O 2 (g)]

      = 2(–370.4) – [2(–300.4) + 0] = –140.0 kJ, spontaneous

    2. ΔG° = 3ΔG° NO(g) – [ΔG° NO 2 (g) + ΔG° N 2 O(g)]

      = 3(86.71) – [51.84 + 103.59] = +104.70 kJ, nonspontaneous

    3. ΔG° = 4ΔG° FeCl 3 (s) + 3ΔG° O 2 (g) – [6ΔG° Cl 2 (g) + 2ΔG° Fe 2 O 3 (s)]

      = 4(–334) + 3(0) – [6(0) + 2(–740.98)] = +146 kJ, nonspontaneous

    4. ΔG° = ΔG° S(s) + 2ΔG° H 2 O(g) – [ΔG° SO 2 (g) + 2ΔG° H 2 (g)]

      = 0 + 2(–228.57) – [(–300.4) + 2(0)] = –156.7 kJ, spontaneous

    19.60

    1. ΔG° = 2ΔG° AgCl(s) – [2ΔG° Ag(s) + ΔG° Cl 2 (g)]

      = 2(–109.7) – 2(0) – 0 = –219.4 kJ, spontaneous

    2. P 4 O 10 (s) + 16 H 2 (g) → 4 PH 3 (g) + 10 H 2 O(g)

      ΔG° = 4ΔG° PH 3 (g) + 10ΔG° H 2 O(g) – [ΔG° P 4 O 10 (s) + 16ΔG° H 2 (g)]

      = 4(13.4) + 10(–228.57) – [–2675.2] – 16(0) = 443.1 kJ, nonspontaneous

    3. ΔG° = ΔG° CF 4 (g) + 4ΔG° HF(g) – [ΔG° CH 4 (g) + 4ΔG° F 2 (g)]

      = –635.1 + 4(–270.70) – (–50.8) –4(0) = –1667.1 kJ, spontaneous

    4. ΔG° = 2ΔG° H 2 O(l) + ΔG° O 2 (g) – 2ΔG° H 2 O 2 (l)

      = 2(–237.13) + 0 – 2(–120.4) = –233.5 kJ, spontaneous

    19.61 Analyze/Plan .  Follow the logic in Sample Exercise 19.8(a). Solve .

    1. 2 C 8 H 18 (l) + 25 O 2 (g) → 16 CO 2 (g) + 18 H 2 O(l)
    2. Because there are more moles of gas in the reactants, ΔS° is negative, which makes –TΔS positive. ΔG° is less negative than ΔH°. (This argument is true for the reaction as written. If the products are all in the gas phase, there are more moles of gas in the products and ΔG° is more negative than ΔH°.)

    19.62

    1. ΔG° should be less negative than ΔH°. Products contain fewer moles of gas, so ΔS° is negative. ΔG° = ΔH° – TΔS°; –TΔS° is positive, so ΔG° is less negative than ΔH°.
    2. We can estimate ΔS° using a similar reaction and then use ΔG° = ΔH° – TΔS° (estimate) to get a ballpark figure. There are no sulfite salts listed in Appendix C, so use a reaction such as CO 2 (g) + CaO(s) → CaCO 3 (s) or CO 2 (g) + BaO(s) → BaCO 3 (s). Or, calculate both ΔS° values and use the average as your estimate.

    19.63 Analyze/Plan .  Based on the signs of ΔH and ΔS for a particular reaction, assign a category from Table 19.3 to each reaction. Solve .

    1. (iii) ΔG is negative at low temperatures, positive at high temperatures. That is, the forward reaction is spontaneous at lower temperatures but not spontaneous at higher temperatures.
    2. (ii) ΔG is positive at all temperatures. The forward reaction is nonspontaneous at all temperatures.
    3. (iv) ΔG is positive at low temperatures, negative at high temperatures. That is, the forward reaction will proceed spontaneously at high temperature, but not at low temperature.

    19.64  ΔG° = ΔH° – TΔS°

    1. ΔG° = –844 kJ – 298 K(–0.165 kJ/K) = –795 kJ, spontaneous
    2. ΔG° = +572 kJ – 298 K(0.179 kJ/K) = +519 kJ, nonspontaneous

      To be spontaneous, ΔG must be negative (ΔG < 0).

      Thus, ΔH° – TΔS° < 0; ΔH° < TΔS°; T > ΔH°/ΔS°; T > 572 kJ 0 .179 kJ/K = 3.20 × 10 3 K

    19.65 Analyze/Plan .  We are told that the reaction is barely spontaneous and endothermic, and asked to estimate the sign and magnitude of ΔS. If a reaction is spontaneous, ΔG < 0. Use this information with Equation 19.11 to solve the problem. Solve .

    At 390 K, ΔG < 0; ΔG = ΔH – TΔS < 0

    23.7 kJ – 390 K (ΔS) < 0; 23.7 kJ < 390 K (ΔS); ΔS > 23.7 kJ/390 K

    ΔS > 0.06077 kJ/K or ΔS > 60.8 J/K

    The reaction is spontaneous and endothermic, so the sign of ΔS must be positive. Because the reaction is “barely” spontaneous, the magnitude will not be much greater than 61 J/K.

    19.66  At 45 °C or 318 K, ΔG > 0. ΔG = ΔH – TΔS > 0

    ΔH – 318 K (72 J/K) > 0; ΔH > +2.3 × 10 4 J; ΔH > +23 kJ

    The reaction is nonspontaneous and has a positive ΔS, so it must be endothermic. Because it is “barely” nonspontaneous, the magnitude will not be much greater than 23 kJ.

    19.67 Analyze/Plan .  Use Equation 19.11 to calculate T when ΔG = 0. This is similar to calculating the temperature of a phase transition in Sample Exercise 19.10. Use Table 19.3 to determine whether the reaction is spontaneous or nonspontaneous above this temperature. Solve .


    1. ΔG = ΔH – TΔS; 0 = –32 kJ – T(–98 J/K); 32 × 10 3 J = T(98 J/K)

      T = 32 × 10 3 J/(98 J/K) = 326.5 = 330 K

    2. Nonspontaneous. The sign of ΔS is negative, so as T increases, ΔG becomes more positive.

    19.68  ΔG is negative when TΔS > ΔH or T > ΔH/ΔS.

    ΔH° = ΔH° CH 3 OH + ΔH° CO(g) – ΔH° CH 3 COOH(l)

    = –201.2 – 110.5 – (–487.0) = 175.3 kJ

    ΔS° = S° CH 3 OH + S° CO(g) - S° CH 3 COOH(l) = 237.6 + 197.9 – 159.8 = 275.7 J/K

    T > 175.3 kJ 0.2757 kJ/K = 635.8 K

    The reaction is spontaneous above 635.8 K (363 °C).

    19.69 Analyze/Plan .  Given a chemical equation and thermodynamic data (values of Δ H f ° , Δ G f ° , and S°) for reactants and products, predict the variation of ΔG° with temperature and calculate ΔG° at 800 K and 1000 K. Use Equations 5.31 and 19.8 to calculate ΔH° and ΔS°, respectively; use these values to calculate ΔG° at various temperatures, using Equation 19.12. The signs of ΔH° and ΔS° determine the variation of ΔG° with temperature. Solve .

    1. Calculate ΔH° and ΔS° to determine the sign of TΔS°.

      ΔH° = 3ΔH° NO(g) – ΔH° NO 2 (g) – ΔH° N 2 O(g)

      = 3(90.37) – 33.84 – 81.6 = 155.7 kJ

      ΔS° = 3S° NO(g) – S° NO 2 (g) – S° N 2 O(g)

      = 3(210.62) – 240.45 – 220.0 = 171.4 J/K

      ΔG° = ΔH° – TΔS°. Because ΔS° is positive, –TΔS° becomes more negative as T increases and ΔG° becomes more negative.

    2. ΔG° = ΔH° – TΔS° = 155.7 kJ – (800 K)(0.1714 kJ/K)

      ΔG° = 155.7 kJ – 137 kJ = 19 kJ

      Because ΔG° is positive at 800 K, the reaction is not spontaneous at this temperature.

    3. ΔG° = 155.7 kJ – (1000 K)(0.1714 kJ/K) = 155.7 kJ – 171.4 kJ = –15.7 kJ

      ΔG° is negative at 1000 K and the reaction is spontaneous at this temperature.

    19.70

    1. Δ H o = Δ H f o CH 3 OH(g ) Δ H f o CH 4 ( g ) 1/2 Δ H f o O 2 ( g )

      = –201.2 – (–74.8) – (1/2)(0) = –126.4 kJ

      ΔS° = S° CH 3 OH(g) – S° CH 4 (g) – 1/2 S° O 2 (g)

      = 237.6 – 186.3 – 1/2(205.0) = –51.2 J/K = –0.0512 kJ/K

    2. ΔG° = ΔH° – TΔS°. –TΔS° is positive, so ΔG° will increase (becomes more positive) as temperature increases.

    1. ΔG° = ΔH° – TΔS° = –126.4 kJ – 298 K(–0.0512 kJ/K) = –111.1 kJ

      The reaction is spontaneous at 298 K because ΔG° is negative at this temperature. In this case, ΔG° could have been calculated from Δ G f ° values in Appendix C, because these values are tabulated at 298 K.

    1. No. The reaction is at equilibrium when ΔG° = 0.

      ΔG° = ΔH° – TΔS° = 0. ΔH° = TΔS°, T = ΔH°/ΔS°

      T = –126.4 kJ/–0.0512 kJ/K = 2469 = 2470 K

      This temperature is so high that the reactants and products are likely to decompose. At standard conditions, equilibrium is functionally unattainable for this reaction.

    19.71 Analyze/Plan .  Follow the logic in Sample Exercise 19.10. Solve .

    1. Δ S vap o = Δ H vap o / T b ; T b = Δ H vap o / Δ S vap o

      Δ H vap o = Δ H o C 6 H 6 ( g ) Δ H o C 6 H 6 ( l ) = 82.9 49.0 = 33.9 kJ Δ S vap o = S o C 6 H 6 ( g ) S o C 6 H 6 ( l ) = 269.2 172.8 = 96.4 J/K

      T b = 33.9 × 10 3 J/96.4 J/K = 351.66 = 352 K = 79 °C

    2. From the Handbook of Chemistry and Physics , 74th edition, T b = 80.1 °C. The values are remarkably close; the small difference is due to deviation from ideal behavior by C 6 H 6 (g) and experimental uncertainty in both the boiling point measurement and the thermodynamic data.

    19.72

    1. The transformation of I 2 (s) to I 2 (g) is sublimation. The temperature at which the free-energy change for this process is zero is the sublimation temperature. According to Sample Exercise 19.10, T sub = Δ H sub o / Δ S sub o .

      Use data from Appendix C to calculate Δ H sub o and Δ S sub o for I 2 (s).

      I 2 ( s ) I 2 ( l ) melting I 2 ( l ) I 2 ( g ) boiling I 2 ( s ) I 2 ( g ) sublimation ¯

      Δ H sub o = Δ H f o I 2 ( g ) Δ H f o I 2 ( s ) = 62.25 0 = 62.25 kJ Δ S sub o = S o I 2 ( g ) S o I 2 ( s ) = 260.57 116.73 = 143.84 J/K = 0.14384 kJ/K

      T sub = Δ H sub o Δ S sub = 62.25 kJ 0 .14384 kJ/K = 432.8 K = 159 .6 o C

      [Note that some assumptions were necessary to make this estimate, that both I 2 (s) and I 2 (g) are present in their standard states and, consequently, Δ G sub = Δ G sub o = 0. We also assume that the values of Δ H sub o and Δ S sub o are the same at 298 K and at the sublimation temperature.]

    2. T m for I 2 (s) = 386.85 K = 113.7 °C; T b = 457.4 K = 184.3 °C (from WebElements TM 2013)

    1. The boiling point of I 2 is closer to the sublimation temperature. Both boiling and sublimation begin with molecules in a condensed phase (little space between molecules) and end in the gas phase (large intermolecular distances). Separation of the molecules is the main phenomenon that determines both ΔH and ΔS, so it is not surprising that the ratio of ΔH/ΔS is similar for sublimation and boiling.

    19.73 Analyze/Plan .  We are asked to write a balanced equation for the combustion of acetylene, calculate ΔH° for this reaction, and calculate maximum useful work possible by the system. Combustion is combination with O 2 to produce CO 2 and H 2 O. Calculate ΔH° using data from Appendix C and Equation 5.31. The maximum obtainable work is ΔG (Equation 19.18), which can be calculated from data in Appendix C and Equation 19.14. Solve .

    1. C 2 H 2 (g) + 5/2 O 2 (g) → 2 CO 2 (g) + H 2 O(l)
    2. ΔH° = 2ΔH° CO 2 (g) + ΔH° H 2 O(l) – ΔH° C 2 H 2 (g) – 5/2ΔH° O 2 (g)

      = 2(–393.5) – 285.83 – 226.77 – 5/2(0)

      = –1299.6 kJ produced/mol C 2 H 2 burned

    3. w max = ΔG° = 2ΔG° CO 2 (g) + ΔG° H 2 O(l) – ΔG° C 2 H 2 (g) – 5/2 ΔG° O 2 (g)

      = 2(–394.4) – 237.13 – 209.2 – 5/2(0) = –1235.1 kJ

      The negative sign indicates that the system does work on the surroundings; the system can accomplish a maximum of 1235.1 kJ of work on its surroundings.

    19.74

    1. CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)

      Δ H o = Δ H f o CO 2 ( g ) + 2 Δ H f o H 2 O(l) Δ H f o CH 4 ( g ) 2 Δ H f o O 2 ( g )

      = (–393.5) + 2(–285.83) – (–74.8) –2(0) = –890.4 kJ/mol CH 4 burned

    2. w max = Δ G o = Δ G f o CO 2 ( g ) + 2 Δ G f o H 2 O(l) Δ G f o CH 4 ( g ) 2 Δ G f o O 2 ( g )

      = (–394.4) + 2(–237.13) – (–50.8) –2(0) = –817.9 kJ

      The system can accomplish at most 817.86 kJ of work per mole of CH 4 on the surroundings.

    Free Energy and Equilibrium (Section 19.6)

    19.75 Analyze/Plan .  We are given a chemical reaction and asked to predict the effect of the partial pressure of O 2 (g) on the value of ΔG for the system. Consider the relationship ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. Solve .

    1. ΔG decreases. O 2 (g) appears in the denominator of Q for this reaction. An increase in pressure of O 2 decreases Q and ΔG gets smaller or becomes more negative. Increasing the concentration or partial pressure of a reactant increases the tendency for a reaction to occur.
    2. ΔG increases. O 2 (g) appears in the numerator of Q for this reaction. Increasing the pressure of O 2 increases Q and ΔG becomes more positive. Increasing the concentration or partial pressure of a product decreases the tendency for the reaction to occur.

    1. ΔG increases. O 2 (g) appears in the numerator of Q for this reaction. An increase in pressure of O 2 increases Q and ΔG becomes more positive. Because pressure of O 2 is raised to the third power in Q, an increase in pressure of O 2 will have the largest effect on ΔG for this reaction. Increasing the concentration or partial pressure of a product decreases the tendency for the reaction to occur.

    19.76  Consider the relationship ΔG = ΔG° + RT ln Q, where Q is the reaction quotient.

    1. ΔG decreases. H 2 (g) appears in the denominator of Q for this reaction. An increase in pressure of H 2 decreases Q and ΔG becomes smaller or more negative. Increasing the concentration or partial pressure of a reactant increases the tendency for a reaction to occur.
    2. ΔG increases. H 2 (g) appears in the numerator of Q for this reaction. Increasing the pressure of H 2 increases Q and ΔG becomes more positive. Increasing the concentration or partial pressure of a product decreases the tendency for the reaction to occur.
    3. ΔG decreases. H 2 (g) appears in the denominator of Q for this reaction. An increase in pressure of H 2 decreases Q and ΔG becomes smaller or more negative. Increasing the concentration or partial pressure of a reactant increases the tendency for a reaction to occur.

    19.77 Analyze/Plan .  Given a chemical reaction, we are asked to calculate ΔG° from Appendix C data and ΔG for a given set of initial conditions. Use Equation 19.14 to calculate ΔG° and Equation 19.19 to calculate ΔG. Follow the logic in Sample Exercise 19.11 when calculating ΔG. Solve .

    1. ΔG° = ΔG° N 2 O 4 (g) – 2ΔG° NO 2 (g) = 98.28 – 2(51.84) = –5.40 kJ
    2. Δ G = Δ + RT ln P N 2 O 4 / P NO 2 2 = 5.40 kJ + 8.314 × 10 3 kJ mol-K × 298 K × ln[1 .60/(0 .40) 2 ] = 0.3048 = 0.30 kJ

    19.78

    1. ΔG° = ΔG° C 3 H 8 (g) + 2ΔG° H 2 (g) → 3ΔG° CH 4 (g)

      = –23.47 + 2(0) – 3(–50.8) = 128.9 kJ

    2. ΔG = ΔG° + RT ln [ P C 3 H 8 × P H 2 2 /P CH 4 3 ]

      = 128.9 + 8.314 × 10 3 kJ mol-K × 298 K × ln[(0.0100) × (0.0180) 2 /(40.0) 3 ]

      = 128.9 – 58.735 = 70.165 = 70.2 kJ

    19.79 Analyze/Plan .  Given a chemical reaction, we are asked to calculate K using Δ G f ° data from Appendix C. Calculate ΔG° using Equation 19.14. Then, ΔG°= –RT ln K, Equation 19.20; ln K = –ΔG°/RT Solve .

    1. ΔG° = 2ΔG° HI(g) – ΔG° H 2 (g) – ΔG° I 2 (g)

      = 2(1.30) – 0 – 19.37 = –16.77 kJ

      ln K = ( 16.77 kJ) × 10 3 J/kJ 8.314 J/K × 298 K = 6.76876 = 6.769 ; K = 870


    1. ΔG° = ΔG° C 2 H 4 (g) + ΔG° H 2 O(g) – ΔG° C 2 H 5 OH(g)

      = 68.11 – 228.57 – (–168.5) = 8.04 = 8.0 kJ

      ln K = ( 8.04 kJ) × 10 3 J/kJ 8.314 J/K × 298 K = 3.24511 = 3.25 ; K = 0 .039

    1. ΔG° = ΔG° C 6 H 6 (g) – 3ΔG° C 2 H 2 (g) = 129.7 – 3(209.2) = –497.9 kJ

      ln K = Δ G o RT = ( 497.9 kJ) × 10 3 J/KJ 8.314 J/K × 298 K = 200.963 = 201.0 ; K = 2 × 10 87

    19.80  ΔG° = –RT ln K; ln K = –ΔG° / RT; at 298 K, RT = 2.4776 = 2.478 kJ

    1. ΔG° = ΔG° NaOH(s) + ΔG° CO 2 (g) – ΔG° NaHCO 3 (s)

      = –379.5 + (–394.4) – (–851.8) = +77.9 kJ

      ln K = Δ G o RT = 77.9 kJ 2 .478 kJ = 31.442 = 31.4 ; K = 2 × 10 14 K = P CO 2 = 2 × 10 14

    2. ΔG° = 2ΔG° HCl(g) + ΔG° Br 2 (g) – 2ΔG° HBr(g) – ΔG° Cl 2 (g)

      = 2(–95.27) + 3.14 – 2(–53.22) – 0 = –80.96 kJ

      ln K = ( 80.96 ) 2.4776 = + 32.68 ; K = 1 .6 × 10 14 K = P HCl 2 × P Br 2 P HBr 2 × P Cl 2 = 1.6 × 10 14

    3. From Solution 19.59(a), ΔG° at 298 K = –140.0 kJ.

      ln K = Δ G o RT = ( 140.0 ) 2.4776 = 56.51 ; K = 3 .5 × 10 24 K = P SO 3 2 P SO 2 2 × P O 2 = 3.5 × 10 24

    19.81 Analyze/Plan .  Given a chemical reaction and thermodynamic data in Appendix C, calculate the equilibrium pressure of CO 2 (g) at two temperatures. K = P CO 2 . Calculate ΔG° at the two temperatures using ΔG° = ΔH° – TΔS° and then calculate K and P CO 2 . Solve .

    ΔH° = ΔH° BaO(s) + ΔH° CO 2 (g) – ΔH° BaCO 3 (s)

    = –553.5 + –393.5 – (–1216.3) = +269.3 kJ

    ΔS° = S° BaO(s) + S° CO 2 (g) – S° BaCO 3 (s)

    = 70.42 + 213.6 – 112.1 = 171.92 J/K = 0.1719 kJ/K

    1. ΔG at 298 K = 269.3 kJ – 298 K (0.17192 kJ/K) = 218.07 = 218.1 kJ

      ln K = Δ G o RT = 218.07 × 10 3 J 8.314 J/K × 298 K = 88.017 = 88.02 K = 6 .0 × 10 39 ; P CO 2 = 6.0 × 10 39 atm


    1. ΔG at 1100 K = 269.3 kJ – 1100 K (0.17192 kJ) = 80.19 = +80.2 kJ

      ln K = Δ G o RT = 80. 19 × 10 3 J 8.314 J/K × 1100 K = 8.768 = 8.77 K = 1 .6 × 10 4 ; P CO 2 = 1.6 × 10 4 atm

    19.82 K = P CO 2 . Calculate ΔG° at the two temperatures using ΔG° = ΔH° – TΔS° and then calculate K and P CO 2 .

    ΔH° = ΔH° PbO(s) + ΔH° CO 2 (g) – ΔH° PbCO 3 (s)

    = –217.3 – 393.5 + 699.1 = 88.3 kJ

    ΔS° = S° PbO(s) + S° CO 2 (g) – S° PbCO 3 (s)

    = 68.70 + 213.6 – 131.0 = 151.3 J/K or 0.1513 kJ/K

    1. ΔG° = ΔH° – TΔS°. At 673 K, ΔG° = 88.3 kJ – 673 K(0.1513 kJ/K) = –13.525 = –13.5 kJ

      ln K = Δ G o RT = ( 13.525 × 10 3 ) J 8.314 J/K × 673 K = 2.4172 = 2.42 K = P CO 2 = 11.214 = 11 atm

    2. ΔG° = ΔH° – TΔS°. At 453 K, ΔG° = 88.3 kJ – 453 K (0.1513 kJ) = 19.7611 = 19.8 kJ

      ln K = ( 19.7611 × 10 3 J) 8.314 J/K × 453 K = 5.2469 = 5.25 ; K = P CO 2 = 5.3 × 10 3 atm

    19.83 Analyze/Plan .  Given an acid dissociation equilibrium and the corresponding K a value, calculate ΔG° and ΔG for a given set of concentrations. Use Equation 19.20 to calculate ΔG° and Equation 19.19 to calculate ΔG. Solve .

    1. HNO 2 (aq)  ⇌  H + (aq) + NO 2 (aq)
    2. ΔG° = –RT ln K a = –(8.314 × 10 –3 )(298) ln (4.5 × 10 –4 ) = 19.0928 = 19.1 kJ
    3. ΔG = 0 at equilibrium
    4. ΔG = ΔG° + RT ln Q

      = 19.09 kJ + (8 .314 × 10 3 ) ( 298 ) ln (5 .0 × 10 2 ) ( 6.0 × 10 4 ) 0.20 = 2.725 = 2.7 kJ

    19.84

    1. CH 3 NH 2 (aq) + H 2 O(l)  ⇌  CH 3 NH 3 + (aq) + OH (aq)
    2. ΔG° = –RT ln K b = –(8.314 × 10 –3 )(298) ln (4.4 × 10 –4 ) = 19.148 = 19.1 kJ
    3. ΔG = 0 at equilibrium
    4. ΔG = ΔG° + RT ln Q; [OH ] = 1 × 10 –14 /6.7 × 10 –9 = 1.4925 × 10 –6 = 1.5 × 10 –6

      = 19.148 + ( 8.314 × 10 3 ) ( 298 ) ln (2 .4 × 10 3 ) ( 1.4925 × 10 6 ) 0.098 = 23.28 = 23.3 kJ


    Additional Exercises

    19.85

    1. The thermodynamic quantities T, E, and S are state functions. T is directly related to the distribution of molecular speeds, which does not depend on the path from one state to another.
    2. The quantities q and w do depend on the path taken from one state to another.
    3. There is only one reversible path between states.
    4. Isothermal processes occur at constant T. As the process is reversible, q is q rev and w is w max .

      Δ E = q rev + w max ; Δ S = q rev T

    19.86

    1. The sign of ΔS° at room temperature is positive, because there are more moles of gas in the products than the reactants. (Also, because the process is spontaneous and ΔH° is positive at room temperature, ΔS° must be positive so that ΔG is negative.)
    2. ΔS° is defined as the entropy change for the reaction with all reactants and products in their standard states, calculated at 298 K. The value of ΔS° does not change with a change in reaction conditions. The value of ΔG will change, depending on the partial pressure of water vapor in the container.

    19.87

    1. False. The essential question is whether the reaction proceeds far to the right before arriving at equilibrium. The position of equilibrium, which is the essential aspect, is dependent not only on ΔH but on the entropy change as well.
    2. True.
    3. True.
    4. False. Non spontaneous processes in general require that work be done to force them to proceed. Spontaneous processes occur without application of work.
    5. False. Such a process might be spontaneous, but would not necessarily be so. Spontaneous processes are those that are exothermic and/or that lead to increased disorder in the system.

    19.88

    Process ΔH ΔS
    (a) + +
    (b)
    (c) + +
    (d) + +
    (e) +

    19.89  There is no mistake in the calculation. The second law states that in any spontaneous process there is an increase in the entropy of the universe. Although there may be a decrease in entropy of the system, as in the present case, this decrease is more than offset by an increase in entropy of the surroundings.


    19.90

    1. Microstates are possible arrangements for the system. For each die, there are six possibilities for the top face, resulting in (6)(6) = 36 possible arrangements or microstates. (The face that appears on top of one die is not related to or determined by the face on top of the other die.) The two arrangements of top faces shown in the exercise are two of the 36 possible microstates.
    2. The left pair of dice belongs to state III; the right pair belongs to state VII.
    3. There are eleven possible states (II through XII; I is not a possibility).
    4. The state with the most microstates has the highest entropy. State VII has six microstates and the highest entropy. The microstates are (1+6), (2+5), (3+4), (4+3), (5+2), and (6+1). States VI and VIII, on either side of VII, have five microstates. Moving farther away from VII, the number of microstates decreases until we reach the two extremes, II and XII, which each have one microstate.
    5. States II and XII, with one microstate each, have the lowest entropy.
    6. The Boltzmann relationship is S = k ln W, where S is the absolute entropy, k is the Botzmann constant, and W is the total number of microstates of the system.

      S = 1.38 × 10 –23 J/K (ln 36) = 4.9453 × 10 –23 = 4.95 × 10 –23 J/K

    19.91  If NH 4 NO 3 (s) dissolves spontaneously in water, ΔG = ΔH – TΔS. If ΔG is negative and ΔH is positive, the sign of ΔS must be positive. Furthermore, TΔS > ΔH at room temperature.

    19.92

    1. The sign of q for expansion is (+). Vaporization is an endothermic process; the enthaphy of the system increases and q is positive. Our system is the refrigerant. Because the expansion does not occur at constant pressure, q is not exactly equal to ΔH, but its sign is positive.
    2. The sign of q for compression is (–). Compression is the reverse of expansion, and it has the opposite sign.
    3. The expansion chamber is inside the house and the compression chamber is outside. During expansion, q sys increases and q sur decreases. The air surrounding the expansion chamber is cooled and then distributed throughout the house to cool it. If expansion occurred outside, the cool air would be wasted. Compression releases heat to the surroundings; it occurs outside so that the released heat can be dissipated by the outside air.
    4. No. Heat can flow reversibly between a system and its surroundings only if the two have an infinitesimally small difference in temperature and the amount of heat transferred is infinitesimally small. There is no mechanism in our system to regulate the amount of heat transferred. When the liquid flows into the low pressure chamber, all the liquid vaporizes, not an infinitesimally small amount.
    5. A spontaneous process occurs without outside intervention. In an air conditioner, expansion (vaporization) of the refrigerant is spontaneous, but compression (condensation) to the liquid state is nonspontaneous. Cooling the house from 31 o C to 24 o C is nonspontaneous. [Note that all spontaneous processes are irreversible, but not all irreversible processes are spontaneous.]

    19.93  At the normal boiling point of a liquid, ΔG = 0 and ΔH vap = TΔS vap ; T = ΔH vap /ΔS vap . By Trouton’s rule, ΔS vap = 88 J/mol-K. The process of vaporization is:

    1. Br 2 (l)  ⇌  Br 2 (g)

      Δ H vap = Δ H f o Br 2 ( g ) Δ H f o Br 2 ( l ) = 30.71 kJ 0 = 30 .71 kJ T b = Δ H vap Δ S vap = 30.71 kJ 88 J/mol-K × 1000 J kJ = 349 = 3 .5 × 10 2 K

    2. According to WebElements™ 2013, the normal boiling point of Br 2 (l) is 332 K. Trouton’s rule provides a good ”ballpark” estimate.

      Trouton’s rule assumes that the entropy of vaporization of most molecules is due to the much greater motional freedom of the gaseous state relative to the liquid state. This assumption is incorrect for liquids with strong intermolecular forces (usually hydrogen bonding) in either the liquid or the gaseous state.

      There are also the usual experimental uncertainties in the measurement of Δ H f o for Br 2 ( g ) and the normal boiling point of Br 2 (l).

    19.94

    1. 1/2 N 2 (g) + 3/2 H 2 (g) → NH 3 (g)

      C(s) + 1/2 O 2 (g) → CO(g)

    2. In the first reaction, there are fewer moles of gas in the products than reactants, so ΔS is negative. If Δ G f o = Δ H f o T Δ S f o and Δ S f o is negative, T Δ S f o is positive and Δ G f o is more positive than Δ H f o .
    3. Condition (iii), when Δ S f o is negative.

    19.95

      1. Ti(s) + 2 Cl 2 (g) → TiCl 4 (g)

        ΔH° = ΔH° TiCl 4 (g) – ΔH° Ti(s) – 2ΔH° Cl 2 (g)

        = –763.2 – 0 – 2(0) = –763.2 kJ

        ΔS° = 354.9 – 30.76 – 2(222.96) = –121.78 = –121.8 J/K

        ΔG° = –726.8 – 0 – 2(0) = –726.8 kJ

        ln K = Δ G o RT = ( 726.8 ) kJ 2 .4777 kJ = 293.337 = 293.3 ; K = 2 × 10 127

      2. C 2 H 6 (g) + 7 Cl 2 (g) → 2 CCl 4 (g) + 6 HCl(g)

        ΔH° = 2ΔH° CCl 4 (g) + 6ΔH° HCl(g) – ΔH° C 2 H 6 (g) – 7ΔH° Cl 2 (g)

        = 2(–106.7) +6(–92.30) – (–84.68) – 7(0) = –682.52 = –682.5 kJ

        ΔS° = 2(309.4) + 6(186.69) – 229.5 – 7(222.96) = –51.28 = –51.4 J/K

        ΔG° = 2(–64.0) + 6( –95.27) – (–32.89) – 7(0) = –666.73 = –666.7 kJ

        l n K = Δ G o RT = ( 666.73 ) kJ 2 .4777 kJ = 269.0923 = 269.1 ; K = 7 × 10 116


    1. (iii).     BaO(s) + CO 2 (g) → BaCO 3 (s)

      ΔH° = ΔH° BaCO 3 (s) – ΔH° BaO(s) – ΔH° CO 2 (g)

      = –1216.3 – (– 553.5) – (–393.5) = –269.3 kJ

      ΔS° = 112.1 – 70.42 – 213.6 = –171.9 J/K

      ΔG° = –1137.6 – (–525.1) – (–394.4) = –218.1 kJ

      l n K = Δ G o RT = ( 218.1 ) kJ 2 .4777 kJ = 88.0252 = 88.03 ; K = 1.7 × 10 38

    1. (i), (ii), and (iii) all have negative ΔG° values and are spontaneous at standard conditions and 25 °C. Essentially, these reactions all go to completion; they have unimaginably large K values.
    1. ΔG° = ΔH° – TΔS°. All three reactions have –ΔH° and –ΔS°. They all have –ΔG° at 25 °C, and ΔG° becomes more positive as T increases.

    19.96  ΔG = ΔG° + RT ln Q; ln K = –ΔG°/RT

    1. Q = P NH 3 2 P N 2 × P H 2 3 = ( 1.2 ) 2 ( 2.6 ) ( 5.9 ) 3 = 2.697 × 10 3 = 2.7 × 10 3

      ΔG° = 2ΔG° NH 3 (g) – ΔG° N 2 (g) – 3ΔG° H 2 (g)

      = 2(–16.66) – 0 – 3(0) = –33.32 kJ

      ln K = Δ G o RT = ( 33.32 ) kJ 2 .4777 kJ = 13.448 = 13.45 ; K = 6.9 × 10 5

      Δ G = 33.32 kJ + 8 .314 × 10 3 kJ mol-K × 298 K × ln ( 2.69 × 10 3 )

      ΔG = –33.32 – 14.66 = –47.98 = –48.0 kJ

    2. Q = P N 2 3 × P H 2 O 4 P N 2 H 4 2 × P NO 2 2 = ( 0.5 ) 3 ( 0.3 ) 4 ( 5.0 × 10 2 ) 2 ( 5.0 × 10 2 ) 2 = 162 = 2 × 10 2

      ΔG° = 3ΔG° N 2 (g) + 4ΔG° H 2 O(g) – 2ΔG° N 2 H 4 (g) – 2ΔG° NO 2 (g)

      = 3(0) + 4(–228.57) – 2(159.4) – 2(51.84) = –1336.8 kJ

      ln K = Δ G o RT = ( 1336.8 ) kJ 2 .4777 kJ = 539.533 = 539.53 ; K = 2.1 × 10 234

      ΔG = –1336.8 kJ + 2.478 ln 162 = –1324.2 = –1.32 x 10 3 kJ

    3. Q = P N 2 × P H 2 2 P N 2 H 4 = ( 1.5 ) ( 2.5 ) 2 0.5 = 18.75 = 2 × 10 1

      ΔG° = ΔG° N 2 (g) + 2ΔG° H 2 (g) – ΔG° N 2 H 4 (g)

      = 0 + 2(0) – 159.4 = –159.4 kJ

      ln K = Δ G o RT = ( 159.4 ) kJ 2 .4777 kJ = 64.334 = 64.33 ; K = 8.7 × 10 27

      ΔG = –159.4 kJ + 2.478 ln 18.75 = –152.1 = –152 kJ


    19.97

    Reaction (a) Sign of ΔH° (a) Sign of ΔS° (b) K > 1? (c) Variation in K as Temp. Increases
    (i) yes decrease
    (ii) + + no increase
    (iii) + + no increase
    (iv) + + no increase
    1. Note that at a particular temperature, positive ΔH° leads to a smaller value of K, whereas positive ΔS° increases the value of K.

    19.98

    1. K = X CH 3 COOH X CH 3 OH P CO

      ΔG° = –RT ln K; ln K = –ΔG/RT

      ΔG° = ΔG° CH 3 COOH(l) – ΔG° CH 3 OH(l) – ΔG° CO(g)

      = –392.4 – (–166.23) – (–137.2) = –89.0 kJ

      ln K = ( 89.0 kJ) ( 8.314 × 10 3 kJ/K) (298 K) = 35.922 = 35.9 ; K = 4 × 10 15

    2. ΔH° = ΔH° CH 3 COOH(l) – ΔH° CH 3 OH(l) – ΔH° CO(g)

      = –487.0 – (–238.6) – (–110.5) = –137.9 kJ

      The reaction is exothermic, so the value of K will decrease with increasing temperature, and the mole fraction of CH 3 COOH will also decrease. Elevated temperatures must be used to increase the speed of the reaction. Thermodynamics cannot predict the rate at which a reaction reaches equilibrium.

    3. ΔG° = –RT ln K; K = 1, ln K = 0, ΔG° = 0

      ΔG° = ΔH° – TΔS°; when ΔG° = 0, ΔH° = TΔS°

      ΔS° = S° CH 3 COOH(l) – S° CH 3 OH(l) – S° CO(g)

      = 159.8 – 126.8 – 197.9 = –164.9 J/K = –0.1649 kJ/K

      –137.9 kJ = T(–0.1649 kJ/K), T = 836.3 K

      The equilibrium favors products up to 836 K or 563 °C, so the elevated temperatures to increase the rate of reaction can be safely employed.

    19.99

    1. First, calculate ΔG° for each reaction:

      For C 6 H 12 O 6 (s) + 6 O 2 (g)  ⇌  6 CO 2 (g) + 6 H 2 O(l)       (A)

      ΔG° = 6(–237.13) + 6(–394.4) – (–910.4) + 6(0) = –2878.8 kJ

      For C 6 H 12 O 6 (s)  ⇌  2 C 2 H 5 OH(l) + 2 CO 2 (g)          (B)

      ΔG° = 2(–394.4) + 2(–174.8) – (–910.4) = –228.0 kJ

      For (A), ln K = 2879 × 10 3 /(8.314)(298) = 1162; K = 5 × 10 504

      For (B), ln K = 228 × 10 3 /(8.314)(298) = 92.026 = 92.0; K = 9 × 10 39


    1. Both these values for K are unimaginably large. However, K for reaction (A) is larger, because ΔG° is more negative. The magnitude of the work that can be accomplished by coupling a reaction to its surroundings is measured by ΔG. According to these calculations, considerably more work can in principle be obtained from reaction (A), because ΔG° is more negative.

    19.100

    1. ΔG° = –RT ln K (Equation 19.20); ln K = –ΔG°/RT

      Use ΔG° = ΔH° – TΔS° to get ΔG° at the two temperatures. Calculate ΔH° and ΔS° using data in Appendix C.

      2 CH 4 (g) → C 2 H 6 (g) + H 2 (g)

      ΔH° = ΔH° C 2 H 6 (g) + ΔH° H 2 (g) – 2ΔH° CH 4 (g) = –84.68 + 0 – 2(–74.8) = 64.92 = 64.9 kJ

      ΔS° = S° C 2 H 6 (g) + S° H 2 (g) – 2S° CH 4 (g) = 229.5 + 130.58 – 2(186.3) = –12.52 = –12.5 J/K

      at 298 K, ΔG = 64.92 kJ – 298 K(–12.52 × 10 –3 kJ/K) = 68.65 = 68.7 kJ

      ln K = 68.65 kJ ( 8.314 × 10 3 kJ/K) (298 K) = 27.709 = 27.7 , K = 9 .25 × 10 13 = 9 × 10 13

      at 773 K, ΔG = 64.9 kJ – 773 K(–12.52 × 10 –3 J/K) = 74.598 = 74.6 kJ

      ln K = 74.598 kJ ( 8.314 × 10 3 kJ/K) (773 K) = 11.607 = 11.6 , K = 9 .1 × 10 6

      Because the reaction is endothermic, the value of K increases with an increase in temperature.

      2 CH 4 (g) + 1/2 O 2 (g) → C 2 H 6 (g) + H 2 O(g)

      ΔH° = ΔH° C 2 H 6 (g) + ΔH° H 2 O(g) – 2ΔH° CH 4 (g) – 1/2 ΔH° O 2 (g)

      = –84.68 + (–241.82) – 2(–74.8) – 1/2 (0) = –176.9 kJ

      ΔS° = S° C 2 H 6 (g) + S° H 2 O(g) – 2S° CH 4 (g) – 1/2 S° O 2 (g)

      = 229.5 + 188.83 – 2(186.3) – 1/2 (205.0) = –56.77 = –56.8 J/K

      at 298 K, ΔG = –176.9 kJ – 298 K(–56.77 × 10 –3 kJ/K) = –159.98 = –160.0 kJ

      ln K = ( 159.98 kJ) ( 8.314 × 10 3 kJ/K) (298 K) = 64.571 = 64.57 ; K = 1.1 × 10 28

      at 773 K, ΔG = –176.9 kJ – 773 K (–56.77 × 10 –3 kJ/K) = –133.02 = –133.0 kJ

      ln K = ( 133.02 kJ) ( 8.314 × 10 3 kJ/K) (773 K) = 20.698 = 20.70 ; K = 9.750 × 10 8 = 9.8 × 10 8

      Because this reaction is exothermic, the value of K decreases with increasing temperature.

    2. The difference in ΔG° for the two reactions is primarily enthalpic; the first reaction is endothermic and the second exothermic. Both reactions have –ΔS°, which inhibits spontaneity.

    1. This is an example of coupling a useful but nonspontaneous reaction with a spontaneous one to spontaneously produce a desired product.

      2 CH 4 ( g ) C 2 H 6 ( g ) + H 2 ( g ) ΔG 298 o = + 68.7 kJ, nonspontaneous H 2 ( g ) + 1 / 2 O 2 ( g ) H 2 O(g) ΔG 298 o = 228.57 kJ, spontaneous 2 CH 4 ( g ) + 1 / 2 O 2 ( g ) C 2 H 6 ( g ) + H 2 O(g) ΔG 298 o = 159.9 kJ, spontaneous

    1. CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g)

    19.101  ΔG° for the metabolism of glucose is:

    6ΔG° CO 2 (g) + 6ΔG° H 2 O(l) – ΔG° C 6 H 12 O 6 (s) – 6ΔG° O 2 (g)

    ΔG° = 6(–394.4) + 6(–237.13) – (–910.4) + 6(0) = –2878.8 kJ

    mol ATP = –2878.8 kJ × 1 mol ATP/(–30.5 kJ) = 94.4 mol ATP/mol glucose

    Note that this calculation is done at standard conditions, not metabolic conditions. A more accurate answer would be obtained using ΔG values that reflect actual concentration, partial pressure, and pH in a cell.

    19.102

    1. The equilibrium of interest here can be written as:

      K + (plasma)  ⇌  K + (muscle)

      Because the fluid is aqueous on both sides of the cell membrane, assume that the equilibrium constant for this process is exactly 1; that is, ΔG° = 0. However, ΔG is not zero because the concentrations are not the same on both sides of the membrane. Use Equation 19.19 to calculate ΔG:

      Δ G = Δ G o + RT ln [K + (muscle)] [ K + (plasma)] = 0 + ( 8.314 ) ( 310 ) ln (0 .15) (5 .0 × 10 3 ) = 8766 J = 8 .8 kJ

    2. Note that ΔG is positive. This means that work must be done on the system (blood plasma plus muscle cells) to move the K + ions “uphill,” as it were. The minimum amount of work possible is given by the value for ΔG. This value represents the minimum amount of work (8.8 kJ) required to transfer one mole of K + ions from the blood plasma at 5 × 10 –3 M to muscle cell fluids at 0.15 M , assuming constancy of concentrations. In practice, a larger than minimum amount of work is required.

    Integrative Exercises

    19.103

    1. At the boiling point, vaporization is a reversible process, so Δ S vap o = Δ H vap o / T .

      acetone: Δ S vap o = Δ H vap o / T = (29 .1 kJ/mol)/329 .25 K = 88 .4 J/mol-K

      dimethyl ether: Δ S vap o = ( 21.5 kJ/mol)/248 .35 K = 86 .6 J/mol -K

      ethanol: Δ S vap o = ( 38.6 kJ/mol)/351 .6 K = 110 J/mol- K

      octane: Δ S vap o = ( 34.4 kJ/mol)/398 .75 K = 86 .3 J/mol-K

      pyridine: Δ S vap o = ( 35.1 kJ/mol)/388 .45 K = 90. 4 J/mol-K


    1. Ethanol is the only liquid listed that does not follow Trouton’s rule and it is also the only substance that exhibits hydrogen bonding in the pure liquid. Hydrogen bonding leads to more ordering in the liquid state and a greater than usual increase in entropy upon vaporization. The rule appears to hold for liquids with London dispersion forces (octane) and ordinary dipole-dipole forces (acetone, dimethyl ether, pyridine), but not for those with hydrogen bonding.
    1. Owing to strong hydrogen bonding interactions, water probably does not obey Trouton’s rule.

      From Appendix B, Δ H vap o at 100 °C = 40.67 kJ/mol.

      Δ S vap o = ( 40.67 kJ/mol)/373 .15 K = 109 .0 J/mol - K

    1. Use Δ S vap o = 88 J/mol-K, the middle of the range for Trouton’s rule, to estimate Δ H vap o for chlorobenzene.

      Δ H vap o = Δ S vap o × T = 88 J/mol-K × 404.95 K = 36 kJ/mol

    19.104  The entropy of activation for a bimolecular process is usually negative. By definition, the activated complex of a bimolecular process is composed of two molecules. Forming the activated complex reduces the number of particles in the system from two to one and the entropy of this process will usually be negative.

    19.105  Calculate ΔG°, ΔH°, and ΔS° using data from Appendix C. Use ΔG° = ΔH° – TΔS° to calculate the temperature at which ΔG° changes sign. Couple this temperature with the sign of ΔG° at 298 K to state a temperature range over which the reaction is spontaneous.

    ΔG° = 2ΔG° CO 2 (g) + 3ΔG° Fe(s) – 2ΔG° Fe 3 O 4 (s) – 2ΔG° C(s, graphite)

    = 2(–394.4) + 3(0) – (–1014.2) – 2(0) = 225.4 kJ

    ΔH° = 2ΔH° CO 2 (g) + 3ΔH° Fe(s) – 2ΔH° Fe 3 O 4 (s) – 2ΔH° C(s, graphite)

    = 2(–393.5) + 3(0) – (–1117.1) – 2(0) = 330.1 kJ

    ΔS° = 2ΔS° CO 2 (g) + 3ΔS° Fe(s) – 2ΔS° Fe 3 O 4 (s) – 2ΔS° C(s, graphite)

    = 2(213.6) + 3(27.15) – 146.4 – 2 (5.69) = 350.87 = 350.9 J/K = 0.3509 kJ/K

    ΔH° = TΔS°; T = ΔH°/ΔS° = 330.1 kJ/0.3509 J/K = 940.7 K = 667.6 °C

    The reaction will be spontaneous at temperatures above 940.7 K or 667.6 °C. Even though ΔS° is large and positive, a very high temperature is required to overcome the large positive enthalpy of reaction.

    19.106

    1. O 2 (g) h ν 2 O(g); S increases because there are more moles of gas in the products.
    2. O 2 (g) + O(g) → O 3 (g); S decreases because there are fewer moles of gas in the products.

    1. S increases as the gas molecules diffuse into the larger volume of the stratosphere; there are more possible positions and, therefore, more motional freedom.
    1. NaCl(aq) → NaCl(s) + H 2 O(l); ΔS decreases as the mixture (seawater, greater disorder) is separated into pure substances (fewer possible arrangements, more order).

    19.107  The heat lost by the hot water (iv) in the cup equals the heat gained by the ice cube. The ice cube is heated from –20 °C to 0 °C (i), the ice melts at 0 °C (ii), the new liquid heats to the final temperature (iii). (iv) = (i) + (ii) + (iii) . The 500 mL of hot water weights 500 g; the ice cube weighs 20 g.

    1. 20 g × 2.03 J 1 g- o C × ( 20 o C) = 812 = 8.1 × 10 2 J
    2. 20 g × 1 mol 18 .02 g × 6.01 kJ mol × 1000 J kJ = 6670.4 = 6.7 × 10 3 J
    3. 20 g × 4.184 J 1 g- o C × ( T f 0 o C) = 83 .60 ( T f 0 o C) = 84 ( T f )
    4. 500 g × 4.184 J 1 g- o C × ( 83 o C T f ) = 2,092 ( 83 o C T f ) = (2 .09 × 10 3 ) ( 83 o C T f )

    2092 ( 83 o C T f ) = 812 J + 6670.4 J + 83 .60 (T f ) ;

    173 , 636 [2092 (T f )] = 7482.4 + 83 .60 (T f )

    166,153.6 = 2175.6(T f ); T f = 76.37 = 76 °C

    19.108

    1. 16 e , 8 e pairs. The C–S bond order is approximately 2.
      image
    2. 2 e domains around C, linear e domain geometry, linear molecular structure.
    3. CS 2 (l) + 3 O 2 (g) → CO 2 (g) + 2 SO 2 (g)
    4. ΔH° = ΔH° CO 2 (g) + 2 ΔH° SO 2 (g) –ΔH° CS 2 (l) – 3 ΔH° O 2 (g)

      = –393.5 + 2(–296.9) – (89.7) – 3(0) = –1077.0 kJ

      ΔG° = ΔG° CO 2 (g) + 2 ΔG° SO 2 (g) –ΔG° CS 2 (l) – 3 ΔG° O 2 (g)

      = –394.4 + 2(–300.4) – (65.3) – 3(0) = –1060.5 kJ

      The reaction is exothermic (–ΔH°) and spontaneous (–ΔG°) at 298 K.

    5. vaporization: CS 2 (l) → CS 2 (g)

      Δ G vap o = Δ H vap o T Δ S vap o ; Δ S vap o = ( Δ H vap o Δ G vap o ) /T Δ G vap o = Δ G o CS 2 ( g) Δ G o CS 2 ( l ) = 67.2 65.3 = 1.9 kJ Δ H vap o = Δ H o CS 2 ( g ) Δ H o CS 2 ( l ) = 117.4 89.7 = 27.7 kJ Δ S vap o = ( 27.7 1.9 ) kJ/298 K = 0 .086577 = 0 .0866 kJ/K = 86 .6 J/K

      ΔS vap is always positive, because the gas phase occupies a greater volume and has more motional freedom and a larger absolute entropy than the liquid.


    1. At the boiling point, ΔG = 0 and ΔH vap = T b ΔS vap .

      T b = ΔH vap /ΔS vap = 27.7 kJ/0.086577 kJ/K = 319.9 = 320 K

      T b = 320 K = 47 °C. CS 2 is a liquid at 298 K, 1 atm

    19.109

    1. Ag(s) + 1/2 N 2 (g) + 3/2 O 2 (g) → AgNO 3 (s); S decreases because there are fewer moles of gas in the product.
    2. Δ G f o = Δ H f o T Δ S o ; Δ S o = ( Δ G f o Δ H f o ) / ( T) = ( Δ H f o Δ G f o ) /T

      Δ S o = 124.4 kJ ( 33.4 kJ)/298 K = 0.305 kJ/K = 305 J/K

      Δ S o is relatively large and negative, as anticipated from part (a).

    3. Dissolving of AgNO 3 can be expressed as

      AgNO 3 (s) → AgNO 3 (aq, 1 M )

      ΔH° = ΔH° AgNO 3 (aq) – ΔH° AgNO 3 (s) = –101.7 – (–124.4) = +22.7 kJ

      ΔH° = ΔH° MgSO 4 (aq) – ΔH° MgSO 4 (s) = –1374.8 – (–1283.7) = –91.1 kJ

      Dissolving AgNO 3 (s) is endothermic (+ΔH°), but dissolving MgSO 4 (s) is exothermic (–ΔH°).

    4. AgNO 3 : Δ G o = Δ G f o AgNO 3 ( aq ) Δ G f o AgNO 3 ( s ) = 34.2 ( 33.4 ) = 0.8 kJ

      ΔS° = (ΔH° – ΔG°) / T = [22.7 kJ – (–0.8 kJ)] / 298 K = 0.0789 kJ/K = 78.9 J/K

      MgSO 4 : Δ G o = Δ G f o MgSO 4 (aq ) Δ G f o MgSO 4 ( s ) = 1198.4 ( 1169.6 ) = 28.8 kJ

      ΔS° = (ΔH° – ΔG°) / T = [–91.1 kJ – (–28.8 kJ)] / 298 K = –0.209 kJ/K = –209 J/K

    5. In general, we expect dissolving a crystalline solid to be accompanied by an increase in positional disorder and an increase in entropy; this is the case for AgNO 3 (ΔS°= +78.9 J/K). However, for dissolving MgSO 4 (s), there is a substantial decrease in entropy (ΔS = –209 J/K). According to Section 13.5, ion-pairing is a significant phenomenon in electrolyte solutions, particularly in concentrated solutions where the charges of the ions are greater than 1. According to Table 13.4, a 0.1 m MgSO 4 solution has a van’t Hoff factor of 1.21. That is, for each mole of MgSO 4 that dissolves, there are only 1.21 moles of “particles” in solution instead of 2 moles of particles. For a 1 m solution, the factor is even smaller. Also, the exothermic enthalpy of mixing indicates substantial interactions between solute and solvent. Substantial ion-pairing coupled with ion–dipole interactions with H 2 O molecules lead to a decrease in entropy for MgSO 4 (aq) relative to MgSO 4 (s).

    19.110

    1. K = P NO 2 2 / P N 2 O 4

      Assume equal amounts means equal number of moles. For gases, P = n(RT/V). In an equilibrium mixture, RT/V is a constant, so moles of gas are directly proportional to partial pressure. Gases with equal partial pressures will have equal moles of gas present. The condition P NO 2 = P N 2 O 4 leads to the expression K = P NO 2 . The value of K then depends on P t for the mixture. For any particular value of P t , the condition of equal moles of the two gases can be achieved at some temperature. For example, P NO 2 = P N 2 O 4 = 1.0 atm, P t = 2.0 atm .


      K = (1 .0) 2 1.0 = 1.0 ; ln K = 0 ; Δ G o = 0 = Δ H o T Δ S o ;  T = Δ H o / Δ S o

      ΔH° = 2ΔH° NO 2 (g) – ΔH° N 2 O 4 (g) = 2(33.84) – 9.66 = +58.02 kJ

      ΔS° = 2S° NO 2 (g) – S° N 2 O 4 (g) = 2(240.45) – 304.3 = 0.1766 kJ/K

      T = 58.02 kJ 0 .1766 kJ/K = 328.5 K or 55 .5 o C

    1. P t = 1.00 atm; P N 2 O 4 = x , P NO 2 = 2 x ; x + 2x = 1.00 atm

      x = P N 2 O 4 = 0.3333 = 0.333 atm;   P NO 2 = 0.6667 = 0 .667 atm K = (0 .6667) 2 0.3333 = 1.334 = 1.33 ; Δ G o = RT ln K = Δ H o T Δ S o

      –(8.314 × 10 –3 kJ/K)(ln 1.334) T = 58.02 kJ – (0.1766 kJ/K) T

      (–0.00239 kJ/K) T + (0.1766 kJ/K) T = 58.02 kJ

      (0.1742 kJ/K) T = 58.02 kJ; T = 333.0 K

    1. P t = 10.00 atm; x + 2x = 10.00 atm

      x = P N 2 O 4 = 3.3333 = 3.333 atm;   P NO 2 = 6.6667 = 6.667 atm K = (6 .6667) 2 3.3333 = 13.334 = 13.33 ; RT ln K = Δ H o T Δ S o

      –(8.314 × 10 –3 kJ/K)(ln 13.334) T = 58.02 kJ – (0.1766 kJ/K) T

      (–0.02154 kJ/K) T + (0.1766 kJ/K) T = 58.02 kJ

      (0.15506 kJ/K) T = 58.02 kJ; T = 374.2 K

    19.111

    1. Δ G o = 3 Δ G f o S(s) + 2 Δ G f o H 2 O(g) Δ G f o SO 2 ( g ) 2 Δ G f o H 2 S(g)

      = 3(0) + 2(–228.57) – (–300.4) – 2(–33.01) = –90.72 = –90.7 kJ

      ln K = Δ G o RT = ( 90.72 kJ) ( 8.314 × 10 3 kJ/K) (298 K) = 36.6165 = 36.6 ; K = 7 .99 × 10 15 = 8 × 10 15

    2. The reaction is highly spontaneous at 298 K and feasible in principle. However, use of H 2 S(g) produces a severe safety hazard for workers and the surrounding community.
    3. P H 2 O = 25 torr 760 torr/atm = 0.033 atm

      K = P H 2 O 2 P SO 2 × P H 2 S 2 ; P SO 2 = P H 2 S = x atm

      K = 7 .99 × 10 15 = ( 0.033 ) 2 x ( x ) 2 ; x 3 = ( 0.033 ) 2 7.99 × 10 15

      x = 5 × 10 –7 atm


    1. Δ H o = 3 Δ H f o S(s) + 2 Δ H f o H 2 O(g) Δ H f o SO 2 ( g ) 2 Δ H f o H 2 S(g)

      = 3(0) + 2(–241.82) – (–296.9) – 2(–20.17) = –146.4 kJ

      ΔS° = 3S° S(s) + 2S° H 2 O(g) – S° SO 2 (g) – 2S° H 2 S(g)

      = 3(31.88) + 2(188.83) – 248.5 – 2(205.6) = –186.4 J/K

      The reaction is exothermic (–ΔH), so the value of K eq will decrease with increasing temperature. The negative ΔS° value means that the reaction will become nonspontaneous at some higher temperature. The process will be less effective at elevated temperatures.

    19.112

    1. When the rubber band is stretched, the molecules become more ordered, so the entropy of the system decreases, ΔS sys is negative.
    2. ΔS sys = q rev /T. Because ΔS sys is negative, q rev is negative and heat is emitted by the system.
    3. The unstretched rubber band feels cooler. This confirms our answer to (b). If heat is emitted by the system when it is stretched, the surroundings feel warmer. Upon return to the initial state, heat is absorbed by the system (the rubber band) and the surroundings (your lip) feel cooler.

     

    20 Electrochemistry

    Visualizing Concepts

    20.1   In this analogy, the electron is analogous to the proton (H + ). Just as acid–base reactions can be viewed as proton-transfer reactions, redox reactions can be viewed as electron-transfer reactions.

    Oxidizing agents are reduced; they gain electrons. A strong oxidizing agent would be analogous to a strong base.

    20.2   An antioxidant is a reducing agent that reacts with certain oxidizing agents in the body. The objective is that the antioxidant will be oxidized in preference to human cells or DNA.

    20.3 Analyze/Plan . Apply the definitions of oxidation, reduction, anode and cathode to the diagram. Recall relationship between atomic and ionic size from Chapter 7 . Solve .

    1. Oxidation. The gray spheres are uniformly sized and closely aligned; they represent an elemental solid. The diagram shows atoms from the surface of the solid going into solution. In a voltaic cell, this happens when metal atoms on an electrode surface are oxidized. They lose electrons, form cations, and move into solution.
    2. Anode. Oxidation occurs at the anode.
    3. When a neutral atom loses a valence electron, Z eff for the remaining electrons increases, and the radius of the resulting cation is smaller than the radius of the neutral atom. The neutral atoms in the electrode are represented by larger spheres than the cations moving into solution.

    20.4 Analyze/Plan . Consider the voltaic cell pictured in Figure 20.5 as a model. The reaction in a voltaic cell is spontaneous. To generate a standard emf, substances must be present in their standard states.

    1. A concentration of 1 M is the standard state for ions in solution. Ions, but not solution, must be able to flow between compartments to complete the circuit, so that the cell can develop an emf. Add 1 M A 2+ (aq) to the beaker with the A(s) electrode. Add 1 M B 2+ (aq) to the beaker with the B(s) electrode. Add a salt bridge to enable the flow of ions from one compartment to the other.
    2. Reduction occurs at the cathode. For the reaction to occur spontaneously (and thus generate an emf), the half-reaction with the greater E red o will be the reduction half-reaction. In this cell, it is the half-reaction involving A(s) and A 2+ (aq). The A electrode functions as the cathode.

    1. According to Figure 20.5, electrons flow through the external circuit from the anode to the cathode. In this example, B is the anode and A is the cathode, so electrons flow from B to A through the external circuit.
    1. E cell o = E red o ( cathode ) E red o ( anode )

      E cell o = 0.10 V ( 1.10 V ) = 1.00 V

    20.5   A(aq) + B(aq) → A (aq) + B + (aq)

    1. Reduction occurs at the cathode; oxidation occurs at the anode.

      A(aq) + 1e → A (aq) occurs at the cathode.

      B(aq) → B + (aq) + 1e occurs at the anode.

    2. In a voltaic cell, the anode is at higher potential energy than the cathode. The anode reaction, B(aq) → B + (aq) + 1e , is higher in potential energy.
    3. ΔG° = –nFE°; the signs of ΔG° and E° (or ΔG and E) are opposite. Because this is a spontaneous reaction, ΔG° is negative and E° is positive.

    20.6 Analyze . Given a series of reduction half-reactions and their standard electrode potentials ( E red o ), draw conclusions about their relative strengths as oxidizing and reducing agents. Plan . The reactant with the largest E red o is the easiest to reduce and the strongest oxidizing agent. The reduced form of this substance, the product of the reduction half-reaction, is the most difficult to oxidize and the weakest reducing agent. Conversely, the reactant with the smallest E red o is the hardest to reduce and the weakest oxidizing agent. The reduced form of this substance, the product of the reduction half-reaction, is the easiest to oxidize and the strongest reducing agent. Solve .

    1. A + (aq) is the strongest oxidizing agent, and D 3+ is the weakest oxidizing agent.
    2. D(s) is the strongest reducing agent, and A(s) is the weakest reducing agent.
    3. Reactants with more positive E red o than C 3+ (aq) will oxidize C 2+ (aq). Both A + (aq) and B 2+ (aq) will oxidize C 2+ (aq).

    20.7 Analyze . Given a redox reaction with a negative E°, answer questions regarding Δ , the equilibrium constant (K), and work (w) Plan . ΔG° = –nFE°; ΔG° = –RT ln K; w max = –nFE. Solve .

    1. The signs of ΔG° and E° are opposite. If E° is negative, ΔG° is positive. (The reaction is not spontaneous in the forward direction.)
    2. If ΔG° is positive, ln K is negative and K < 1. Also, K is less than one for a nonspontaneous reaction.
    3. No. If E° is negative, the sign of w is positive. A positive value for w means that work is done on the system by the surroundings. An electrochemical cell based on this reaction cannot accomplish work on its surroundings.

    20.8 Analyze . Given the voltaic cell shown in the diagram, answer questions about the cell and the effect of solution concentration on cell potential, E.

    Plan . Use the definition of a voltaic cell and standard emf, along with the Nernst equation, E = ΔE° – (0.0592 V/n)log Q, to answer the questions. Solve .


    1. A voltaic cell involves a spontaneous redox reaction, one with positive E cell o . To achieve a positive E cell o , the half-reaction with the more positive E red o occurs at the cathode. For this cell the two half reactions are

      Ag + ( aq) + e Ag ( s),   E red o = 0.80 V Fe 2 + ( aq) + 2e Fe(s) , E red o = 0 .44 V

      The Ag(s) electrode is the cathode.

    2. The standard emf is just E cell o .

      E cell o = E red o (cathode) E red o ( anode ) = 0.80 V ( 0.44 V ) = 1.24 V

      The cell in the diagram is at standard conditions, with solid metal electrodes and 1 M aqueous solutions, so the potential on the meter in the circuit is the standard emf.

    3. 2 Ag + (aq) + Fe(s) → 2 Ag(s) + Fe 2+ (aq); n = 2; E = E ° 0.0592 2 log [Fe 2 + ] [ Ag + ] 2

      The solution in the cathode half-cell is Ag + (aq). If [Ag + (aq)] increases by a factor of 10, the change in cell voltage is E E ° = 0.0592 2 log [1] [ 10 ] 2 = 0.059 V .

    4. The solution in the anode half-cell is Fe 2+ (aq). If [Fe 2+ (aq)] increases by a factor of 10, the change in cell voltage is E E ° = 0.0592 2 log [10] [ 1 ] 2 = 0.030 V .

    20.9 Analyze/Plan . Consider the Nernst equation, which describes the variation of potential (emf) with respect to changes in concentration.

    Solve . E = 0.0592 n log Q .

    1. For this half-reaction, E red o = 0.80 V; Q = 1/[Ag + ]

      E = 0 .80 0.0592 1 log 1 [ Ag + ] ; E = 0 .80 + 0.0592 1 log [ Ag + ]

      The y-intercept of the graph is E°. The slope of the line is +0.0592. So, as [Ag + ] and log[Ag + ] increase, E increases. The line that describes this behavior is line 1.

    2. When log[Ag + ] = 0, [Ag + ] = 1 M ; this is the standard state for Ag + (aq), so E red = E red o = 0.80 V .

    20.10

    1. Zinc is the anode. Metallic zinc cannot be reduced; it must be oxidized. Oxidation occurs at the anode.
    2. The energy density of the silver oxide battery is most similar to the nickel-cadmium battery. Energy density (see Figure 20.23) is related to the molar masses of the electrode materials and voltage per cell of the battery. The molar mass of (Ag 2 O + Zn) is most like that of (Ni + Cd). The cell potential for a Ni-Cd battery is 1.3 V; for the reaction in the silver oxide battery, the cell potential is about 1.1 V. The Li-ion (3.7 V) and lead-acid (2.0 V) batteries have larger cell potentials. (The standard reduction potential of Ag 2 O is 0.34 V. From Handbook of Chemistry and Physics , 74th edition)

    20.11  Beaker B. The process of iron corrosion includes Fe being oxidized and O 2 being reduced. The reduction of O 2 requires H + . Beaker B, which contains dilute HCl(aq), provides the H + required to encourage corrosion. Iron in contact with a solution above pH 9 does not corrode.


    20.12

    1. +2
    2. MgCl 2 (l)
    3. Oxidation occurs at the anode. The anode half-reaction is:

      2 Cl (aq) → Cl 2 (g) + 2e

    4. Reduction occurs at the cathode. The cathode half-reaction is:

      Mg 2+ (aq) + 2e → Mg(l)

    Oxidation–Reduction Reactions (Section 20.1)

    20.13

    1. Oxidation is the loss of electrons.
    2. The electrons appear on the products side (right side) of an oxidation half-reaction.
    3. The oxidant is the reactant that is reduced; it gains the electrons that are lost by the substance being oxidized.
    4. An oxidizing agent is the substance that promotes oxidation. That is, it gains electrons that are lost by the substance being oxidized. It is the same as the oxidant.

    20.14

    1. Reduction is the gain of electrons.
    2. The electrons appear on the reactants side (left side) of a reduction half-reaction.
    3. The reductant is the reactant that is oxidized; it provides the electrons that are gained by the substance being reduced.
    4. A reducing agent is the substance that promotes reduction. It donates the electrons gained by the substance that is reduced. It is the same as the reductant.

    20.15

    1. True.
    2. False. Fe 3+ is reduced to Fe 2+ , so it is the oxidizing agent, and Co 2+ is the reducing agent.
    3. True.

    20.16

    1. False. If something is reduced, it gains electrons.
    2. True.
    3. True. Oxidation can be thought of as a gain of oxygen atoms. Looking forward, this view will be useful for organic reactions, Chapter 24 .

    20.17 Analyze/Plan .  Given a chemical equation, we are asked to determine the oxidation number of each reactant and product, and the total number of electrons transferred in the overall reaction. Assign oxidation numbers according to the rules given in Section 4.4. To find electrons transferred, count the electrons gained by an atom that is reduced and multiply by the number of atoms reduced. Check by counting electrons for the atom oxidized. Solve .

      1. Reactants: I, +5; O, –2; C +2; O, –2. Products: I, 0; C, +4; O, –2
      2. The total number of electrons transferred is 10. (Each I atom gains 5 electrons, for a total of 10 electrons gained; each C atom loses 2 electrons for a total of 10 electrons lost; electrons balance.)

      1. Reactants: Hg, +2; N, –2; H, +1. Products: Hg, 0; N, 0; H, +1.
      2. The total number of electrons transferred is 4. (Each Hg gains 2 electrons for a total of 4 electrons gained; each N loses 2 electrons for a total of 4 electrons lost; electrons balance.)
      1. Reactants: H, +1; S, –2; N, +5; O, –2. Products: S, 0; N, +2; O, –2; H, +1; O, –2.
      2. The total electrons transferred is 6. (Each N atom gains 3 electrons for a total of 6 electrons gained; each S atom loses 2 electrons for a total of 6 electrons lost; electrons balance.)

    20.18

      1. Reactants: Mn, +7; O, –2; S, –2; H, +1; O, –2. Products: S, 0; Mn, +4; O, –2; O, –2; H, +1.
      2. The total number of electrons transfered is 6.
      1. Reactants: H, +1; O, –1; Cl, +7; O, –2; H, +1; O, –2; Products: Cl, +4; O, –2; H, +1; O, –2; O, 0.
      2. The total number of electrons transferred is 8. (Two Cl atoms each gain 4 electron, eight O atoms each lose 1 electron by changing oxidation number from –1 to 0.)
      1. Reactants: Ba, +2; O, –2; H, +1; H, +1; O, –1; Cl, +4; O, –2. Products: Ba, +2; Cl, +3; O, –2; H, +1; O, –2; O, 0.
      2. The total number of electrons transferred is 2. (Two Cl atoms each gain 1 electron, two O atoms each lose 1 electron by changing oxidation number from –1 to 0.)

    20.19

    1. No oxidation–reduction.
    2. I is oxidized from –1 to +5; Cl is reduced from +1 to –1.
    3. S is oxidized from +4 to +6; N is reduced from +5 to +2.

    20.20

    1. No oxidation–reduction.
    2. Pb is reduced from +4 to +2; O is oxidized from –2 to 0.
    3. S is reduced from +6 to +4; Br is oxidized from –1 to 0.

    Balancing Oxidation–Reduction Reactions (Section 20.2)

    20.21 Analyze/Plan .  Write the balanced chemical equation and assign oxidation numbers. The substance oxidized is the reductant and the substance reduced is the oxidant. Solve .

    1. TiCl 4 (g) + 2 Mg(l) → Ti(s) + 2 MgCl 2 (l)
    2. Mg(l) is oxidized; TiCl 4 (g) is reduced.
    3. Mg(l) is the reductant; TiCl 4 (g) is the oxidant.

    20.22

    1. 2 N 2 H 4 (g) + N 2 O 4 (g) → 3 N 2 (g) + 4 H 2 O(g)
    2. N 2 H 4 (g) is oxidized; N 2 O 4 (g) is reduced.
    3. N 2 H 4 (g) serves as the reducing agent; it is itself oxidized. N 2 O 4 (g) serves as the oxidizing agent; it is itself reduced.

    20.23 Analyze/Plan .  Follow the logic in Sample Exercises 20.2 and 20.3. If the half-reaction occurs in basic solution, balance as in acid, then add OH to each side. Solve .

    1. Sn 2+ (aq) → Sn 4+ (aq) + 2 e , oxidation

    1. TiO 2 (s) + 4 H + (aq) + 2 e → Ti 2+ (aq) + 2 H 2 O(l), reduction
    1. ClO 3 (aq) + 6 H + (aq) + 6 e → Cl (aq) + 3 H 2 O(l), reduction
    1. N 2 (g) + 8 H + (aq) + 6 e → 2 NH 4 + (aq), reduction
    1. 4 OH (aq) → O 2 (g) + 2 H 2 O(l) + 4 e , oxidation
    1. SO 3 2– (aq) + 2 OH (aq) → SO 4 2– (aq) + H 2 O(l) + 2 e , oxidation
    1. N 2 (g) + 6 H 2 O + 6 e → 2 NH 3 (g) + 6 OH (aq), reduction

    20.24

    1. Mo 3+ (aq) + 3 e → Mo(s), reduction
    2. H 2 SO 3 (aq) + H 2 O(l) → SO 4 2– (aq) + 4 H + (aq) + 2 e , oxidation
    3. NO 3 (aq) + 4 H + (aq) + 3 e → NO(g) + 2 H 2 O(l), reduction
    4. O 2 (g) + 4 H + (aq) + 4 e → 2 H 2 O(l), reduction
    5. O 2 (g) + 2 H 2 O(l) + 4 e → 4 OH (aq), reduction

      (O 2 is reduced to OH , not H 2 O, in basic solution)

    6. Mn 2+ (aq) + 4 OH (aq) → MnO 2 (s) + 2 H 2 O(l) + 2 e , oxidation
    7. Cr(OH) 3 (s) + 5 OH (aq) → CrO 4 2– (aq) + 4 H 2 O(l) + 3 e , oxidation

    20.25 Analyze/Plan .  Follow the logic in Sample Exercises 20.2 and 20.3 to balance the given equations. Use the method in Sample Exercise 20.1 to identify oxidizing and reducing agents. Solve .

    1. Cr 2 O 7 2– (aq) + I (aq) + 8 H + → 2 Cr 3+ (aq) + IO 3 (aq) + 4 H 2 O(l)

      oxidizing agent, Cr 2 O 7 2– ; reducing agent, I

    2. The half-reactions are:

      4 [ MnO 4 (aq) + 8 H + (aq) + 5 e Mn 2 + (aq) + 4 H 2 O(l)] 5 [CH 3 OH(aq) + H 2 O(l) HCOOH(aq) + 4 H + ( aq) + 4 e ] 4 MnO 4 (aq) + 5 CH 3 OH(aq) + 12 H + (aq) 4 Mn 2 + (aq) + 5 HCOOH(aq) + 11 H 2 O(l)

      oxidizing agent, MnO 4 ; reducing agent, CH 3 OH

    3. I 2 ( s ) + 6 H 2 O ( l ) 2 IO 3 (aq) + 12 H + (aq) + 10 e 5 [ OCl ( aq) + 2 H + (aq) + 2 e Cl (aq) + H 2 O(l) I 2 (s) + 5 OCl (aq) + H 2 O(l) 2 IO 3 (aq) + 5 Cl (aq) + 2 H + (aq)]

      oxidizing agent, OCl–; reducing agent, I 2

    4. As 2 O 3 ( s ) + 5 H 2 O(l) 2 H 3 AsO 4 ( aq) + 4 H + (aq) + 4 e 2 NO 3 (aq) + 6 H + (aq) + 4 e N 2 O 3 (aq) + 3 H 2 O(l) As 2 O 3 ( s ) + 2 NO 3 (aq) + 2 H 2 O(l) + 2 H + (aq) 2 H 3 AsO 4 (aq) + N 2 O 3 (aq)

      oxidizing agent, NO 3 ; reducing agent, As 2 O 3


    1. 2 [ MnO 4 (aq) + 2 H 2 O(l) + 3 e MnO 2 ( s ) + 4 OH ] Br (aq) + 6 OH (aq) BrO 3 (aq) + 3 H 2 O(l) + 6 e 2 MnO 4 (aq) + Br (aq) + H 2 O(l) 2 MnO 2 ( s) + BrO 3 (aq) + 2 OH (aq)

      oxidizing agent, MnO 4 ; reducing agent, Br–

    1. Pb(OH) 4 2– (aq) + ClO (aq) → PbO 2 (s) + Cl (aq) + 2 OH (aq) + H 2 O(l)

      oxidizing agent, ClO ; reducing agent, Pb(OH) 4 2–

    20.26

    1. Net: 3 [ NO 2 (aq) + H 2 O(l) NO 3 (aq) + 2 H + (aq) + 2 e ] Cr 2 O 7 2 (aq) + 14 H + (aq) + 6 e 2 Cr 3 + (aq) + 7 H 2 O(l) 3 NO 2 (aq) + Cr 2 O 7 2 (aq) + 8 H + (aq) 3 NO 3 (aq) + 2 Cr 3 + (aq) + 4 H 2 O(l)

      oxidizing agent, Cr 2 O 7 2 ; reducing agent, NO 2

    2. The oxidation half-reaction involves S, and is listed in Appendix E. The reduction half-reaction involves N, and must be written and balanced, according to the procedure in Section 20.2.

      HNO 3 ( aq ) N 2 O(g) 2 HNO 3 ( aq ) N 2 O(g) 2 HNO 3 ( aq ) N 2 O(g) + 5 H 2 O(l) 2 HNO 3 ( aq ) + 8 H + ( aq ) N 2 O(g) + 5 H 2 O(l) 2 HNO 3 ( aq ) + 8 H + + 8 e N 2 O(g) + 5 H 2 O(l)

      2 [S(s) + 3 H 2 O(l) H 2 SO 3 (aq) + 4 H + + 4 e ] 2 HNO 3 (aq) + 8 H + + 8 e N 2 O(g) + 5 H 2 O(l) 2 HNO 3 (aq) + 2 S(s) + H 2 O(l) 2 H 2 SO 3 (aq) + N 2 O(g)

      oxidizing agent, HNO 3 ; reducing agent, S.

    3. Net: 2 [ Cr 2 O 7 2 (aq) + 14 H + (aq) + 6 e 2 Cr 3 + (aq) + 7 H 2 O(l)] 3 [CH 3 OH(aq) + H 2 O(l) HCOOH(aq) + 4 H + (aq) + 4 e ] 2 Cr 2 O 7 2 (aq) + 3 CH 3 OH(aq) + 16 H + (aq) 4 Cr 3 + (aq) + 3 HCOOH(aq) + 11 H 2 O(l)

      oxidizing agent, Cr 2 O 7 2– ; reducing agent, CH 3 OH

    4. The half-reaction involving N 2 H 4 is given in Appendix E in base. We add 4 H + (aq) to each side and reverse the reaction to obtain the oxidation half-reaction shown below.

      Net: 2 [ 2 BrO 3 (aq) + 12 H + (aq) + 1 0 e Br 2 (l) + 6 H 2 O(l)] 5 [N 2 H 4 (aq) N 2 (g) + 4 H + (aq) + 4 e ] 4 BrO 3 (aq) + 5 N 2 H 4 (aq) + 4 H + (aq) 2 Br 2 (l) + 5 N 2 (g) + 12 H 2 O(l)

      oxidizing agent, BrO 3 ; reducing agent, N 2 H 4


    1. Write and balance each half-reaction, and then sum to get the overall reaction. Follow the procedure in Sample Exercise 20.3 for reactions in basic solution.

      oxidation: Al(s) → AlO 2 (aq)

      Al(s) + 2 H 2 O(l) → AlO 2 (aq) + 4 H + (aq) + 3 e

      Al(s) + 4 OH (aq) → AlO 2 (aq) + 2 H 2 O(l) + 3 e

      reduction: NO 2 (aq) → NH 4 + (aq)

      NO 2 (aq) + 8 H + (aq) + 6 e → NH 4 + (aq) )+ 2 H 2 O(l)

      NO 2 (aq) + 6 H 2 O(l) + 6 e → NH 4 + (aq) + 8 OH (aq)

      NO 2 (aq) + 6 H 2 O(l) + 6 e → NH 4 + (aq) + 8 OH (aq)

      2[Al(s) + 4 OH (aq) → AlO 2 (aq) + 2 H 2 O(l) + 3 e ]

      NO 2 (aq) + 6 H 2 O(l) + 6 e → NH 4 + (aq) + 8 OH (aq)

      NO 2 (aq) + 2 Al(s) + 2 H 2 O(l) → NH 4 + (aq) + 2 AlO 2 (aq)

      oxidizing agent, NO 2 ; reducing agent, Al

    1. H 2 O 2 (aq) + 2 e → O 2 (g) + 2 H + (aq)

      Because the reaction is in base, the H + can be “neutralized” by adding 2 OH to each side of the equation: H 2 O 2 (aq) + 2 OH (aq) + 2 e → O 2 (g) + 2 H 2 O(l). The other half reaction is: 2 [ClO 2 (aq) + e → ClO 2 (aq)].

      Net: H 2 O 2 (aq) + 2 ClO 2 (aq) + 2 OH (aq) → O 2 (g) + 2 ClO 2 (aq) + 2 H 2 O(l)

      oxidizing agent, ClO 2 ; reducing agent, H 2 O 2

    Voltaic Cells (Section 20.3)

    20.27

    1. False. Oxidation occurs at the anode; reduction occurs at the cathode.
    2. True. The terms voltaic and galvanic are interchangeable.
    3. True. The oxidation half-reaction at the anode generates free electrons which flow from the anode to the cathode.

    20.28

    1. True. Oxidation occurs at the anode.
    2. True. A voltaic cell utilizes a spontaneous redox reaction and a spontaneous reaction has a positive emf.
    3. True. Ions must be able to migrate between the anode and cathode compartments in order to maintain charge balance and complete the electrical circuit.

    20.29 Analyze/Plan .  Follow the logic in Sample Exercise 20.4. Solve .

    1. Fe(s) is oxidized, Ag + (aq) is reduced.
    2. Ag + (aq) + 1e → Ag(s); Fe(s) → Fe 2+ (aq) + 2e
    3. Fe(s) is the anode, Ag(s) is the cathode.
    4. Fe(s) is negative; Ag(s) is positive.
    5. Electrons flow from the Fe(–) electrode toward the Ag(+) electrode.
    6. Cations migrate toward the Ag(s) cathode; anions migrate toward the Fe(s) anode.

    20.30

    1. Al(s) is oxidized, Ni 2+ (aq) is reduced.
    2. Al(s) → Al 3+ (aq) + 3e ; Ni 2+ (aq) + 2e → Ni(s)
    3. Al(s) is the anode; Ni(s) is the cathode.
    4. Al(s) is negative (–); Ni(s) is positive (+).
    5. Electrons flow from the Al(–) electrode toward the Ni(+) electrode.
    6. Cations migrate toward the Ni(s) cathode; anions migrate toward the Al(s) anode.

    Cell Potentials under Standard Conditions (Section 20.4)

    20.31

    1. One volt is the potential energy difference required to impart 1 J of energy to a charge of 1 coulomb. 1 V = 1 J/C.
    2. Yes. All voltaic cells involve a spontaneous redox reaction that produces a positive cell potential or emf.

    20.32

    1. In a voltaic cell, the anode has the higher potential energy for electrons. To achieve a lower potential energy, electrons flow from the anode to the cathode.
    2. The units of electrical potential are volts. A potential of one volt imparts one joule of energy to one coulomb of charge.

    20.33

    1. 2 H + (aq) + 2 e → H 2 (g)
    2. H 2 (g) → 2 H + (aq) + 2 e
    3. A standard hydrogen electrode is a hydrogen electrode where the components are at standard conditions, 1 M H + (aq) and H 2 (g) at 1 atm, such that E o = 0 V.

    20.34

    1. For a reduction potential to be a standard reduction potential, the substances in the reaction or half-reaction must be at standard conditions, 1 M aqueous solutions and 1 atm gas pressures.
    2. E red o = 0 V for a standard hydrogen electrode.
    3. It is not possible to measure the standard reduction potential of a single half-reaction because each voltaic cell consists of two half-reactions and only the potential of a complete cell can be measured.

    20.35 Analyze/Plan .  Follow the logic in Sample Exercise 20.5. Solve .

    1. The two half-reactions are:

      Tl 3 + ( aq) + 2 e Tl + ( aq)                                                 cathode E red o = ? 2 [ Cr 2 + ( aq) Cr 3 + ( aq) + e ] anode    E red o = 0 .41 V

    2. E cell o = E red o (cathode) E red o (anode) 1 .19 V = E red o ( 0.41 V) E red o = 1.19 V 0.41 V = 0 .78 V

    1. image

      Note that because Cr 2+ (aq) is readily oxidized, it would be necessary to keep oxygen out of the left-hand cell compartment.

    20.36

    1. PdCl 4 2 ( aq) + 2 e Pd(s) + 4 Cl cathode E red o = ? Cd(s) Cd 2 + ( aq) + 2 e anode E red o = 0.40 V
    2. E cell o = E red o (cathode) E red o (anode); 1 .03 V = E red o ( 0.40 V); E red o = 1.03 V 0.40 = 0.63 V
    3. image

    20.37 Analyze/Plan .  Follow the logic in Sample Exercise 20.6. Solve .

    1. Cl 2 ( g) 2 Cl ( aq) + 2 e I 2 ( s) + 2 e 2 I ( aq) E red o = 1.36 V E red o = 0.54 V

      E° = 0.77 V – (–0.44 V) = 1.21 V

    2. Ni(s) Ni 2 + ( aq) + 2 e 2 [Ce 4 + ( aq) + 1 e Ce 3 + (aq)] E red o = 0.28 V E red o = 1.61 V

      E° = 1.61 V – (–0.28 V) = 1.89 V

    3. Fe(s) Fe 2 + ( aq) + 2 e 2 [Fe 3 + ( aq) + 1 e Fe 2 + ( aq)] E red o = 0.44 V E red o = 0.77 V

      E° = 0.96 V – (0.34 V) = 0.62 V

    4. 2 [NO 3 (aq) + 4 H + + 3 e NO(g) + 2 H 2 O(l)] 3 [ Cu ( s ) Cu 2 + ( aq) + 2 e ] E red o = 0.96 V E red o = 0.34 V

      E° = 2.87 V – 0.00 V = 2.87 V


    20.38

    1. F 2 ( g) + 2 e 2 F (aq) H 2 (g) 2 H + ( aq) + 2 e E red o = 2.87 V E red o = 0.00 V

      E° = 0.34 V – (–2.87 V) = 3.21 V

    2. Cu 2 + ( aq) + 2 e Cu(s) Ca(s) Ca 2 + ( aq) + 2 e E red o = 0.34 V E red o = 2.87 V

      E° = 0.34 V – (–2.87 V) = 3.21 V

    3. Fe 2 + ( aq) + 2 e Fe(s) 2[Fe 2 + ( aq) Fe 3 + ( aq) + 1 e ] E red o = 0.44 V E red o = 0.77 V

      E° = –0.44 V – 0.77 V = –1.21 V

    4. 2 ClO 3 (aq) + 12 H + + 1 0 e Cl 2 (g) + 6 H 2 O(l) 5 [2 Br ( aq ) Br 2 ( l ) + 2 e ] E red o = 1.47 V E red o = 1.07 V

      E° = 1.47 V – 1.07 V = 0.40 V

    20.39 Analyze/Plan .  Given four half-reactions, find E red o from Appendix E and combine them to obtain a desired E cell . (a) The largest E cell will combine the half-reaction with the most positive E red o as the cathode reaction and the one with the most negative E red o as the anode reaction. (b) The smallest positive E cell o will combine two half-reactions whose E red o values are closest in magnitude and sign. Solve .

    1. 3 [ Ag + ( aq) + 1 e Ag(s)] Cr(s) Cr 3 + ( aq) + 3 e ¯ 3 Ag + ( aq) + Cr(s) 3 Ag(s) + Cr 3 + (aq) E red o = 0.80 E red o = 0.74 E o = 0.80 ( 0.74 ) = 1.54 V
    2. Two of the combinations have equal E° values.
      2 [Ag + ( aq) + 1 e Ag(s)] Cu(s) Cu 2 + (aq) + 2 e ¯ 2 Ag + ( aq) + Cu(s) 2 Ag(s) + Cu 2 + ( aq) E red o = 0.80 V E red o = 0.34 V E o = 0 .80 V 0.34 V = 0 .46 V
      3 [Ni 2 + ( aq) + 2 e Ni(s)] 2 [Cr(s) Cr 3 + ( aq) + 3 e ] ¯ 3 Ni 2 + ( aq) + 2 Cr(s) 3 Ni(s) + 2 Cr 3 + ( aq) E red o = 0.28 V E red o = 0.74 V E o = 0.28 V ( 0.74 V) = 0 .46 V

    20.40

    1. 2 [ Au(s) + 4 Br ( aq) AuBr 4 ( aq) + 3 e ] E red o = 0.86 V [2e + IO ( aq) + H 2 O(l) I ( aq) + 2 OH ( aq)]      E red o = 0.49 V 2 Au(s) + 8 Br ( aq) + 3 IO ( aq) + 3 H 2 O(l) 2 AuBr 4 ( aq) + 3 I ( aq) + 6 OH ( aq)

      E° = 0.49 – (–0.86) = 1.35 V

    2. 2 [ Eu 2 + ( aq) Eu 3 + ( aq) + 1 e ] E red o = 0.43 V Sn 2 + ( aq) + 2 e Sn(s) E red o = 0.14 V 2 Eu 2 + ( aq) + Sn 2 + ( aq) 2 Eu 3 + ( aq) + Sn(s) ¯ E o = 0.14 ( 0.43 ) = 0.29 V

    20.41 Analyze/Plan .  Given the description of a voltaic cell, answer questions about this cell. Combine ideas in Sample Exercises 20.4 and 20.7. The reduction half-reactions are:

    Cu 2+ (aq) + 2 e → Cu(s)  E° = 0.34 V

    Sn 2+ (aq) + 2 e → Sn(s)   E° = –0.14 V

    Solve .

    1. It is evident that Cu 2+ is more readily reduced. Therefore, Cu serves as the cathode, Sn as the anode.
    2. The copper electrode gains mass as Cu is plated out, the Sn electrode loses mass as Sn is oxidized.
    3. The overall cell reaction is Cu 2+ (aq) + Sn(s) → Cu(s) + Sn 2+ (aq)
    4. E° = 0.34 V – (–0.14 V) = 0.48 V

    20.42

    1. The two half-reactions are:

      Cd 2 + ( aq) + 2 e Cd(s) E ° = 0.40 V Cl 2 ( g) + 2 e 2 Cl ( aq) E ° = 1 .36 V

      Because E° for the reduction of Cl 2 is greater, Cl 2 is reduced at the cathode, the Pt electrode. Cd(s) is oxidized at the anode, the Cd electrode.

    2. The Cd anode loses mass as Cd 2+ (aq) is produced.
    3. Cl 2 (g) + Cd(s) → Cd 2+ (aq) + 2Cl–(aq)
    4. E° = 1.36 V – (–0.40 V) = 1.76 V

    Strengths of Oxidizing and Reducing Agents (Section 20.4)

    20.43 Analyze/Plan .  The more readily a substance is oxidized, the stronger it is as a reducing agent. In each case choose the half-reaction with the more negative reduction potential and the given substance on the right. Solve .

    1. Mg(s) (–2.37 V vs. –0.44 V)
    2. Ca(s) (–2.87 V vs. –1.66 V)
    3. H 2 (g, acidic) (0.00 V vs. 0.14 V)
    4. IO 3 (aq) [Both IO 3 (aq) and BrO 3 (aq) are good oxidizing agents, but IO 3 (aq) has the smaller positive reduction potential. (1.20 V vs. 1.52 V)]

    20.44  Follow the logic in Sample Exercise 20.8. In each case, choose the half-reaction with the more positive reduction potential and with the given substance on the left.

    1. Cl 2 (g) (1.36 V vs. 1.07 V)
    2. Cd 2+ (aq) (–0.40 V vs. –0.76 V)
    3. ClO 3 (aq) (Cl–(aq) is in its minimum oxidation state and cannot act as an oxidizing agent)
    4. O 3 (g) (2.07 V vs. 1.78 V)

    20.45 Analyze/Plan .  If the substance is on the left of a reduction half-reaction, it will be an oxidant; if it is on the right, it will be a reductant. Solve .

    1. Cl 2 (aq): oxidant (on the left, E red o = 1.36 V )
    2. MnO 4 (aq, acidic): oxidant (on the left, E red o = 1.51 V )
    3. Ba(s): reductant (on the right, E red o = 2.90 V )
    4. Zn(s): reductant (on the right, E red o = 0.76 V )

    20.46  If the substance is on the left of a reduction half-reaction, it will be an oxidant; if it is on the right, it will be a reductant.

    1. Ce 3+ (aq): reductant (on the right, E red o = 1.61 V )
    2. Ca(s): reductant (on the right, E red o = –2.87 V)
    3. ClO 3 (aq): oxidant (on the left, E red o = 1.47 V)
    4. N 2 O 5 (g): oxidant (N has maximum oxidation number, +5; can only be reduced and act as oxidant.)

    20.47 Analyze/Plan .  Follow the logic in Sample Exercise 20.8. Solve .

    1. Arranged in order of increasing strength as oxidizing agents (and increasing reduction potential):

      Cu 2+ (aq) < O 2 (g) < Cr 2 O 7 2– (aq) < Cl 2 (g) < H 2 O 2 (aq)

    2. Arranged in order of increasing strength as reducing agents (and decreasing reduction potential):

      H 2 O 2 (aq) < I–(aq) < Sn 2+ (aq) < Zn(s) < Al(s)

    20.48

    1. The strongest oxidizing agent is the species most readily reduced, as evidenced by a large, positive reduction potential. That species is H 2 O 2 . The weakest oxidizing agent is the species that least readily accepts an electron. We expect that it will be very difficult to reduce Zn(s); indeed, Zn(s) acts as a comparatively strong reducing agent. No potential is listed for reduction of Zn(s), but we can safely assume that it is less readily reduced than any of the other species present.
    2. The strongest reducing agent is the species most easily oxidized (the largest negative reduction potential). Zn, E red o = 0.76 V, is the strongest reducing agent and F , E red o = 2.87 V, is the weakest.

    20.49 Analyze/Plan .  To reduce Eu 3+ to Eu 2+ , we need an oxidizing agent, one of the reduced species from Appendix E. It must have a greater tendency to be oxidized than Eu 3+ has to be reduced. That is, E red o must be more negative than –0.43 V. Solve .

    Any of the reduced species in Appendix E from a half-reaction with a reduction potential more negative than –0.43 V will reduce Eu 3+ to Eu 2+ . From the list of possible reductants in the exercise, Al and H 2 C 2 O 4 will reduce Eu 3+ to Eu 2+ .

    20.50  Any oxidized species from Appendix E with a reduction potential greater than 0.59 V will oxidize RuO 4 2– to RuO 4 . From the list of possible oxidants in the exercise, Br 2 (l) and BrO 3 (aq) will definitely oxidize RuO 4 2– to RuO 4 . Sn 2+ (aq) will not, and O 2 (g) depends on conditions. In base, it will not, but in strongly acidic solution, it will.


    Free Energy and Redox Reactions (Section 20.5)

    20.51 Analyze/Plan .  In each reaction, Fe 2+ → Fe 3+ will be the oxidation half-reaction and one of the other given half-reactions will be the reduction half-reaction. Follow the logic in Sample Exercise 20.10 to calculate E°, ΔG° and K for each reaction. Solve .

    1. 2 Fe 2+ (aq) + S 2 O 6 2– (aq) + 4 H + (aq) → 2 Fe 3+ (aq) + 2 H 2 SO 3 (aq)

      E° = 0.60 V – 0.77 V = –0.17 V

      2 Fe 2+ (aq) + N 2 O(aq) + 2 H + (aq) → 2 Fe 3+ (aq) + N 2 (g) + H 2 O(l)

      E° = –1.77 V – 0.77 V = –2.54 V

      Fe 2+ (aq) + VO 2 + (aq) + 2 H + (aq) → Fe 3+ (aq) + VO 2+ (aq) + H 2 O(l)

      E° = 1.00 V – 0.77 V = +0.23 V

    2. ΔG° = –nFE° For the first reaction,

      Δ = 2 mol × 96,485 J 1 V-mol × ( 0 .17 V) = 3 .280 × 10 4 = 3.3 × 10 4 J or 33 kJ

      For the second reaction, ΔG° = –2(96,485)(–2.54) = 4.901 × 10 5 = 4.90 × 10 2 kJ

      For the third reaction, ΔG° = –1(96,485)(0.23) = –2.22 × 10 4 J = –22 kJ

    3. ΔG° = –RT ln K; ln K = –ΔG°/RT

      For the first reaction,

      ln K = 3 .281 × 10 4 J (8 .314 J/mol-K)(298 K) = 13.243 = 13 ; K = e 13.2428 = 1.78 × 10 6 = 2 × 10 6

      [Convert ln to log; the number of decimal places in the log is the number of sig figs in the result.]

      For the second reaction,

      ln K = 4 .902 × 10 5 J 8 .314 J/mol-K × 298 K = 197.86 = 198 ; K = e 198 = 1.23 × 10 86 = 10 86

      For the third reaction,

      ln K = ( 2.22 × 10 4 J) 8 .314 J/mol-K × 298 K = 8.958 = 9.0 ; K = e 9 .0 = 7.77 × 10 3 = 8 × 10 3

      Check .  The equilibrium constants calculated here are indicators of equilibrium position, but are not particularly precise numerical values.

    20.52

    1. 2 I (aq) I 2 ( s ) + 2 e E red o = 0.54 V Hg 2 2 + (aq) + 2 e 2 Hg(l) E red o = 0.79 V 2 I (aq) + Hg 2 2 + (aq) I 2 ( s ) + 2 Hg(l) = 0 .79 0.54 = 0.25 V

      Δ G o = nFE o = 2 mol e _ × 96.5 kJ V-mol e × 0.25 V = 48.829 = 49 kJ ln K = ( 4.8829 × 10 4 J) ( 8.314 J/mol-K)(298 K) = 19.708 = 20 ; K = e 19.708 = 3.62 × 10 8 = 10 8


    1. 3 [ Cu + (aq) Cu 2 + (aq) + 1 e ] E red o = 0.15 V NO 3 (aq) + 4 H + (aq) + 3 e NO(g) + H 2 O(l) E red o = 0.96 V 3 Cu + ( aq ) + NO 3 ( aq ) + 4 H + (aq) 3 Cu 2 + ( aq ) + NO(g) + 2 H 2 O(l)

      E° = 0.96 – 0.15 = 0.81 V; ΔG° = –3(96.5)(0.81) = –2.345 × 10 2 kJ = –2.3 × 10 5 J

      ln K = ( 2.345 × 10 5 J) ( 8.314 J/mol - K)(298 K) = 94.65 = 95 ; K = e 95 = 1.3 × 10 41 = 10 41

    1. 2 [ Cr(OH) 3 ( s ) + 5 OH ( aq ) CrO 4 2 ( aq ) + 4 H 2 O(l) + 3 e ] E red o = 0.13 V 3 [ ClO ( aq ) + H 2 O(l) + 2 e Cl ( aq ) + 2 OH ( aq ) ] E red o = 0.89 V 2 Cr(OH) 3 ( s ) + 3 ClO ( aq ) + 4 OH ( aq ) 2 CrO 4 2 ( aq ) + 3 Cl ( aq ) + 5 H 2 O(l)

      E° = 0.89 – (–0.13) = 1.02 V; ΔG° = –6(96.5)(1.02) = –590.58 kJ = –5.91 × 10 5 J

      ln K = ( 5.9058 × 10 5 J) ( 8.314 J/mol-K)(298 K) = 238.37 = 238 ; K = 3 .3 × 10 103 = 10 103

      This is an unimaginably large number.

    20.53 Analyze/Plan .  Given K, calculate ΔG° and E°. Reverse the logic in Sample Exercise 20.10. According to Equation 19.20, ΔG° = –RT ln K. According to Equation 20.12,

    ΔG° = –nFE°, E° = – ΔG°/nF. Solve .

    K = 1.5 × 10 −4

    ΔG° = –RT ln K = –(8.314 J/mol-K)(298) ln (1.5 × 10 −4 ) = 2.181 × 10 4 J = 21.8 kJ

    E° = –ΔG°/nF; n = 2; F = 96.5 kJ/mol e

    E o = 21.81 kJ 2 mol e × 96.5 kJ/V-mol e = 0.113 V

    Check .  The unit of ΔG° is actually kJ/mol, which means kJ per “mole of reaction,” or for the reaction as written. Because we do not have a specific reaction, we interpret the unit as referring to the overall reaction.

    20.54  K = 8.7 × 10 4 ; ΔG° = –RT ln K; E° = –ΔG°/nF; n = 1; T = 298 K

    ΔG° = –8.314 J/mol-K × 298 K × ln(8.7 × 10 4 ) = –2.818 × 10 4 J = –28.2 kJ

    = Δ / nF = ( 28.18 kJ) 1 e × 96.5 kJ/V-mol e = 0.292 V

    20.55 Analyze/Plan . Given E red o values for half reactions, calculate the value of K for a given redox reaction. Follow the logic in Sample Exercise 20.10. For each reaction, calculate E° from E red o , then use Equation 20.13 to calculate K.

    Solve . E o = RT nF ln K;  ln K = nF RT E o ; F RT = 96,485 J/V-mol 8 .314 J/K-mol × 298 .15 K = 38.924 /V

    1. E° = –0.28 – (–0.44) = 0.16 V, n = 2 (Ni 2+ + 2e → Ni)

      ln K = 2(38.924/V)(0.16 V) = 12.456 = 12; K = 2.57 × 10 5 = 3 × 10 5

    2. E° = 0 – (–0.28) = 0.28 V; n = 2 (2H + + 2e → H 2 )

      ln K = 2(38.924/V)(0.28 V) = 21.797 = 22; K = 2.93 × 10 9 = 3 × 10 9


    1. E° = 1.51 – 1.07 = 0.44 V; n = 10 (2 MnO 4 + 10 e → 2 Mn +2 )

      ln K = 10(38.924/V)(0.44 V) = 171.265 = 170; K = 2.40 × 10 74 = 10 74

    Check. Note that small differences in E° values lead to large changes in the magnitude of K. Sig fig rules limit the precision of K values.

    20.56  At 298 K, ln K = n (38 .924/V)E° . See Solution 20.55 for a more complete development.

    1. E° = 0.80 V – 0.34 V = 0.46 V; n = 2 (2Ag + + 2e → 2Ag)

      ln K = 2 ( 38.924 /V ) 0.46 V = 35.810 = 35 ; K = 3.57 × 10 15 = 4 × 10 15

    2. E° = 1.61 V – 0.32 V = 1.29 V; n = 3 (3 Ce 4+ + 3e → 3 Ce 3+ )

      ln K = 3 ( 38.924 /V ) 1.29 V = 150.635 = 150 ; K = 2.63 × 10 65 = 10 65

    3. E° = 0.36 V – (–0.23 V) = 0.59 V; n = 4 (4 Fe(CN) 6 3– + 4e → 4 Fe(CN) 6 4– )

      ln K = 4 ( 38.924 /V ) 0.59 V = 91.860 = 92 ; K = 7.84 × 10 39 = 10 40

    20.57 Analyze/Plan .  At 298 K, ln K = n ( 38.924 /V ) . See Solution 20.55 for a more complete development. Solve .

    1. ln K = 1 ( 38.924 /V ) 0.177 V = 6.890 = 6.89 ; K = 982 = 9.8 × 10 2
    2. ln K = 2 ( 38.924 /V ) 0.177 V = 13.779 = 13.8 ; K = 9.64 × 10 5 = 1 × 10 6
    3. ln K = 3 ( 38.924 /V ) 0.177 V = 20.669 = 20.7 ; K = 9.47 × 10 8 = 1 × 10 9

    20.58  At 298 K, ln K = n ( 38.924 /V ) E ° . See Solution 20.55 for a more complete development.

    ln K = n (38 .924/V) E o ;  n = ln K ( 38 .924/V)E o ; n = ln 5 .5 × 10 5 ( 38 .924/V) 0 .17 V = 1 .9975 = 2

    20.59 Analyze/Plan .  Given a spontaneous chemical reaction, calculate the maximum possible work for a given amount of reactant at standard conditions. Separate the equation into half-reactions and calculate cell emf. Use Equation 20.14, w max = –nFE, to calculate maximum work. At standard conditions, E = E°. Solve .

    I 2 (s) + 2 e 2 I (aq) E red o = 0.54 V Sn(s) Sn 2 + ( aq ) + 2 e E red o = 0.14 V I 2 (s) + Sn(s) 2 I (aq) + Sn 2 + (aq) E o = 0.54 V ( 0.14 V ) = 0.68 V

    w max = –2(96.5)(0.68) = –131.24 = 1.3 × 10 2 kJ/mol Sn

    131.24 kJ mol Sn(s) × 1 mol Sn 118 .71 g Sn × 75.0 g Sn × 1000 J kJ = 8.3 × 10 4 J

    Check. The (–) sign indicates that work is done by the cell.

    20.60  For this cell at standard conditions, E° = 1.10 V.

    w max = ΔG° = –nFE° = –2(96.5)(1.10) = –212.3 = –212 kJ/mol Cu

    50.0 g Cu × 1 mol Cu 63 .55 g Cu × 212.3 kJ mol Cu = 167 kJ = 1 .67 × 10 5 J


    Cell EMF under Nonstandard Conditions (Section 20.6)

    20.61

    1. In the Nernst equation, Q = 1 if all reactants and products are at standard conditions.
    2. Yes. The Nernst equation is applicable to cell emf at nonstandard conditions, so it must be applicable at temperatures other than 298 K. There are two terms in the Nernst equation. First, values of E° at temperatures other than 298 K are required. Then, in the form of Equation 20.16, there is a variable for T in the second term. In the short-hand form of Equation 20.18, the value 0.0592 assumes 298 K. A different coefficient would apply to cells at temperatures other than 298 K.

    20.62  Decrease. As the spontaneous chemical reaction of the voltaic cell proceeds, the concentrations of products increase and the concentrations of reactants decrease.

    20.63 Analyze/Plan .  Given a circumstance, determine its effect on cell emf. Each circumstance changes the value of Q. An increase in Q reduces emf; a decrease in Q increases emf. Solve .

    Zn(s) + 2H + (aq) Zn 2 + ( aq ) + H 2 (g);  E = 0 .0592 n log Q;   Q = [Zn 2 + ] P H 2 [ H + ] 2

    1. P H 2 increases, Q increases, E decreases
    2. [Zn 2+ ] increases, Q increases, E decreases
    3. [H + ] decreases, Q increases, E decreases
    4. No effect; does not appear in the Nernst equation

    20.64 Al(s) + 3Ag + ( aq ) Al 3 + (aq) + 3Ag(s);    E = 0.0592 n log Q;   Q = [Al 3 + ] [ Ag + ] 3

    Any change that causes the reaction to be less spontaneous (that causes Q to increase and ultimately shifts the equilibrium to the left) will result in a less positive value for E.

    1. Increases E by decreasing [Al 3+ ] on the right side of the equation, which decreases Q.
    2. No effect; the “concentrations” of pure solids and liquids do not influence the value of K for a heterogeneous equilibrium.
    3. No effect; the concentration of Ag + and the value of Q are unchanged.
    4. Decreases E; forming AgCl(s) decreases the concentration of Ag + , which increases Q.

    20.65 Analyze/Plan .  Follow the logic in Sample Exercise 20.11. Solve .

    1. Ni 2 + (aq) + 2 e Ni(s) E red o = 0.28 V Zn(s) Zn 2 + (aq) + 2 e E red o = 0.76 V Ni 2 + ( aq ) + Zn(s) Ni(s) + Zn 2 + ( aq ) E o = 0.28 ( 0.76 ) = 0.48 V
    2. E = E o 0.0592 n log [Zn 2 + ] [ Ni 2 + ] ; n = 2 E = 0.48 0.0592 2 log (0 .100) (3 .00) = 0.48 0.0592 2 log (0 .0333) E = 0 .48 0.0592 ( 1.477 ) 2 = 0.48 + 0 .0437 = 0.5237 = 0.52 V
    3. E = 0.48 0.0592 2 log (0 .900) ( 0.200 ) = 0.48 0.0193 = 0.4607 = 0.46 V

    20.66

    1. 3 [ Ce 4 + (aq) + 1 e Ce 3 + ( aq ) ] E red o = 1.61 V Cr(s) Cr 3 + ( aq ) + 3 e E red o = 0.74 V 3 Ce 4 + ( aq ) + Cr(s) 3 Ce 3 + (aq) + Cr 3 + (aq) E o = 1 .61 ( 0.74 ) = 2.35 V
    2. E = E o 0.0592 n log [Ce 3 + ] 3 [ Cr 3 + ] [ Ce 4 + ] 3 ; n = 3 E = 2.35 0.0592 3 log (0 .10) 3 ( 0.010 ) ( 3.0 ) 3 = 2.35 0.0592 3 log (3 .704 × 10 7 ) E = 2.35 0.0592 ( 6.431 ) 3 = 2.35 + 0.127 = 2.48 V
    3. E = 2.35 0.0592 3 log (2 .0) 3 (1 .5) ( 0.010 ) 3 = 2.35 0.1397 = 2.21 V

    20.67 Analyze/Plan .  Follow the logic in Sample Exercise 20.11. Solve .

    1. 4 [ Fe 2 + ( aq ) Fe 3 + ( aq ) + 1 e ] E red o = 0.77 V O 2 ( g ) + 4 H + ( aq ) + 4 e 2 H 2 O(l) E red o = 1.23 V 4 Fe 2 + ( aq ) + O 2 ( g ) + 4 H + ( aq ) 4 Fe 3 + ( aq ) + 2 H 2 O(l) E o = 1 .23 0.77 = 0.46 V
    2. E = E o 0.0592 n log [Fe 3 + ] 4 [ Fe 2 + ] 4 [ H + ] 4 P O 2 ; n = 4,  [H + ] = 10 3.50 = 3.2 × 10 4 M

      E = 0.46 V 0.0592 4 log (0 .010) 4 (1 .3) 4 ( 3.2 × 10 4 ) 4 ( 0.50 ) = 0.46 0.0592 4 log ( 7.0 × 10 5 ) E = 0.46 0.0592 4 ( 5.845 ) = 0.46 0.0865 = 0.3735 = 0.37 V

    20.68

    1. 2 [ Fe 3 + ( aq ) + 1 e Fe 2+ ( aq ) ] E red o = 0.77 V H 2 ( g ) 2H + ( aq ) + 2 e E red o = 0.00 V 2 Fe 3 + ( aq ) + H 2 ( g ) 2 Fe 2 + ( aq ) + 2 H + ( aq ) E o = 0 .77 0.00 = 0.77 V
    2. E = E o 0.0592 n log [Fe 2 + ] 2 [ H + ] 2 [ Fe 3 + ] 2 P H 2 ; [H + ] = 10 pH = 1.0 × 10 4 , n = 2 E = 0 .77 0.0592 2 log (0 .0010) 2 ( 1.0 × 10 4 ) 2 ( 3.50 ) 2 ( 0.95 ) = 0.77 0.0592 2 log (8 .6 × 10 16 ) E = 0 .77 0.0592 ( 15.066 ) 2 = 0.77 + 0.446 = 1.216 = 1.22 V

    20.69 Analyze/Plan .  We are given a concentration cell with Zn electrodes. Use the definition of a concentration cell in Section 20.6 to answer the stated questions. Use Equation 20.18 to calculate the cell emf. For a concentration cell, Q = [dilute]/[concentrated]. Solve .

    1. The compartment with the more dilute solution will be the anode. That is, the compartment with [Zn 2+ ] = 1.00 × 10 −2 M is the anode.
    2. Because the oxidation half-reaction is the opposite of the reduction half-reaction, E° is zero.

    1. E = E o 0.0592 n log Q; Q = [ Zn 2 + , dilute] / [Zn 2 + , conc .]

      E = 0 0.0592 2 log (1 .00 × 10 2 ) ( 1.8 ) = 0.0668 V

    1. In the anode compartment, Zn(s) → Zn 2+ (aq), so [Zn 2+ ] increases from 1.00 × 10 −2 M . In the cathode compartment, Zn 2+ (aq) → Zn(s), so [Zn 2+ ] decreases from 1.8 M .

    20.70

    1. The compartment with 0.0150 M Cl– (aq) is the cathode.
    2. E° = 0 V
    3. E = E ° 0.0592 n log Q; Q = [ Cl , dilute] / [Cl , conc .]

      E = 0 0.0592 1 log (0 .0150 ) ( 2.55 ) = 0.13204 = 0.1320 V

    4. In the anode compartment, [Cl–] will decrease from 2.55 M . In the cathode, [Cl–] will increase from 0.0150 M .

    20.71 Analyze/Plan .  Follow the logic in Sample Exercise 20.12. Solve .

    E = E ° 0.0592 2 log [P H 2 ] [Zn 2 + ] [ H + ] 2 ; E ° = 0.0 V ( 0.76 V) = 0.76 V

    0.684 = 0.76 0.0592 2 × (log [P H 2 ] [ Zn 2 + ] 2 log [H + ] )

    = 0.76 0.0592 2 × ( 0.5686 2 log [H + ] )

    0.684 = 0.76 + 0.0168 + 0.0592 log [H + ] ; log [H + ] = 0.684 0.0168 0.76 0.0592

    log [H + ] = –1.5676 = – 1.6; [H + ] = 0.0271 = 0.03 M ; pH = 1.6

    20.72

    1. E° = –0.14 V – (–0.13 V) = –0.01 V; n = 2

      0.22 = 0.01 0.0592 2 log [Pb 2 + ] [ Sn 2 + ] = 0.01 0.0592 2 log [Pb 2 + ] 1.00

      log [Pb 2 + ] = 0.23 (2) 0.0592 = 7.770 = 7.8 ; [Pb 2 + ] = 1.7 × 10 8 = 2 × 10 8 M

    2. For PbSO 4 (s), K sp = [Pb 2+ ] [SO 4 2– ] = (1.0)(1.7 × 10 −8 ) = 1.7 × 10 −8

    Batteries and Fuel Cells (Section 20.7)

    20.73 Analyze/Plan .  Given mass of a reactant (Pb), calculate mass of product (PbO 2 ), and coulombs of charge transferred. This is a stoichiometry problem; we need the balanced equation for the chemical reaction that occurs in the lead-acid battery.

    The overall cell reaction is:

    Pb(s) + PbO 2 (s) + 2 H + (aq) + 2 HSO 4 (aq) → 2 PbSO 4 (s) + 2 H 2 O(l) Solve .

    1. g Pb → mol Pb → mol PbO 2 → g PbO 2

      402 g Pb × 1 mol Pb 207 .2 g Pb × 1 mol PbO 2 1 mol Pb × 239.2 g PbO 2 1 mol PbO 2 = 464 g PbO 2


    1. From the half-reactions for the lead-acid battery, 2 mol electrons are transferred for each mol of Pb reacted. From Section 20.5, 96,485 C/mol e .

      402 g Pb × 1 mol Pb 207 .2 g Pb × 2 mol e 1 mol Pb × 96 , 485 C 1 mol e = 374 , 392 = 3.74 × 10 5 C

    20.74

    1. The overall cell reaction is:

      2 MnO 2 (s) + Zn(s) + 2 H 2 O(l) → 2 MnO(OH)(s) + Zn(OH) 2 (s)

      4.50 g Zn × 1 mol Zn 65 .39 g Zn × 2 mol MnO 2 1 mol Zn × 86.94 g MnO 2 1 mol MnO 2 = 12.0 g MnO 2

    2. Two mol e are transferred for every mol of Zn reacted. 96,485 C/mol e

      4.50 g Zn × 1 mol Zn 65 .39 g Zn × 2 mol e 1 mol Zn × 96 , 485 C 1 mol e = 13 , 280 = 1.32 × 10 4 C

    20.75 Analyze/Plan .  We are given a redox reaction and asked to write half-reactions, calculate E°, and indicate whether Li(s) is the anode or cathode. Determine which reactant is oxidized and which is reduced. Separate into half-reactions, find E red o for the half-reactions from Appendix E and calculate E°. Solve .

    1. Li(s) is oxidized at the anode.
    2. Ag 2 CrO 4 (s) + 2 e 2 Ag(s) + CrO 4 2 (aq) E red o = 0.45 V 2 [Li(s) Li + ( aq ) + 1 e ] E red o = 3.05 V Ag 2 CrO 4 (s) + 2 Li(s) 2 Ag(s) + CrO 4 (aq) + 2 Li + (aq)

      E° = 0.45 V – (–3.05 V) = 3.50 V

    3. The emf of the battery, 3.5 V, is exactly the standard cell potential calculated in part (b).
    4. For this battery at ambient conditions, E ≈ E°, so log Q ≈ 0. This makes sense because all reactants and products in the battery are solids and thus present in their standard states. Assuming that E° is relatively constant with temperature, the value of the second term in the Nernst equation is ≈ 0 at 37 °C, and E ≈ 3.5 V.

    20.76

    1. HgO(s) + Zn(s) → Hg(l) + ZnO(s)
    2. E cell o = E red o (cathode) E red o (anode)

      E red o (anode) = E red o E cell o = 0.098 1.35 = 1.25 V

    3. E red o is different from Zn 2+ (aq) + 2e → Zn(s) (–0.76 V) because in the battery the process happens in the presence of base and Zn 2+ is stabilized as ZnO(s). Stabilization of a reactant in a half-reaction decreases the driving force, so E red o is more negative.

    20.77 Analyze/Plan .  (a) Consider the function of Zn in an alkaline battery. What effect would it have on the redox reaction and cell emf if Cd replaces Zn? (b) Both batteries contain Ni. What is the difference in environmental impact between Cd and the metal hydride? Solve .

    1. E red o for Cd (–0.40 V) is less negative than E red o for Zn (–0.76 V), so E cell will have a smaller (less positive) value.
    2. NiMH batteries use an alloy such as ZrNi 2 as the anode material. This eliminates the use and concomitant disposal problems associated with Cd, a toxic heavy metal.

    20.78

    1. NiO(OH)(s) + H 2 O(l) + 1 e → Ni(OH) 2 (s) + OH (aq)
    2. Zn(s) + 2 OH (aq) → Zn(OH) 2 (s) + 2 e
    3. A nickel-zinc battery will produce about 0.94 V. E red o for Cd (–0.40 V) is less negative than E red o for Zn (–0.76 V), so E cell for the nickel-zinc battery will be less positive than that for the nickel-cadmium battery by 0.36 V. That is, the potential of the nickel-zinc battery will be approximately (1.30 V – 0.36 V) = 0.94 V.
    4. Higher. Specific energy density is the amount of energy stored per unit mass of the battery. The potential of the nickel-zinc battery is about 72 % of the nickel-cadmium battery (0.94 V/1.30 V); the molar mass of zinc is about 58 % of the molar mass of cadmium (65.38 g/112.4 g). The reduction in mass by exchanging zinc for cadmium more than compensates for the reduction in cell potential.

    20.79

    1. The oxidation states of the elements in LiCoO 2 are +1, +3 and –2, respectively. At any point in the operation of the battery, the positive and negative charges must balance. On charging, the oxidation state of some of the cobalt ions increases to +4, which releases some of the Li + ions to migrate back to the anode. If approximately half of the Li + ions migrate, the same number of Co 3+ ions are oxidized to the +4 state. When the battery is fully charged, the mole ratios of the elements in the cathode are: 0.5 mol Li + to 0.5 mol Co 3+ to 0.5 mol Co 4+ to 2 mol O 2– . The material in the cathode is LiCo 2 O 4 .
    2. The molar mass of LiCoO 2 is 97.87 g/mol.

      10 g LiCoO 2 × 1 mol LiCoO 2 97 .87 g LiCoO 2 × 1 mol Li 1 mol LiCoO 2 = 0.1022 = 0.10 mol Li +

      Half of the Li + , 0.051 mol, migrates during a full charge.

      0.051 mol Li + × 1 mol e 1 mol Li + × 96 , 485 C 1 mol e = 4 , 921 = 4.9 × 10 3 C

    20.80

    1. LiMn 2 O 4 : FW = 6.941 + 2(54.938) + 4(15.9994) = 180.815 amu

      % Li = 6 .941 g 180.815 g × 100 = 3.8387 = 3.839 %

      LiCoO 2 : FW = 6.941 + 58.9332 + 2(15.9994) = 97.873 amu

      % Li = 6 .941 g 97.873 g × 100 = 7.0918 = 7.092 %

    2. LiCoO 2 has almost twice as great a mass percent Li as LiMn 2 O 4 . This definitely explains why LiMn 2 O 4 cathodes deliver less power on discharging. A Li ion battery produces power by the migration of Li + , but not all the Li + in the cathode material migrates during a full charge. The greater percentage of Li in the cathode material, the more Li + that migrates and the greater the power that can be produced upon discharge.
    3. Assume 100.000 g of cathode material. A LiCoO 2 cathode has 7.092 g of Li; if half of the Li migrates during charging, 3.546 g Li migrates. A LiMn 2 O 4 cathode contains 3.839 g of Li. To deliver the same amount of Li to the graphite anode,

      3 .546 g 3.839 g × 100 = 92.368 = 92 % of the Li in the LiMn 2 O 4 cathode would have to migrate.


    20.81 Analyze/Plan . Write the balanced equation for the spontaneous reaction in a hydrogen fuel cell. Separate the overall reaction into half-reactions in acid solution and basic solution, find the matching standard reduction potentials in Appendix E and calculate standard voltages. Solve .

    1. The spontaneous reaction in the hydrogen fuel cell is hydrogen gas plus oxygen gas makes water.
    2. O 2 ( g ) + 4 H + ( aq ) + 4 e 2 H 2 O(l) E red o = 1.23 V 2 H 2 ( g ) 4 H + ( aq ) + 4 e E red o = 0.00 V O 2 ( g ) + 2 H 2 ( g ) 2 H 2 O(l) E o = 1 .23 0.00 = 1.23 V

    20.82

    1. Both batteries and fuel cells are electrochemical power sources. Both take advantage of spontaneous oxidation–reduction reactions to produce a certain voltage. The difference is that batteries are self-contained (all reactants and products are present inside the battery casing), whereas fuel cells require continuous supply of reactants and exhaust of products.
    2. No. The fuel in a fuel cell must be fluid, either gas or liquid. Because fuel must be continuously supplied to the fuel cell, it must be capable of flow; the fuel cannot be solid.

    Corrosion (Section 20.8)

    20.83 Analyze/Plan .  (a) Decide which reactant is oxidized and which is reduced. Write the balanced half-reactions and assign the appropriate one as anode and cathode. (b) Write the balanced half-reaction for Fe 2+ (aq) → Fe 2 O 3 · 3H 2 O. Use the reduction half-reaction from part (a) to obtain the overall reaction. Solve .

    1. anode: Fe(s) → Fe 2+ (aq) + 2 e

      cathode: O 2 (g) + 4 H + (aq) + 4 e → 2 H 2 O(l)

    2. 2 Fe 2+ (aq) + 6 H 2 O(l) → Fe 2 O 3 · 3H 2 O(s) + 6 H + (aq) + 2 e

      O 2 (g) + 4 H + (aq) + 4 e → 2 H 2 O(l)

      (Multiply the oxidation half-reaction by two to balance electrons and obtain the overall balanced reaction.)

    20.84

    1. Calculate E cell o for the given reactants at standard conditions.

      O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O(l) E red o = 1.23 V 2 [Cu(s) Cu 2 + ( aq ) + 2 e ] E red o = 0.34 V 2 Cu(s) + O 2 (g) + 4 H + (aq) 2 Cu 2 + (aq) + 2 H 2 O(l) E ° = 1.23 0.34 = 0.89 V

      At standard conditions with O 2 (g) and H + (aq) present, the oxidation of Cu(s) has a positive E° value and is spontaneous. Cu(s) will oxidize (corrode) in air in the presence of acid.

    2. Fe 2+ has a more negative reduction potential (–0.44 V) than Cu 2+ (+0.34 V), so Fe(s) is more readily oxidized than Cu(s). If the two metals are in contact, Fe(s) would act as a sacrificial anode and oxidize (corrode) in preference to Cu(s); this would weaken the iron support skeleton of the statue. The teflon spacers prevent contact between the two metals and insure that the iron skeleton doesn’t corrode when the Cu(s) skin comes in contact with atmospheric O 2 (g) and H + (aq).

    20.85

    1. A “sacrificial anode” is a metal that is oxidized in preference to another when the two metals are coupled in an electrochemical cell; the sacrificial anode has a more negative E red o than the other metal. In this case, Mg acts as a sacrificial anode because it is oxidized in preference to the pipe metal; it is sacrificed to preserve the pipe.
    2. E red o for Mg 2+ is –2.37 V, more negative than most metals present in pipes, including Fe (E red o = 0.44 V) and Zn (E red o = 0.76 V) .

    20.86  No. To afford cathodic protection, a metal must be more difficult to reduce (have a more negative reduction potential) than Fe 2+ . E red o Co 2 + = 0.28 V, E red o Fe 2 + = 0.44 V .

    20.87 Analyze/Plan .  Positive and negative ion charges must balance in neutral compounds. Use that fact along with oxidation numbers to answer the questions. Solve .

    1. +3
    2. +8/3 or +2.67
    3. 2 Fe(III) and 1 Fe(II)

    20.88

    1. +1
    2. +2
    3. CuO 2 . Peroxide ion is O 2 2– , which balances charge with Cu 2+ .
    4. Cu 2 O 3 .

    Electrolysis; Electrical Work (Section 20.9)

    20.89

    1. Electrolysis is an electrochemical process driven by an outside energy source.
    2. Electrolysis reactions are, by definition, nonspontaneous.
    3. 2 Cl (l) → Cl 2 (g) + 2 e
    4. When an aqueous solution of NaCl undergoes electrolysis, sodium metal is not formed because H 2 O is preferentially reduced to form H 2 (g).

    20.90

    1. An electrolytic cell is the vessel in which electrolysis occurs. It consists of a power source and two electrodes in a molten salt or aqueous solution.
    2. It is the cathode. In an electrolysis cell, as in a voltaic cell, electrons are consumed (via reduction) at the cathode. Electrons flow from the negative terminal of the voltage source and then to the cathode.
    3. A small amount of H 2 SO 4 (aq) present during the electrolysis of water acts as a change carrier, or supporting electrolyte. This facilitates transfer of electrons through the solution and at the electrodes, speeding up the reaction. [Considering H + (aq) as the substance reduced at the cathode changes the details of the half-reactions, but not the overall E° for the electrolysis. SO 4 2– (aq) cannot be oxidized.]
    4. If the active metal salt is present as an aqueous solution during electrolysis, water is reduced [to H 2 (g)] rather than the metal ion being reduced to the metal. This is true for any active metal with an E red o value more negative than –0.83 V.

    20.91 Analyze/Plan .  Follow the logic in Sample Exercise 20.14, paying close attention to units. Coulombs = amps-s; because this is a 3e reduction, each mole of Cr(s) requires

    3 Faradays. Solve .

    1. 7.60 A × 2 .00 d × 24 h 1 d × 60 min 1 h × 60 s 1 min × 1 C 1 amp-s × 1 F 96,485 C × 1 mol Cr 3 F × 52.00 g Cr 1 mol Cr = 236 g Cr(s)
    2. 0.250 mol Cr × 3 F 1 mol Cr × 96 , 485 C F × 1 amp-s 1 C × 1 8 .00 h × 1 h 60 min × 1 min 60 s = 2.51 A

    20.92  Coulombs = amps-s; because this is a 2e reduction, each mole of Mg(s) requires

    2 Faradays.

    1. 4.55 A × 4 .50 d × 24 h 1 d × 60 min 1 h × 60 s 1 min × 1 C 1 amp-s × 1 F 96,485 C × 1 mol Mg 2 F × 24.31 g Mg 1 mol Mg = 223 g Mg
    2. 25.00 g Mg × 1 mol Mg 24 .31 g Mg × 2 F 1 mol Mg × 96 , 485 C F × 1 amp-s C × 1 min 60 s × 1 3.50 A = 945 min

    20.93 Analyze/Plan . Follow the logic in Sample Exercise 20.14, paying close attention to units. Take the 85% efficiency into account. Li + is reduced at the anode; Cl is oxidized at the anode. Solve .

    1. If the cell is 85% efficient, 96 , 485 C F × 1 F 0 .85 mol = 1.13512 × 10 5 = 1.1 × 10 5 C/mol Li is required.

      7.5 × 10 4 A × 24 h × 3600 s 1 h × 1 C 1 amp-s × 1 mol Li 1 .13512 × 10 5 C × 6.94 g Li 1 mol Li = 3.962 × 10 5 = 4.0 × 10 5 g Li

    2. E cell o = E red o (cathode) E red o ( anode) = –3.05 V – (1.36 V) = –4.41 V

      The minimum voltage required to drive the reaction is the magnitude of E cell o , 4.41 V.

    20.94

    1. 7.5 × 10 3 A × 48 h × 3600 s 1 h × 1 C 1 amp-s × 1 F 96,485 C × 1 mol Ca 2 F × 0.68 × 40.0 g Ca 1 mol Ca = 1.830 × 10 5 = 1.8 × 10 5 g Ca
    2. E cell o = E red o (cathode) E red o ( anode) = –2.87 V – (1.36 V) = –4.23 V

      The minimum voltage required to drive the reaction is the magnitude of E red o 4.23 V.


    20.95  Refer to Table 4.5, “The Activity Series of the Metals.” Gold is the least active metal on this table, less active than copper. This means that gold is more difficult to oxidize than copper (and that E red o for Au 3+ is more positive than E red o for Cu 2+ ). When a mixture of copper and gold is refined by electrolysis, Cu is oxidized from the anode, but any metallic gold present in the mixture is not oxidized, so it accumulates near the anode.

    20.96  The standard reduction potential for Te 4+ , 0.57 V, is more positive than that of Cu 2+ , 0.34 V. This means the Te 4+ is “easier” to reduce than Cu 2+ , but Te is harder to oxidize and less active than Cu. During electrorefining, although Cu is oxidized from the crude anode, Te will not be oxidized. It is likely to accumulate along with other impurities less active than Cu, in the so-called anode sludge.

    Additional Exercises

    20.97

    1. Ni + ( aq ) + 1 e Ni ( s ) Ni + ( aq ) Ni 2 + (aq) + 1 e 2 Ni + (aq) Ni(s) + Ni 2 + (aq)
    2. MnO 4 2 (aq) + 4 H + (aq) + 2 e MnO 2 ( s ) + 2 H 2 O(l) 2 [MnO 4 2 (aq) MnO 4 (aq) + 1 e ] 3 MnO 4 2 (aq) + 4 H + (aq) 2 MnO 4 (aq) + MnO 2 (s) + 2 H 2 O(l)
    3. H 2 SO 3 (aq) + 4 H + (aq) + 4 e S ( s ) + 3 H 2 O(l) 2 [H 2 SO 3 (aq) + H 2 O(l) HSO 4 (aq) + 3 H + (aq) + 2 e ] 3 H 2 SO 3 (aq) S(s) + 2 HSO 4 (aq) + 2 H + (aq) + H 2 O(l)
    4. Cl 2 (aq) + 2 H 2 O(l) 2 ClO (aq) + 4 H + (aq) + 2 e 4 OH (aq) + 4 OH (aq) Cl 2 (aq) + 4 OH (aq) 2 ClO (aq) + 2 H 2 O(l) + 2 e Cl 2 (aq) + 2 e 2 Cl (aq) 1/2 [2 Cl 2 (aq) + 4 OH (aq) 2 Cl (aq) + 2 ClO (aq) + 2 H 2 O(l)] Cl 2 (aq) + 2 OH (aq) Cl (aq) + ClO (aq) + H 2 O(l)

    20.98

    1. Fe(s) Fe 2 + (aq) + 2 e 2Ag + (aq) + 2 e 2 Ag(s) Fe(s) + 2 Ag + (aq) Fe 2 + (aq) + 2 Ag(s)

      E cell o = E red o (cathode) E red o ( anode) = 0.80 V – (–0.44 V) = 0.36 V

    2. Zn(s) Zn 2 + (s) + 2 e 2 H + (aq) + 2 e H 2 (g) Zn(s) + 2 H + (aq) Zn 2 + (aq) + H 2 (g)

      E cell o = E red o (cathode) E red o ( anode) = 0.00 V – (–0.76 V) = 0.76 V


    1. Cu | Cu 2 + | | ClO 3 , Cl | Pt . Here, both the oxidized and reduced forms of the cathode solution are in the same phase, so we separate them by a comma and then indicate an inert electrode.

      E cell o = E red o (cathode) E red o ( anode) = 1.45 V – (0.34 V) = 1.11 V

    20.99  We need in each case to determine whether E° is positive (spontaneous) or negative (nonspontaneous).

    1. I 2 (s) + 2 e 2 I (aq) E red o = 0.54 V Sn(s) Sn 2 + (aq) + 2 e E red o = 0.14 V Sn(s) + I 2 (s) Sn 2 + (aq) + 2 I (aq) E ° = 0.54 ( 0.14 ) = 0.68 V, spontaneous
    2. Ni 2 + (aq) + 2 e Ni(s) E red o = 0.28 V 2 I (aq) I 2 (s) + 2 e E red o = 0.54 V Ni 2 + (aq) + 2 I (aq) Ni(s) + I 2 (s) E ° = 0.28 0.54 = 0.82 V, nonspontaneous
    3. 2 [Ce 4 + (aq) + 1 e Ce 3 + (aq) E red o = 1.61 V H 2 O 2 (aq) O 2 (g) + 2 H + (aq) + 2 e E red o = 0.68 V 2 Ce 4 + (aq) + H 2 O 2 (aq) 2 Ce 3 + (aq) + O 2 (g) + 2 H + (aq) E ° = 1.61 0.68 = 0.93 V, spontaneous
    4. Cu 2 + (aq) + 2 e Cu(s) E red o = 0.34 V Sn 2 + (aq) Sn 4 + (aq) + 2 e E red o = 0.15 V Cu 2 + (aq) + S  n 2 + (aq) Cu(s) + Sn 4 + (aq) E ° = 0.34 0.15 = 0.19 V, spontaneous

    20.100

    1. The reduction potential for O 2 (g) in the presence of acid is 1.23 V. O 2 (g) cannot oxidize Au(s) to Au + (aq) or Au 3+ (aq), even in the presence of acid.
    2. The possible oxidizing agents need a reduction potential greater than 1.50 V. These include Co 3+ (aq), F 2 (g), H 2 O 2 (aq), and O 3 (g). Marginal oxidizing agents (those with reduction potential near 1.50 V) from Appendix E are BrO 3 (aq), Ce 4+ (aq), HClO(aq), MnO 4 (aq), and PbO 2 (s).
    3. 4 Au(s) + 8 NaCN(aq) + 2 H 2 O(l) + O 2 (g) → 4 Na[Au(CN) 2 ](aq) + 4 NaOH(aq)

      Au(s) + 2 CN (aq) → [Au(CN) 2 ] + 1 e

      O 2 (g) + 2 H 2 O(l) + 4 e → 4 OH (aq)

      Au(s) is being oxidized and O 2 (g) is being reduced.

    4. 2 [ Na[Au(CN) 2 ] (aq) + 1 e Au(s) + 2 CN (aq) + Na + (aq)] Zn(s) Zn 2 + (aq) + 2 e 2 Na[Au(CN) 2 ] (aq) + Zn(s) 2 Au(s) + Zn 2 + (aq) + 2 Na + (aq) + 4 CN (aq)

      Zn(s) is being oxidized and [Au(CN) 2 ] (aq) is being reduced. Although OH (aq) is not included in this redox reaction, its presence in the reaction mixture probably causes Zn(OH) 2 (s) to form as the product. This increases the driving force (and E°) for the overall reaction.


    20.101

    1. 2 [ Ag + (aq) + 1 e Ag(s)] E red o = 0.80 V Ni(s) Ni 2 + (aq) + 2 e E red o = 0.28 V 2 Ag + (aq) + Ni(s) 2 Ag(s) + Ni 2 + (aq) E ° = 0.80 ( 0.28 ) = 1.08 V
    2. As the reaction proceeds, Ni 2+ (aq) is produced, so [Ni 2+ ] increases as the cell operates.
    3. E = E ° 0.0592 n log Q; 1 .12 = 1.08 0.0592 2 log [Ni 2 + ] [ Ag + ] 2

      0.04 ( 2 ) 0.0592 = log(0 .0100) log[Ag + ] 2 ; log[Ag + ] 2 = log(0 .0100) + 0 .04(2) 0 .0592

      log[Ag + ] 2 = –2.000 + 1.351 = –0.649; [Ag + ] 2 = 0.255 M ; [Ag + ] = 0.474 = 0.5 M

      [Strictly speaking, [E – E°] having only 1 sig fig leads (after several steps) to the answer having only 1 sig fig. This is not a very precise or useful result.]

    20.102

    1. I 2 ( s ) + 2 e 2 I (aq) E red o = 0.54 V 2 [ Cu(s) Cu + (aq) + 1 e ] E red o = 0.52 V I 2 (s) + 2 Cu(s) 2 Cu + (aq) + 2 I (aq) E ° = 0.54 0.52 = 0.02 V

      E = E ° 0.0592 n log Q = 0.02 0.0592 2 log [Cu + ] 2 [ I ] 2

      E = 0.02 0.0592 2 log (0 .25) 2 (3 .5) 2 = 0.02 + 0.0034 = 0.0234 = 0.02 V

    2. Because the cell potential is positive at these concentration conditions, the reaction as written in part (a) is spontaneous in the forward direction. Cu is oxidized and Cu(s) is the anode.
    3. Yes. E° is positive, so Cu is oxidized and Cu(s) is the anode at standard conditions.
    4. E = 0 ; + 0.02 = 0.0592 2 log (0 .15) 2 [ I ] 2 ; 2 ( 0.02 ) 0.0592 = log (0 .15) 2 + 2 log [ I ] ;

      log[I–] = 1.1617 = 1.2; [I–] = 10 1.1617 = 14.71 = 1 × 10 1 M I

      Strictly speaking, E° having only 1 sig fig leads, after several steps, to the [I ] having only 1 sig fig. This is not a very precise or useful result. The result does indicate that, at low [Cu + ], the cell potential is relatively insensitive to [I ].

    20.103  Assume that room temperature is 298 K. Use the relationship developed in Solution 20.55 to calculate K from E° at 298 K. Use data from Appendix E to calculate E° for the disproportionation.

    Cu + ( aq ) + 1 e Cu(s) E red o = 0.52 V Cu + (aq) Cu 2 + (aq) + 1 e E red o = 0.15 V 2 Cu + (aq) Cu(s) + Cu 2 + (aq) E ° = 0.52 V 0.15 V = 0.37 V

    E ° = 0.0592 n log K, log K = nE ° 0 .0592 = 1 × 0.37 0.0592 = 6.25 = 6.3

    K=10 6.25 = 1.778 × 10 6 = 2 × 10 6


    20.104

    1. In discharge, Cd(s) + 2 NiO(OH)(s) + 2 H 2 O(l) → Cd(OH) 2 (s) + 2 Ni(OH) 2 (s).

      In charging, the reverse reaction occurs.

    2. E° = 0.49 V – (–0.76 V) = 1.25 V
    3. The 1.25 V calculated in part (b) is the standard cell potential, E°. The concentrations of reactants and products inside the battery are adjusted so that the cell output is greater than E°. Note that most of the reactants and products are pure solids or liquids, which do not appear in the Q expression. It must be [OH ] that is other than 1.0 M , producing an emf of 1.30 rather than 1.25.
    4. = 0.0592 n log K; log k = nE° 0 .0592

      log K = 2 × 1.30 0.0592 = 43.92 = 43.9 ; K = 8.3 × 10 43 = 8 × 10 43

    20.105

    1. The battery capacity expressed in units of mAh indicates the total amount of electrical charge that can be delivered by the battery.
    2. Quantity of electrical charge is measured in coulombs, C. C = A-s

      2850 mAh × 1 A 1000 mA × 3600 s h × 1 C 1 A-s = 10 , 260 C

      The battery can deliver 10,260 C. Work, electrical or otherwise, is measured in J. J = V × C. If the battery voltage decreases linearly from 1.55 V to 0.80 V, assume an average voltage of 1.175 = 1.2 V.

      10, 260 C × 1.175 V = 12,055.5 = 12 × 10 3 J = 12 kJ

      The total maximum electrical work of the battery is 12 kJ.

      (This is 3.3 × 10 −3 kWh or 3.3 Wh.)

    20.106

    1. By analogy to O in hydroxide (OH ) and in alcohol (R–OH), the oxidation state of S in a thiol (R–SH) is –2.
    2. By analogy to O in peroxide (–O–O–), the oxidation state of S in a disulfide (–S–S–) is –1.
    3. When two thiols react to form a disulfide, the oxidation number of S changes from –2 to –1. The oxidation number of S becomes more positive and the thiols are oxidized.
    4. Converting a disulfide to two thiols is the reverse of the process described in part (c). In order to reduce the disulfide, a reducing agent must be added to the solution.
    5. When two thiols (R–SH) react to form a disulfide (R–S–S–R), the H atoms are probably removed by base before the oxidizing agent does its work.

    20.107

    1. Total volume of Cr = 2.5 × 10 −4 m × 0.32 m 2 = 8.0 × 10 −5 m 3

      mol Cr = 8.0 × 10 5 m 3 Cr × 100 3 cm 3 1 m 3 × 7.20 g Cr 1 cm 3 × 1 mol Cr 52 .0 g Cr = 11.077 = 11 mol Cr

      The electrode reaction is:

      CrO 4 2– (aq) + 4 H 2 O(l) + 6 e → Cr(s) + 8 OH (aq)

      Coulombs required = 11.077 mol Cr × 6 F 1 mol Cr × 96 , 485 C 1 F = 6.41 × 10 6 = 6.4 × 10 6 C


    1. 6.41 × 10 6 C × 1 amp-s 1 C × 1 10.0 s = 6.4 × 10 5 amp
    1. If the cell is 65% efficient, (6.41 × 10 6 /0.65) = 9.867 × 10 6 = 9.9 × 10 6 C is required to plate the bumper.

      6.0 V × 9 .867 × 10 6 C × 1 J 1 C-V × 1 kWh 3 .6 × 10 6 J = 16.445 = 16 kWh

    20.108

    1. The standard reduction potential for H 2 O(l) is much greater than that of Mg 2+ (aq)(–0.83 V vs. –2.37 V). In aqueous solution, H 2 O(l) would be preferentially reduced and no Mg(s) would be obtained.
    2. 97 , 000 A × 24 h × 3600 s 1 h × 1 C 1 A-s × 1 F 96,485 C × 1 mol Mg 2 F × 24.31 g Mg 1 mol Mg × 0.96 = 1.0 × 10 6 g Mg = 1.0 × 10 3 kg Mg

    20.109 Analyze . Given mass of aluminum desired, applied voltage, and electrolysis efficiency, calculate kWh of electricity required. Plan . Beginning with mass Al and paying attention to units, calculate coulombs required if the process is 100% efficient. Then, take efficiency into account and use V, C, and the relationship between J and kWh to calculate kWh required. Solve .

    1.0 × 10 3 kg Al × 1 000 g 1 kg × 1 mol Al 2 6.98 g Al × 3 F 1 mol Al × 96 , 485 C F = 1.073 × 10 10 = 1.1 × 10 10 C

    If the cell is 45% efficient, (1.073 × 10 10 /0.45) = 2.384 × 10 10 = 2.4 × 10 10 C is required to plate the bumper.

    4.50 V × 2 .384 × 10 10 C × 1 J 1 C-V × 1 kWh 3 .6 × 10 6 J = 29 , 801 = 3.0 × 10 4 kWh

    20.110

    1. 7 × 10 8 mol H 2 × 2 F 1 mol H 2 × 96 , 485 C 1 F = 1.35 × 10 14 = 1 × 10 14 C
    2. 2 [ 2 H + (aq) + 2 e H 2 (g)] E red o = 0.00 V 2 H 2 O(l) O 2 (g) + 4 H + (aq) + 4 e E red o = 1.23 V 2 H 2 O(l) O 2 (g) + 2 H 2 (g) E o = 0.00 1.23 = 1.23 V

      P t = 300 atm = P O 2 + P H 2 . Because H 2 (g) and O 2 (g) are generated in a 2:1 mole ratio, P H 2 = 200 atm and P O 2 = 100 atm .

      E = 0.0592 4 log (P O 2 × P H 2 2 ) = 1.23 V 0.0592 4 log [100 × (200) 2 ]

      E = –1.23 V – 0.100 V = –1.33 V; E min = 1.33 V

    3. Energy = nFE = 2 ( 7 × 10 8 mol) (1 .33 V) 96,485 J V-mol = 1.80 × 10 14 = 2 × 10 14 J
    4. 1.80 × 10 14 J × 1 kWh 3 .6 × 10 6 J × $ 0.85 kWh = $ 4.24 × 10 7 = $ 4 × 10 7

      It would cost more than $40 million for the electricity alone.


    Integrative Exercises

    20.111  N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

    1. The oxidation number of H 2 (g) and N 2 (g) is 0. The oxidation number of N in NH 3 is –3, H in NH 3 is +1. H 2 is being oxidized and N 2 is being reduced.
    2. Calculate ΔG° from Δ G f o values in Appendix C. Use ΔG° = –RT ln K to calculate K.

      Δ = 2 Δ G f o NH 3 (g) Δ G f o N 2 (g) 3 Δ G f o H 2 (g)

      ΔG° = 2(–16.66 kJ) – 0 – 3(0) = –33.32 kJ

      Δ = RT ln K, ln K = Δ RT = ( 33.32 × 10 3 J) ( 8.314 J/mol-K) (298 K) = 13.4487 = 13.45

      K = e 13.4487 = 6.9 × 10 5

    3. Δ = nfE°; = Δ nF ; n = ?

      2 N atoms change from 0 to –3 or 6 H atoms change from 0 to +1.

      Either way, n = 6.

      = ( 33.32 kJ) 6 × 96.5 kJ/V = 0.05755 V

    20.112  The redox reaction is: 2 Ag + (aq) + H 2 (g) → 2 Ag(s) + 2 H + (aq). n = 2 for this reaction.

    E cell o = E red o cathode E red o anode = 0.80 V 0.00 V = 0.80 E = 0.0592 n log [H + ] 2 [ Ag + ] 2 P H 2

    [H + ] in the cell is held essentially constant by the benzoate buffer.

    C 6 H 5 COOH(aq)  ⇌  H + (aq) + C 6 H 5 COO (aq)    K a = ?

    K a = [ H + ] [ C 6 H 5 COO ] [ C 6 H 5 COOH] ; [ H + ] = K a [ C 6 H 5 COOH] [C 6 H 5 COO ] = 0.10 M 0 .050 M × K a = 2 K a

    Solve the Nernst expression for [H + ] and calculate K a and pK a as shown above.

    1.030 V = 0.80 V 0.0592 n log [H + ] 2 ( 1.00 ) 2 ( 1.00 ) 0.23 × 2 0.0592 = log [ H + ] 2 = 2 log[H + ] 0.23 0.0592 = log [H + ] = pH; pH = 3.885 = 3.9 ; [H + ] = 10 3.885 = 1.303 × 10 4 = 1 × 10 4

    [H + ] = 2K a ; K a = [H + ]/2 = 6.515 × 10 −5 = 7 × 10 −5 ; pK a = 4.186 = 4.2

    Check .  According to Appendix D, K a for benzoic acid is 6.3 × 10 −5 .

    20.113

    1. +1
    2. –1 (While we don't know the oxidation number of N in chloramine, we can assume the most common oxidation number for Cl.)
    3. Reduced. The oxidation number of Cl becomes more negative, so Cl is reduced.
    4. +3
    5. Oxidized. The oxidation number of N goes from –3 to +3, so N is oxidized.

    20.114

    1. Ag + (aq) + e Ag(s) E red o = 0.80 V Fe 2 + (aq) Fe 3 + (aq) + 1 e E red o = 0.77 V Ag + (aq) + Fe 2 + (aq) Ag(s) + Fe 3 + (aq) E ° = 0.80 V 0.77 V = 0.03 V
    2. Ag + (aq) is reduced at the cathode and Fe 2+ (aq) is oxidized at the anode.
    3. ΔG° = –nFE o = –(1)(96.5)(0.03) = –2.895 = 3 kJ

      ΔS° = S° Ag(s) + S° Fe 3+ (aq) – S°Ag + (aq) – S° Fe 2+ (aq)

      = 42.55 J + 293.3 J – 73.93 J – 113.4 J = 148.5 J

      ΔG° = ΔH° – TΔS°. Because ΔS° is positive, ΔG° will become more negative and E° will become more positive as temperature is increased.

    20.115

    1. ΔH° = 2ΔH° H 2 O(l) – 2ΔH° H 2 (g) – ΔH° O 2 (g) = 2(–285.83) – 2(0) – 0 = –571.66 kJ

      ΔS° = 2S° H 2 O(l) – 2S° H 2 (g) – ΔS° O 2 (g)

      = 2(69.91) – 2(130.58) – (205.0) = –326.34 J

    2. Because ΔS° is negative, –TΔS is positive and the value of ΔG will become more positive as T increases. The reaction will become nonspontaneous at a fairly low temperature, because the magnitude of ΔS° is large.
    3. ΔG = w max . The larger the negative value of ΔG, the more work the system is capable of doing on the surroundings. As the magnitude of ΔG decreases with increasing temperature, the usefulness of H 2 as a fuel decreases.
    4. The combustion method increases the temperature of the system, which quickly decreases the magnitude of the work that can be done by the system. Even if the effect of temperature on this reaction could be controlled, only about 40% of the energy from any combustion can be converted to electrical energy, so combustion is intrinsically less efficient than direct production of electrical energy via a fuel cell.

    20.116  First, balance the equation:

    4 CyFe 2+ (aq) + O 2 (g) + 4 H + (aq) → 4 CyFe 3+ (aq) + 2 H 2 O(l); E = +0.60 V; n = 4

    1. From Equation 20.12, we can calculate ΔG for the process under the conditions specified for the measured potential E:

      Δ G = nFE = ( 4 mol e ) × 96.485 kJ 1 V-mol e (0 .60 V) = 231.6 = 232 kJ

    2. The moles of ATP synthesized per mole of O 2 is given by:

      231.6 kJ O 2 molecule × 1 mol ATP formed 37 .7 kJ = approximately 6 mol ATP/mol O 2


    20.117 AgSCN(s) + e Ag(s) + SCN (aq) E red o = 0.09 V Ag(s) Ag + (aq) + e E red o = 0.80 V AgSCN(s) Ag + (aq) + SCN (aq) E ° = 0.09 0.80 = 0.71 V

    = 0.0592 n log K sp ; log K sp = ( 0.71 ) ( 1 ) 0.0592 = 11.993 = 12

    K sp = 10 −11.993 = 1.02 × 10 −12 = 10 −12

    20.118  The reaction can be written as a sum of the steps:

    Pb 2 + (aq) + 2 e Pb(s) E red o = 0.13 V PbS(s) Pb 2 + (aq) + S 2 (aq) “E o = ? PbS(s) + 2 e Pb(s) + S 2 (aq) E red o = ?

    “E o” for the second step can be calculated from K sp .

    = 0.0592 n log K sp = 0.0592 2 log (8 .0 × 10 28 ) = 0.0592 2 ( 27.10 ) = 0.802 V

    E o for the half-reaction = –0.13 V + (–0.802 V) = –0.93 V

    Calculating an imaginary E o for a nonredox process like step 2 may be a disturbing idea. Alternatively, one could calculate K for step 1 (5.4 × 10 −5 ), K for the reaction in question (K = K 1 × K sp = 4.4 × 10 −32 ), and then E o for the half-reaction. The result is the same.

    20.119  The two half-reactions in the electrolysis of H 2 O(l) are:

    2 [ 2 H 2 O(l) + 2 e H 2 (g) + 2 OH ] 2 H 2 O(l) O 2 (g) + 4 H + + 4 e 2 H 2 O(l) 2 H 2 (g) + O 2 (g)

    4 mol e /2 mol H 2 (g) or 2 mol e /mol H 2 (g)

    Using partial pressures and the ideal-gas law, calculate the mol H 2 (g) produced, and the current required to do so.

    P t = P H 2 + P H 2 O . From Appendix B, P H 2 O at 25.5 °C is approximately 24.5 torr.

    P H 2 = 768 torr 24.5 torr = 743 .5 = 744 torr

    n = PV/RT = ( 743.5 / 760 ) atm × 0 .0123 L 298.5 K × 0 .08206 L-atm/mol-K = 4.912 × 10 4 = 4.91 × 10 4 mol H 2

    4.912 × 10 4 mol H 2 × 2 mol e mol H 2 × 96 , 485 C 1 mol e × 1 amp-s 1 C × 1 min 60 s × 1 2.00 min = 0.790 amp


     

    21 Nuclear Chemistry

    Visualizing Concepts

    21.1 Analyze .  Given the name and mass number of a nuclide, decide if it lies within the belt of stability. If not, suggest a process that moves it toward the belt.

    Plan .  Calculate the number of protons and neutrons in each nuclide. Locate this point on Figure 21.1. If the point is above the belt, beta decay increases protons and decreases neutrons, decreasing the neutron-to-proton ratio. If the point is below the belt, either positron emission or neutron capture decreases protons and increases neutrons, increasing the neutron-to-proton ratio. Solve.

    1. 24 Ne: 10 p, 14 n, just above the belt of stability. Reduce the neutron-to-proton ratio via beta decay.
    2. 32 Cl: 17 p, 15 n, just below the belt of stability. Increase the neutron-to-proton ratio via positron emission or orbital electron capture.
    3. 108 Sn: 50 p, 58 n, just below the belt of stability. Increase the neutron-to-proton ratio via positron emission or orbital electron capture.
    4. 216 Po: 84 p, 132 n, just beyond the belt of stability. Nuclei with atomic numbers ≥ 84 tend to decay via alpha emission, which decreases both protons and neutrons.

    21.2 Analyze/Plan. From the diagram, determine the atomic number (number of protons) and mass number (number of protons plus neutrons) of the two nuclides involved. Based on the relationship between the two nuclides, decide whether the reaction is alpha or beta decay, positron emission or electron capture. Complete the nuclear reaction, balancing atomic numbers and mass numbers.

    Solve .  The two nuclides in the diagram are 46 109 Pd and 47 109 Ag, so the second product is a beta particle. The balanced reaction is:

    46 109 Pd 47 109 Ag + 1 0 e

    Check. Atomic number and mass number balance.

    21.3 Analyze/Plan. Determine the number of protons and neutrons present in the two heavy nuclides in the reaction. Draw a graph with appropriate limits and plot the two points. Draw an arrow from reactant to product. Solve.


    Bi: 83 p, 128 n; Tl: 81 p, 126 n.

    image

    Check. An alpha particle has 2 p and 2 n. The diagram shows a decrease in 2 p and 2 n for the reaction.

    21.4 Analyze/Plan. Count the protons and neutrons in each particle. Use a periodic table and Table 21.2 to identify the particles. Refer to Sample Exercise 21.4 when writing the reaction using condensed notation. Use Figure 21.1 to determine stability of the product nucleus.

    1. From left to right, the particles are 5 10 B , n 0 1 , 3 7 Li , and 2 4 He
    2. 5 10 B (n, α) 3 7 Li
    3. The 3 7 Li product nucleus is probably stable. It appears to be in the belt of stability in Figure 21.1. The product nucleus has an odd number of protons but an even number of neutrons. There are 50 stable “odd, even” nuclei.

    21.5 Analyze/Plan. Use the information in Table 21.3 to identify the type of radioactive decay for each step.

      1. Alpha decay. Mass number decreases by 4 and atomic number decreases by 2.
      2. Beta decay. Mass number is the same, atomic number increases by 1.
      3. Beta decay. Mass number is the same, atomic number increases by 1.
    1. 89 228 Ac has the highest activity; it has the shortest half-life.
    2. 90 232 Th has the lowest activity; it has the longest half-life.
    3. 88 224 Rn . Mass number decreases by 4 and atomic number decreases by 2, relative to the last isotope shown.

    21.6 Analyze/Plan. Write the balanced equation for the decay. Nuclear decay is a first-order process; use appropriate relationships for first-order processes to determine t 1/2 , k and remaining 88 Mo after 12 minutes. Solve.

    1. t 1/2 is the time required for half of the original nuclide to decay. Relative to the graph, this is the time when the amount of 88 Mo is reduced from 1.0 to 0.5. This time is 7 minutes.
    2. For a first-order process, t 1/2 = 0.693/k or k = 0.693/t 1/2 .

      k = 0.693/7 min = 0.0990 = 0.1 min –1


    1. From the graph, the fraction of 88 Mo remaining after 12 min is 0.3/1.0 = 0.3

      Check. In(N t /N o ) = –kt = –(0.099)(12) = –1.188; N t /N o = e 1.188 = 0.30.

    1. 42 88 Mo 41 88 Nb + e + 1 0

    21.7 Analyze/Plan. Atomic number is number of protons, mass number is (protons + neutrons). Chemical symbol is determined by atomic number. Beta decay increases the number of protons, whereas mass number stays constant. Positron emission decreases the number of protons, whereas mass number stays constant. Half-life, t 1/2 , is the time required to reduce the amount of radioactive material by half. Solve.

    1. 5 10 B , 5 11 B ; 6 12 C , 6 13 C ; 7 14 N , 7 15 N ; 8 16 O , 8 17 O , 8 18 O ; 9 19 F
    2. On the diagram, 6 14 C is the only radioactive nuclide above and left of the band of stable red nuclides. 6 14 C will reduce its neutron-to-proton ratio by beta decay.

      6 14 C 7 14 N + e 1 0

    3. On the diagram, radioactive nuclides right and below the band of stable red nuclides are likely to increase their neutron-to-proton ratio via positron emission. To be useful for positron emission tomography, the nuclides must have a half-life on the order of minutes. (Nuclides with very fast decay disappear before they can be imaged. Those with longer half-lives linger in the patient.) Four radioactive nuclides fit these criteria: 6 11 C , 7 13 N , 8 15 O , 9 18 F
    4. If 12.5% of the isotope remains, this amounts to three half-lives of decay. If the total decay time is 1 hour, 1/3 of this time is 20 min. The radionuclide with a half-life of 20 min is 6 11 C .

    21.8 Analyze/Plan. Express the particles in the diagram as a nuclear reaction. Determine the mass number and atomic number of the unknown particle by balancing these quantities in the nuclear reaction. Solve.

    1. 94 239 Pu + 0 1 n 40 95 Zr + ? + 2 0 1 n

      The unidentified particle has an atomic number of (94–40) = 54; it is Xe. The mass number of the nuclide is [(239 + 1) – (95 + 2)] = 143. The unknown particle is 54 143 Xe .

    2. 95 Zr: 40 p, 55 n is stable. 143 Xe: 54 p, 89 n is above the belt of stability and is not stable; it will probably undergo beta decay.

    Radioactivity and Nuclear Equations (Section 21.1)

    21.9 Analyze/Plan. Given various nuclide descriptions, determine the number of protons and neutrons in each nuclide. The left superscript is the mass number, protons plus neutrons. If there is a left subscript, it is the atomic number, the number of protons. Protons can always be determined from the chemical symbol; all isotopes of the same element have the same number of protons. A number following the element name, as in part (c) is the mass number. Solve.

    p = protons, n = neutrons, e = electrons; number of protons = atomic number; number of neutrons = mass number – atomic number

    1. 24 56 Cr: 24p, 32n
    2. 193 Tl: 81p, 112n
    3. 38 Ar: 18p, 20n

    21.10  p = protons, n = neutrons, e = electrons; number of protons = atomic number; number of neutrons = mass number – atomic number

    1. 53 129 I: 53p, 76n
    2. 138 Ba: 56p, 82n
    3. 237 Np: 93p, 144n

    21.11 Analyze/Plan. See definitions in Section 21.1. In each case, the left superscript is mass number, the left subscript is related to atomic number. Solve.

    1. 0 1 n
    2. 2 4 He or α
    3. 0 0 γ or γ

    21.12

    1. 1 1 p or 1 1 H
    2. e 1 0 or β
    3. e + 1 0 or β +

    21.13 Analyze/Plan. Follow the logic in Sample Exercises 21.1 and 21.2. Pay attention to definitions of decay particles in Table 21.2 as well as conservation of mass and charge. Solve .

    1. 37 90 Rb 38 90 Sr + 1 0 e
    2. 34 72 Se + 1 0 e (orbital electron) 33 72 As
    3. 36 76 Kr 35 76 Br + + 1 0 e
    4. 88 226 Ra 86 222 Rn + 2 4 He

    21.14

    1. 83 213 Bi 81 209 Tl + 2 4 He
    2. 7 13 N + e 1 0 (orbital electron) 6 13 C
    3. 43 98 Tc + e 1 0 (orbital electron) M 42 98 o
    4. 79 188 Au 78 188 Pt + + 1 0 e

    21.15 Analyze/Plan. Using definitions of the decay processes and conservation of mass number and atomic number, work backwards to the reactants in the nuclear reactions. Solve .

    1. 82 211 Pb 83 211 Bi + 1 0 e
    2. 25 50 Mn 24 50 Cr + + 1 0 e
    3. 74 179 W + 1 0 e 73 179 Ta
    4. 90 230 Th 88 226 Ra + 2 4 He

    21.16

    1. 11 24 Na 12 24 Mg + 1 0 e; a beta particle is produced
    2. 80 188 Hg 79 188 Au + + 1 0 e; a positron is produced
    3. 53 122 I 54 122 Xe + 1 0 e; a beta particle is produced
    4. 94 242 Pu 92 238 U + 2 4 He; an alpha particle is produced

    21.17 Analyze/Plan. Given the starting and ending nuclides in a nuclear decay sequence, we are asked to determine the number of alpha and beta emissions. Use the total change in A and Z, along with definitions of alpha and beta decay, to answer the question. Solve.

    The total mass number change is (235–207) = 28. As each alpha particle emission decreases the mass number by four, whereas emission of a beta particle does not correspond to a mass change, there are 7 alpha particle emissions. The change in atomic number in the series is 10. Each alpha particle results in an atomic number lower by two. The 7 alpha particle emissions alone would cause a decrease of 14 in atomic number. Each beta particle emission raises the atomic number by one. To obtain the observed lowering of 10 in the series, there must be 4 beta emissions.


    21.18  This decay series represents a change of (232–208 =) 24 mass units. Because only alpha emissions change the nuclear mass, and each changes the mass by four, there must be a total of 6 alpha emissions. Each alpha emission causes a decrease of two in atomic number.

    Therefore, the 6 alpha emissions, by themselves, would cause a decrease in atomic number of 12. The series as a whole involves a decrease of 8 in atomic number. Thus, there must be a total of 4 beta emissions, each of which increases atomic number by one. Overall, there are 6 alpha emissions and 4 beta emissions.

    Patterns of Nuclear Stability (Section 21.2)

    21.19 Analyze/Plan. Follow the logic in Sample Exercise 21.3, paying attention to the guidelines for neutron-to-proton ratio. Solve.

    (a) 5 8 B - low neutron/proton ratio, positron emission (for low atomic numbers, positron emission is more common than orbital electron capture)
    (b) 29 68 Cu - high neutron/proton ratio, beta emission
    (c) 15 32 P - slightly high neutron/proton ratio, beta emission
    (d) 17 39 Cl - high neutron/proton ratio, beta emission

    21.20

    (a) 1 3 H - high neutron/proton ratio, beta emission
    (b) S 38 89 r - (slightly) high neutron/proton ratio, beta emission
    (c) 53 120 I - low neutron/proton ratio, positron emission
    (d) 4 7 102 Ag - low neutron/proton ratio, positron emission

    21.21 Analyze/Plan. Use the criteria listed in Table 21.4. Solve.

    1. 19 39 K ; stable, odd proton, even neutron; 20 neutrons is a magic number.

      19 40 K ; radioactive, odd proton, odd neutron

    2. 83 209 Bi ; stable, odd proton, even neutron; 126 neutrons is a magic number.

      83 208 Bi ; radioactive, odd proton, odd neutron

    3. 28 58 Ni ; stable, even proton, even neutron

      28 65 Ni ; radioactive, high neutron/proton ratio

    21.22  Use criteria listed in Table 21.4.

    1. C 20 40 a ; stable, magic numbers of protons and neutrons

      C 20 45 a , radioactive, high neutron/proton ratio.

    2. C 6 12 , stable, even proton, even neutron

      C 6 14 , radioactive, high neutron/proton ratio


    1. 8 2 206 Pb , stable, magic number of protons, even proton, even neutron

      9 0 230 Th , radioactive, atomic number greater than 84

    21.23 Analyze/Plan. For each nuclide, determine the number of protons and neutrons and decide if they are magic numbers. Solve.

    1. 2 4 He , both
    2. 8 18 O has a magic number of protons, but not neutrons
    3. 20 40 Ca , both
    4. 30 66 Zn has neither
    5. P 82 208 b , both

    21.24  Statement (d). Evidence shows that pairs of protons and neutrons have special stability, so elements with even atomic numbers have more abundant isotopes.

    Regarding statements (a) and (b), elements with odd atomic numbers can lie above, below, or within the belt of stability. Regarding statement (c), none of the elements on the graph has a magic number of protons.

    21.25  Statement (a). Because of its magic numbers, the alpha particle is a very stable emitted particle, which makes alpha emission a favorable process. The proton has no magic numbers, and is an odd proton–even neutron particle. Its instability does not encourage proton emission as a process.

    21.26  The criterion employed in judging whether the nucleus is likely to be radioactive is the position of the nucleus on the plot shown in Figure 21.1. If the neutron/proton ratio is too high or low, or if the atomic number exceeds 83, the nucleus will be radioactive.

    Radioactive: 27 60 Co —odd proton, odd neutron

    41 92 Nb —odd proton, odd neutron

    226 Ra —high atomic number

    Stable: 26 58 Fe —even proton, even neutron, stable neutron/proton ratio

    202 Hg —even proton, even neutron, stable neutron/proton ratio

    Nuclear Transmutations (Section 21.3)

    21.27  Statement (b) is the best explanation. Because they are electrically neutral, neutrons are not repelled by the positively charged nucleus being bombarded.

    Statement (a) does not apply and statements (c) and (d) are false. Neutrons are not attracted to the nucleus at long distances.

    21.28  The element is technetium, Tc. 42 96 Mo + H 1 2 43 98 Tc

    21.29 Analyze/Plan. Determine A and Z for the missing particle by conservation principles. Find the appropriate symbol for the particle. Solve.

    1. C 98 252 f + B 5 10 3 n 0 1 + L 103 259 r
    2. H 1 2 + H 2 3 e H 2 4 e + H 1 1
    3. H 1 1 + B 5 11 3 H 2 4 e
    4. I 53 122 X 54 122 e + e 1 0
    5. F 26 59 e e 1 0 + C 27 59 o

    21.30

    1. N 7 14 + H 2 4 e O 8 17 + H 1 1
    2. 19 40 K + e 1 0 (orbital electron) 18 40 Ar
    3. A 13 27 l + H 2 4 e S 14 30 i + H 1 1
    4. F 26 58 e + 2 n 0 1 e 1 0 + C 27 60 o
    5. U 92 235 + n 0 1 X 54 135 e + S 38 99 r + 2 n 0 1

    21.31 Analyze/Plan. Follow the logic in Sample Exercise 21.4, paying attention to conservation of A and Z. Solve.

    1. U 92 238 + H 2 4 e P 94 241 u + n 0 1
    2. N 7 14 + H 2 4 e O 8 17 + H 1 1
    3. F 26 56 e + H 2 4 e C 29 60 u + e 1 0

    21.32

    1. U 92 238 + n 0 1 U 92 239 + γ 0 0
    2. O 8 16 + H 1 1 N 7 13 + H 2 4 e
    3. O 8 18 + n 0 1 F 9 19 + e 1 0

    Rates of Radioactive Decay (Section 21.4)

    21.33

    1. True. k = 0.693/t 1/2 . The decay rate constant, k, and half-life, t 1/2 are inversely related.
    2. False. If X is not radioactive, it does not spontaneously decay and its half-life is essentially infinity.
    3. True. Changes in the amount of A would be measurable over the 40-year time frame, whereas changes in the amount of X would be very small and difficult to detect.

    21.34

    1. The suggestion is not reasonable. The energies of nuclear states are very large relative to ordinary temperatures. Thus, merely changing the temperature by less than 100 K would not be expected to significantly affect the behavior of nuclei with regard to nuclear decay rates.
    2. No. Radioactive decay has no activation energy like a chemical reaction. Activation energy is the minimum amount of energy required to initiate a chemical reaction. Radioactive decay is a spontaneous nuclear transformation from a less stable to a more stable nuclear configuration. Radioisotopes are by definition in a “transition state,” prone to nuclear change or decay. Changes in external conditions such as temperature, pressure or chemical state provide insufficient energy to either excite or relax an unstable nucleus.

    21.35 Analyze/Plan. The half-life is 12.3 yr. Use t 1/2 to calculate k and the mass fraction (N t / N 0 ) remaining after 50 yr.

    Solve. k = 0.693/ t 1/2 = 0.693/12.3 yr = 0.056341 = 0.0563 yr –1

    ln N t N 0 = kt ; ln N t N 0 = 0.056341 yr 1 ( 50 yr ) = 2.81707 = 2.8 ; N t N 0 = 0.05978 = 0.06

    When the watch is 50 years old, only 6% (or 6.0%) of the tritium remains. The dial will be dimmed by 94%.


    21.36  Calculate the decay constant, k, and then t 1/2 . t = 4 h 39 min = 279 min

    k = 1 t ln N t N o = 1 279 min ln 0 .25 mg 2 .000 mg = 0.007453 = 0.0.0075 min 1

    Using Equation 21.19, t 1/2 = 0.693/k = 0.693/0.007453 min –1 = 92.980 = 93 min

    21.37 Analyze/Plan. We are given half-life of cobalt-60, and replacement time when the activity of the sample is 75% of the initial value. Consider the rate law for (first-order) nuclear decay: ln(N t /N o ) = –kt. Solve.

    k = 0.693 / t 1/2 = 0.693/5.26 yr = 0.1317 = 0.132 yr –1 ; N t /N o = 0.75

    t = 1 k ln N t N o = ( 1 / 0.1317 yr 1 ) ln (0 .75) = 2.18 yr = 26.2 mo = 797 d.

    The source will have to be replaced sometime in the summer of 2018, probably in August.

    21.38  Follow the logic in Sample Exercise 21.6. In this case, we are given initial sample mass as well as mass at time t, so we can proceed directly to calculate k (Equation 21.19), and then t (Equation 21.20). Solve.

    k = 0.693 / t 1/2 = 0.693/27.8 d = 0.02493 = 0.0249 d –1

    t = 1 k ln N t N o = 1 0.02493 d 1 ln 0 .75 6 .25 = 85.06 = 85 d

    21.39

    1. Analyze/Plan. R 88 226 a R 86 222 n + H 2 4 e

      1 alpha particle is produced for each 226 Ra that decays. Rate = kN. Calculate the number of 226 Ra particles in the 10.0 mg sample. Calculate t 1/2 and k in min, then rate in dis/min, then the number of disintegrations in 5.0 min. Solve.

      10.0 mg Ra × 1 g 1000 mg × 1 mol Ra 226 g Ra × 6.022 × 10 23 Ra atoms 1 mol Ra = 2.6646 × 10 19 = 2.66 × 10 19 atoms

      Calculate k in min –1 . 1600 yr × 365 d 1 yr × 24 h 1 d × 60 min 1 h = 8.410 × 10 8 m i n 1

      k = 0 .693 t 1 / 2 = 0.693 8.410 × 10 8 min = 8.241 × 10 10 min 1

      Rate = kN = (8.241 × 10 –10 min –1 )(2.6646 × 10 19 atoms) = 2.20 × 10 10 atoms/min

      (2.20 × 10 10 atoms/min)(5.0 min) = 1.1 × 10 11 226 Ra atoms decay in 5.0 min

      1.1 × 10 11 alpha particles emitted in 5.0 min

    2. Plan. From part (a), the rate is 2.20 × 10 10 disintegrations/min. Change this to dis/s and apply the definition 1 Ci = 3.7 × 10 10 dis/s.

      2.20 × 10 10 dis 1 min × 1 min 60 s × 1 Ci 3 .7 × 10 10 dis / s × 1000 mCi Ci = 9.891 = 9.9 mCi

    21.40

    1. Proceeding as in Solution 21.39, calculate number of 60 Co atoms and k in s –1 .

      3.75 mg Co × 1 g 1000 mg × 1 mol Co 60 g Co × 6.022 × 10 23 Co atoms 1 mol Co = 3.76375 × 10 19 = 3.76 × 10 19


      5.26 yr × 365 d 1 yr × 24 h 1 d × 3600 s 1 h = 1.659 × 10 8 = 1.66 × 10 8 s

      k = 0 .693 t 1 / 2 = 0.693 1.659 × 10 8 = 4.178 × 10 9 = 4.18 × 10 9 s 1

      Rate = kN = (4.178 × 10 –9 s –1 )(3.76375 × 10 19 atoms) = 1.57 × 10 11 atoms/s

      (1.57 × 10 11 atoms/s )(600 s) = 9.43 × 10 13 60 Co atoms decay in 600 s

      9.43 × 10 13 beta particles emitted by a 3.75 mg sample in 600 s

    1. 1.57 × 10 11 dis s × 1 Bq 1 dis / s = 1.57 × 10 11 Bq

      The activity of the sample is 1.57 × 10 11 Bq.

    21.41 Analyze/Plan. Calculate k in yr –1 and solve Equation 21.20 for t. N o = 16.3/min/g, N t = 9.7/min/g. Solve.

    k = 0.693/t 1/2 = 0.693/5715 yr = 1.213 × 10 –4 = 1.21 × 10 –4 yr –1

    t = 1 k ln N t N o = 1 1.213 × 10 4 yr 1 ln 9 .7 16 .3 = 4.280 × 10 3 = 4.3 × 10 3 yr

    21.42  Follow the logic in Sample Exercise 21.6.

    t = 1 k ln N t N o ; k = 0.693 / 5715 yr = 1 .213 × 10 4 yr 1 t = 1 1.213 × 10 4 yr 1 ln 38 .0 58 .2 = 3.52 × 10 3 yr

    21.43 Analyze/Plan. Follow the procedure outlined in Sample Exercise 21.6. If the mass of 40 Ar is 4.2 times that of 40 K, the original mass of 40 K must have been 4.2 + 1 = 5.2 times the amount of 40 K present now. Solve.

    k = 0.693/1.27 × 10 9 yr = 5.457 × 10 –10 = 5.46 × 10 –10 yr –1

    t = 1 5.457 × 10 10 yr 1 × ln 1 (5 .2) = 3.0 × 10 9 yr

    21.44  Follow the procedure outlined in Sample Exercise 21.6. The original quantity of 238 U is 75.0 mg plus the amount that gave rise to 18.0 mg of 206 Pb. This amount is 18.0(238/206) = 20.8 mg.

    k = 0.693/4.5 × 10 9 yr = 1.54 × 10 –10 = 1.5 × 10 –10 yr –1

    t = 1 k ln N t N o = 1 1.54 × 10 10 yr 1 ln 75 .0 95 .8 = 1.59 × 10 9 yr

    Energy Changes in Nuclear Reactions (Section 21.6)

    21.45 Analyze/Plan. Use Equation 21.22 to change 0.1 mg to energy. Solve.

    Δ E = c 2 Δ m = (2 .9979246 × 10 8 m / s) 2 × 0.1 mg × 1 g 1000 mg × 1 kg 1000 g × 1 kJ 1000 J = 9 × 10 6 kJ


    21.46 Analyze/Plan. Given a particle reaction, calculate Δm and ΔE. An alpha particle is 2 4 He . Use Equation 21.22, E = mc 2 . Compare the calculated energy to the energy given off by the thermite reaction. Solve.

    2 1 1 p + 2 0 1 n 2 4 He

    Δm = mass of individual protons and neutrons – mass of 2 4 He

    Δm = 2(1.0072765 amu) + 2(1.0086649 amu) – 4.0015 amu = 0.030383 = 0.0304 amu

    The mass change for the formation of 1 mol of 2 4 He can be expressed as 0.0304 g.

    ΔE = c 2 Δm; 1 J = kg-m 2 /s 2

    Δ E = 0 .030383 g × 1 kg 1000 g × ( 2.9979 × 10 8 m / s ) 2 = 2.73 × 10 12 kg-m 2 s 2 = 2.73 × 10 12 J

    The energy released when one mole of Al 2 O 3 is produced is 851.5 kJ or 8.515 × 10 5 J. The energy released when one mole of 2 4 He is formed from protons and neutrons is 2.73 × 10 12 J. This is 3 × 10 6 or 3 million times as much energy as the thermite reaction.

    21.47 Analyze/Plan. Given the mass of an 27 Al atom, subtract the mass of 13 electrons to get the mass of an 27 Al nucleus. Calculate the mass difference between the 27 Al nucleus and the separate nucleons, convert this to energy using Equation 21.22. Use the molar mass of 27 Al to calculate the energy required for 100 g of 27 Al . Solve .

    The mass of an electron is 5.485799 × 10 –4 amu (inside back cover of the text). The mass of a 27 Al nucleus is then 26.9815386 amu –13(5.485799 × 10 –4 amu ) = 26.9744071 amu. Δm = 13(1.0072765 amu) + 14(1.0086649 amu) – 26.9744071 amu = 0.2414960 amu.

    ΔE = (2.9979246 × 10 8 m/s) 2 × 0.2414960 amu × 1 g 6.0221421 × 10 23 amu × 1 kg 1 × 10 3 g

    = 3.604129 × 10 –11 = 3.604129 × 10 –11 J/ 27 Al nucleus required

    If the mass change for a single 27 Al nucleus is 0.2414960 amu, the mass change for 1 mole of 27 Al is 0.2414960 g.

    Δ E = 100 g 27 Al × 1 mol 27 Al 26 .9815386 g 27 Al × 0 .241960 mol 27 Al × 1 kg 1000 g × ( 2.9979246 × 10 8 m/s ) 2

    = 8.044234 × 10 13 J = 8.044234 × 10 10 kJ/100 g 27 Al

    21.48  Δm = mass of individual protons and neutrons – mass of nucleus

    Δm = 10(1.0072765 amu) + 11(1.0086649 amu) – 20.98846 amu = 0.1796189 = 0.17962 amu

    Δ E = (2 .9979246 × 10 8 m / s) 2 × 0.1796189 amu × 1 g 6 .0221421 × 10 23 amu × 1 kg 1000 g = 2 .680664 × 10 11 = 2.6807 × 10 11 J/ 21 Ne nucleus required

    2.680664 × 10 11 J nucleus × 6.0221421 × 10 23 nuclei mol = 1.6143 × 10 13 J / mol 21 Ne binding energy


    21.49 Analyze/Plan. Given atomic mass, subtract mass of the electrons to get nuclear mass. Calculate the nuclear binding energy by finding the mass difference between the nucleus and the separate nucleons and converting this to energy using Equation 21.22. Divide by the total number of nucleons to find binding energy per nucleon. Solve.

    1. Nuclear mass

      2 H: 2.014102 amu – 1(5.485799 × 10 –4 amu) = 2.013553 amu

      4 He: 4.002602 amu – 2(5.485799 × 10 –4 amu) = 4.001505 amu

      6 Li: 6.0151228 amu – 3(5.4857991 × 10 –4 amu) = 6.0134771 amu

    2. Nuclear binding energy

      2 H: Δm = 1(1.0072765) + 1(1.0086649) – 2.013553 = 0.002388 amu

      Δ E = 0.002388 amu × 1 g 6 .022142 × 10 23 amu × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2

      = 3.564490 × 10 –13 = 3.564 × 10 –13 J

      4 He: Δm = 2(1.0072765) + 2(1.0086649) – 4.001505 = 0.030378 amu

      Δ E = 0.030378 amu × 1 g 6 .022142 × 10 23 amu × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2

      = 4.533636 × 10 –12 = 4.5336 × 10 –12 J

      6 Li: Δm = 3(1.0072765) + 3(1.0086649) – 6.0134771 = 0.0343471 amu

      Δ E = 0.0343471 amu × 1 g 6 .022142 × 10 23 amu × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2

      = 5.126021 × 10 –12 = 5.12602 × 10 –12 J

    3. Binding energy per nucleon

      2 H: 3.564490 × 10 –13 J/2 nucleons = 1.782 × 10 –13 J/nucleon

      4 He: 4.533636 × 10 –12 J/4 nucleons = 1.1334 × 10 –12 J/nucleon

      6 Li: 5.126021 × 10 –12 J/6 nucleons = 8.54337 × 10 –13 J/nucleon

    4. This trend in binding energy/nucleon agrees with the curve in Figure 21.12, which shows an irregular increase in binding energy/nucleon up to atomic number 56. The anomalously high value for 4 He calculated above is also apparent on the figure.

    21.50  Nuclear mass is atomic mass minus mass of electrons. Nuclear binding energy is nuclear mass minus mass of the separate nucleons, converted to energy using Equation 21.22. Divide by the total number of nucleons to find binding energy per nucleon.

    1. Nuclear mass

      14 N: 13.999234 – 7(5.485799 × 10 –4 amu) = 13.995394

      48 Ti: 47.935878 – 22(5.485799 × 10 –4 amu) = 47.923809

      129 Xe: 128.904779 – 54(5.485799 × 10 –4 amu) = 128.875156 amu


    1. Nuclear binding energy

      14 N: Δm = 7(1.0072765) + 7(1.0086649) – 13.995394 = 0.1161959 = 0.116196 amu

      Δ E = 0.1161959 amu × 1 g 6 .0221421 × 10 23 amu × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2

      = 1.734127 × 10 –11 = 1.73413 × 10 –11 J

      48 Ti: Δm = 22(1.0072765) + 26(1.0086649) – 47.923809 = 0.4615614 = 0.461561 amu

      Δ E = 0.4615614 amu × 1 g 6 .0221421 × 10 23 amu × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2

      = 6.888428 × 10 –11 = 6.88843 × 10 –11 J

      129 Xe: Δm = 54(1.0072765) + 75(1.0086649) – 128.875156 = 1.1676425 = 1.167643 amu

      Δ E = 1.1676425 amu × 1 g 6 .0221421 × 10 23 amu × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2

      = 1.742610 × 10 –10 J

    1. Binding energy per nucleon

      14 N: 1.73413 × 10 –11 J / 14 nucleons = 1.23866 × 10 –12 J/nucleon

      48 Ti: 6.888428 × 10 –11 J / 48 nucleons = 1.43509 × 10 –12 J/nucleon

      129 Xe: 1.742610 × 10 –10 J/129 nucleons = 1.350860 × 10 –12 J/nucleon

    21.51 Analyze/Plan. Use Equation 21.22 to calculate the mass equivalence of the solar radiation. Solve.

    1. 1.07 × 10 16 kJ 1 min × 60 min 1 h × 24 h 1 d = 1.541 × 10 19 kJ d = 1.54 × 10 22 J / d

      Δ m = 1.541 × 10 22 kg-m 2 / s 2 / d ( 2.998 × 10 8 m / s ) 2 = 1.714 × 10 5 = 1.71 × 10 5 kg / d

    2. Analyze/Plan. Calculate the mass change in the given nuclear reaction, then a conversion factor for g 235 U to mass equivalent. Solve.

      Δm = 140.8833 + 91.9021 + 2(1.0086649) – 234.9935 = –0.19077 = –0.1908 amu

      Converting from atoms to moles and amu to grams, it requires 1.000 mol or

      235.0 g 235 U to produce energy equivalent to a change in mass of 0.1908 g.

      0.10% of 1.714 × 10 5 kg is 1.714 × 10 2 kg = 1.714 × 10 5 g

      1.714 × 10 5 g × 235.0 g 235 U 0 .1908 g = 2.111 × 10 8 = 2.1 × 10 8 g 235 U

      (This is about 230 tons of 235 U per day .)

    21.52  First, calculate nuclear masses from atomic masses. Then, the calculated Δm is for one group of single nuclides involved in a reaction, labeled Δm/‘atomic reaction.’ Multiplying by Avogadro’s number changes the quantity to ‘mol of reaction.’ Because energy is released, the sign of ΔE is negative.


    1 H: 1.00782 amu – 1(5.485799 × 10 –4 amu) = 1.0072714 = 1.00727 amu

    2 H: 2.01410 amu – 1(5.485799 × 10 –4 amu) = 2.0135514 = 2.01355 amu

    3 H: 3.10605 amu – 1(5.485799 × 10 –4 amu) = 3.1055014 = 3.10550 amu

    3 He: 3.10603 amu – 2(5.485799 × 10 –4 amu) = 3.1049328 = 3.10493 amu

    4 He: 4.00260 amu – 2(5.485799 × 10 –4 amu) = 4.0015028 = 4.00150 amu

    1. Δm = 4.0015028 + 1.0086649 – 3.1055014 – 2.0135514 = –0.1088851 = –0.10889 amu

      Δ E = 0 .1088851 amu ‘atomic reaction’ × 1 g 6 .022 × 10 23 amu × 6.022 × 10 23 ‘atomic reaction’ mol of reaction × 1 kg 10 3 g × ( 2.99792458 × 10 8 m / s ) 2 = 9.7861 × 10 12 J / mol

    2. Δm = 3.1049328 + 1.0086649 – 2(2.0135514) = –0.0864949 = –0.08649 amu

      ΔE = –7.774 × 10 12 J/mol

    3. Δm = 4.0015028 + 1.0072714 – 2.0135514 – 3.1049328 = –0.1097100 = –0.10971 amu

      ΔE = –9.8602 × 10 12 J/mol

    21.53  Nucleus (b), V 51 , should possess the greatest mass defect per nucleon. Figure 21.12 shows that the binding energy per nucleon (which gives rise to the mass defect) is greatest for nuclei with mass numbers around 50.

    21.54  Nuclear mass = 61.928345 amu – 28(5.485799 × 10 –4 amu) = 61.912985

    Binding energy = 28(1.0072765) + 34(1.0086649) – 61.912985 = 0.585363 amu

    Δ E = 0.585363 amu × 1 g 6 .0221421 × 10 23 amu × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2

    = 8.73606 × 10 –11 J

    Binding energy/nucleon = 8.73606 × 10 –11 J / 62 = 1.40904 × 10 –12 J/nucleon

    The value given for iron-56 in Table 21.7 is 1.41 × 10 –12 J/nucleon. These values are the same, to three significant figures.

    Nuclear Power and Radioisotopes (Sections 21.7, 21.8, and 21.9)

    21.55

    1. NaI is a good source of iodine, because it is a strong electrolyte and completely dissociated into ions in aqueous solution. The I (aq) are mobile and immediately available for bio-uptake. They do not need to be digested or processed in the body before uptake can occur. Also, iodine is a large percentage of the total mass of NaI.
    2. After ingestion, I (aq) must enter the bloodstream, travel to the thyroid and then be absorbed. This requires a finite amount of time. A Geiger counter placed near the thyroid immediately after ingestion will register background, then gradually increase in signal until the concentration of I (aq) in the thyroid reaches a maximum. Then, over time, iodine-131 decays, and the signal decreases.
    3. Analyze/Plan. The half-life of iodine-131 is 8.02 days. Use t 1/2 to calculate the decay rate constant, k. Then solve Equation 21.20 for t. N o = 0.12 (12% of ingested iodine absorbed); N t = 1.2 × 10 –5 (0.01% of the original amount taken up by the thyroid).

      Solve. k = 0.693 t 1/2 = 0.693/8.02 d = 0.086409 = 0.0864 d –1

      ln(N t /N o ) = –kt; t = –ln(N t /N o )/k

      t = ln ( 1.2 × 10 5 / 0.12 ) 0.086409 d 1 = 106.59 = 107 d

      Check. N t is given to 1 sig fig, so 1 × 10 2 days may be a more correct representation of the time frame for decay.

    21.56  Radioisotopes used as diagnostic tools are introduced into the body and carried to the point where imaging or some other diagnostic data is needed. We want the decay products of these radioisotopes to leave the body and do as little damage as possible on the way. Gamma rays are penetrating radiation and can escape the body more easily than other radioactive decay products. Also, gamma rays leaving the body can be easily detected using scintillation counters. This is particularly important when imaging is the goal of the procedure.

    Alpha emitters are never used as diagnostic tools because alpha particles are ionizing but do not move easily through the body. Trapped inside the body, alpha particles initiate the ionization of water, which ultimately produces free radicals that disrupt the normal operation of cells.

    21.57

    1. Characteristics (ii) and (iv) are required for a fuel in a nuclear power plant.

      Two or more neutrons (ii) are required so that a nuclear chain reaction occurs. Fission after absorption of a slow neutron (iv) is the nuclear process that produces energy in all current nuclear power plants. Gamma radiation (i) is produced by most nuclear decay processes and is a non-specific characteristic. Short half-life (iii) would require that fuel be replaced too frequently.

    2. 235 U

    21.58  Statements (i) and (iv) are true. Natural uranium contains about 0.7% 235 U; it must be enriched to about 3–5% for use as a fuel. 238 U undergoes neutron-induced fission according to Equation 21.12; no neutrons are produced.

    21.59  The control rods in a nuclear reactor regulate the flux of neutrons to keep the reaction chain self-sustaining and also prevent the reactor core from overheating. They are composed of materials such as boron or cadmium that absorb neutrons.

    21.60

    1. A moderator slows neutrons, so that they are more easily captured by fissioning nuclei.
    2. Water is the moderator in a pressurized water generator.
    3. Graphite is used as a moderator in gas-cooled reactors, and D 2 O is used in heavy-water reactors.

    21.61

    1. H 1 2 + H 1 2 H 2 3 e + n 0 1
    2. U 92 239 + n 0 1 S 51 133 b + N 41 98 b + 9 n 0 1

    21.62 Analyze/Plan. Use conservation of A and Z to complete the equations, keeping in mind the symbols and definitions of various decay products. Solve.

    1. U 92 235 + n 0 1 S 62 160 m + Z 30 72 n + 4 n 0 1
    2. P 94 239 u + n 0 1 C 58 144 e + K 36 94 r + 2 n 0 1

    21.63 Analyze/Plan. At these temperatures, assume the reaction occurs between nuclei rather than atoms. From Table 21.7, the nuclear mass of H 2 4 e is 4.00150. The nuclear mass of H 1 1 is simply the mass of a proton, 1.007276467 amu. Note that + 1 0 e is a positron, which has the same mass as an electron, 5.4857991 × 10 –4 amu. Calculate the difference in mass between product and reactant nuclei, and the energy released by this mass change. Do the calculation in terms of moles and grams, rather than nuclei and amu. 1 mol amu = 1 g. Solve.

    Δm = 4.00150 + 2(5.4857991 × 10 –4 ) – 4(1.007276467) = –0.0265087 = –0.02651 g

    If the reaction is run with 1 mol of H 1 1 , the mass change is (0.0265087/4) = 0.0066272 = 0.006627 g

    Δ E = c 2 Δ m = 0.0066272 g × 1 kg 1000 g × 8.987551 × 10 16 m 2 s 2 = 5.95621 × 10 11 = 5.956 × 10 11 J = 5.956 × 10 8 kJ

    21.64

    1. If the spent fuel rods are more radioactive than the original rods, the products of fission must lie outside the belt of stability and be radioactive themselves.
    2. The heavy (Z > 83) nucleus has a high neutron/proton ratio. The lighter radioactive fission products (for example, barium-142 and krypton-91) also have high neutron/proton ratios, because only 2 or 3 free neutrons are produced during fission. The preferred decay mode to reduce the neutron/proton ratio is beta decay, which has the effect of converting a neutron into a proton. Both barium-142 (86 n, 56 p) and krypton-91 (55 n, 36 p) undergo beta decay.

    21.65

    1. A boiling water reactor does not use a secondary coolant.
    2. A fast breeder reactor creates more fissionable material than it consumes.
    3. A gas-cooled reactor uses a gas as a primary coolant.

    21.66

    1. A heavy water reactor and a gas-cooled rector can use natural uranium as a fuel.
    2. A fast breeder reactor does not use a moderator.
    3. A high-temperature pebble-bed reactor can be refueled without shutting down.

    21.67 Analyze/Plan. Hydroxyl radical is electrically neutral but has an unpaired electron, ⋅ OH. Hydroxide is an anion, OH . Solve.

    Hydrogen abstraction: RCOOH + ⋅ OH → RCOO ⋅ + H 2 O

    Deprotonation:    RCOOH + OH → RCOO + H 2 O


    Hydroxyl radical is more toxic to living systems, because it produces other radicals when it reacts with molecules in the organism. This often starts a disruptive chain of reactions, each producing a different free radical.

    Hydroxide ion, OH , on the other hand, will be readily neutralized in the buffered cell environment. Its most common reaction is ubiquitous and innocuous:

    H + + OH → H 2 O. The acid–base reactions of OH are usually much less disruptive to the organism than the chain of redox reactions initiated by ⋅ OH radical.

    21.68  X-rays, alpha particles and gamma rays are classified as ionizing radiation.

    21.69 Analyze/Plan. Use definitions of the various radiation units and conversion factors to calculate the specified quantities. Pay particular attention to units. Solve.

    1. 1 Ci = 3.7 × 10 10 disintegrations(dis)/s; 1 Bq = 1 dis/s

      1 4 .3 mCi × 1 Ci 1000 m Ci × 3.7 × 10 10 dis / s Ci = 5.29 × 10 8 = 5.3 × 10 8 dis / s = 5 .3 × 10 8 Bq

    2. 1 rad = 1 × 10 –2 J/kg; 1 Gy = 1 J/kg = 100 rad. From part (a), the activity of the source is 5.3 × 10 8 dis/s.

      5.29 × 10 8 dis / s × 14.0 s × 0.35 × 9.12 × 10 13 J dis × 1 0.385 kg = 6.14 × 10 3 = 6.1 × 10 3 J/kg 6 .1 × 10 3 J / kg × 1 rad 1 × 10 2 J / kg × 1000 mrad rad = 6.1 × 10 2 mrad 6 .1 × 10 3 J / kg × 1 Gy 1 J/kg = 6.1 × 10 3 Gy

    3. rem = rad (RBE); Sv = Gy (RBE) , where 1 Sv = 100 rem

      mrem = 6.14 × 10 2 mrad (9.5) = 5.83 × 10 3 = 5.8 × 10 3 mrem (or 5.8 rem)

      Sv = 6.14 × 10 –3 Gy (9.5) = 5.83 × 10 –2 = 5.8 × 10 –2 Sv

    21.70

    1. 1 Ci = 3.7 × 10 10 dis/s; 1 Bq = 1 dis/s

      15 mCi × 1 Ci 1000 mCi × 3.7 × 10 10 dis/s = 5.55 × 10 8 = 5.6 × 10 8 dis/s = 5.6 × 10 8 Bq

    2. 1 Gy = 1 J/kg; 1 Gy = 100 rad

      5.55 × 10 8 dis/s × 240 s × 0 .075 × 8 .75 × 10 14 J dis × 1 65 kg = 1.345 × 10 5 = 1.3 × 10 5 J/kg

      1.3 × 10 5 J/kg × 1 Gy 1 J/kg = 1.3 × 10 5 Gy; 1 .3 × 10 5 Gy × 100 rad 1 Gy = 1.3 × 10 3 rad

    3. rem = rad (RBE); Sv = Gy (RBE)

      1.3 × 10 3 rad (1 .0) = 1.3 × 10 3 rem × 1000 mrem 1 rem = 1.3 mrem

      1.3 × 10 5 Gy ( 1.0 ) = 1.3 × 10 5 Sv

    4. The radiation dose (1.3 mrem) is much less than that for a typical mammogram.

    Additional Exercises

    21.71 Analyze/Plan. Atomic number is number of protons; mass number is number of (protons + neutrons). The element symbol is determined by atomic number. Check the list of magic numbers, Figure 21.1 and Table 21.4 to determine which isotope is unstable. Use information in Table 21.3 about positron emission to determine which isotope will produce potassium-39.

      1. K 19 38
      2. K 19 40
      3. C 20 39 a
      4. C 20 40 a
    1. K 19 38 is most likely to be unstable. Both K 19 38 and K 19 40 are odd proton/odd neutron isotopes. K 19 38 is more likely to be unstable because it has a lower neutron/proton ratio. (The two isotopes of Ca have magic numbers of protons and are thus more likely to be stable.)
    2. C 20 39 a has a magic number of protons. C 20 40 a has magic numbers of protons and neutrons.
    3. C 20 39 a 19 39 K + + 1 0 e

    21.72 86 222 Rn X + 3 2 4 He + 2 1 0 e

    This corresponds to a reduction in mass number of (3 × 4 =) 12 and a reduction in atomic number of (3 × 2 – 2) = 4. The stable nucleus is 82 210 Pb . (This is part of the sequence in Figure 21.1.)

    21.73 H 2 3 e + H 2 3 e H 2 4 e + 2 H 1 1 . Use nuclear masses calculated in Solution 21.52.

    Δm = 4.00150 + 2(1.00727) – 2(3.10493) = –0.19382 amu

    Δ E = 0 .19382 amu ‘atomic reaction’ × 1 g 6 .022 × 10 23 amu × 6.022 × 10 23 ‘atomic reaction’ mol of reaction × 1 kg 10 3 g × ( 2.99792458 × 10 8 m / s ) 2 = 1.7420 × 10 13 J / mol

    21.74

    1. 17 36 Cl 18 36 Ar + 1 0 e
    2. According to Table 21.4, nuclei with even numbers of both protons and neutrons, or an even number of one kind of nucleon, are more stable. 35 Cl and 37 Cl both have an odd number of protons but an even number of neutrons. 36 Cl has an odd number of protons and neutrons (17 p, 19 n), so it is less stable than the other two isotopes. Also, 37 Cl has 20 neutrons, a nuclear closed shell.

    21.75 2 p 1 1 H 1 2 + e 1 0

    21.76

    1. 3 6 Li + 28 56 Ni 31 62 Ga
    2. 20 40 Ca + 96 248 Cm 62 147 Sm + 54 141 X e
    3. 38 88 Sr + 36 84 Kr 46 116 Pd + 28 56 Ni
    4. 20 40 Ca + 92 238 U 30 70 Zn + 4 0 1 n + 2 41 102 Nb

    21.77

    1. Z  is 117 297 Ts ; Q is 20 48 Ca
    2. Isotope Q has 20 protons and 28 neutrons; these are both magic numbers. Even though Q has an unfavorable neutron-to-proton ratio, the special stability associated with magic numbers of protons and neutrons explains its long half-life.
    3. The target isotope was 96 248 Cm . 96 248 Cm + 20 48 Ca 116 296 Lv

    21.78

    Time (h) N t (dis/min) ln N t
    0 180 5.193
    2.5 130 4.868
    5.0 104 4.644
    7.5 77 4.34
    10.0 59 4.08
    12.5 46 3.83
    17.5 24 3.18
    image

    The plot on the left is a graph of activity (disintegrations per minute) vs. time. Choose t 1/2 at the time where N t = 1/2 N o = 90 dis/min. t 1/2 ≈ 6.0 h.

    Rearrange Equation 21.20 to obtain the linear relationship shown on the right.

    ln(N t / N o ) = –kt; ln N t – ln N o = –kt; ln N t = –kt + ln N o

    The slope of this line = –k = –0.11; t 1/2 = 0.693/0.11 = 6.3 h.

    21.79 1 × 10 6 curie × 3 .7 × 10 10 dis/s curie = 3.7 × 10 4 dis/s

    rate = 3.7 × 10 4 nuclei/s = kN

    k = 0.693 t 1/2 = 0.693 28.8 yr × 1 yr 365 × 24 × 3600 s = 7.630 × 10 10 = 7.63 × 10 10 s 1

    3.7 × 10 4 nuclei/s = (7.63 × 10 –10 /s) N; N = 4.849 × 10 13 = 4.8 × 10 13 90 Sr nuclei

    mass 90 Sr = 4.849 × 10 13 nuclei × 89 .907738 g Sr 6 .022 × 10 23 nuclei = 7.2 × 10 9 g Sr

    (mass of 90 Sr from webelements.com)


    21.80

    1. The C OH bond of the acid and the O H bond of the alcohol break in this reaction. Initially, 18 O is present in the C — 18 OH group of the alcohol. In order for 18 O to end up in the ester, the 18 O H bond of the alcohol must break. This requires that the C OH bond in the acid also breaks. The unlabeled O from the acid ends up in the H 2 O product.
    2. No. When TOCH 3 is used to react with CH 3 COOH, T will end up in the H 2 O product, regardless of whether the C OT or O T bond breaks in the reaction.

    21.81

      1. X is C 6 11 .  The long-hand reaction is N 7 14 + p 1 1 C 6 11 + H 2 4 e .
      2. X is n 0 1 . The long-hand reaction is O 8 18 + p 1 1 F 9 18 + n 0 1 .
      1. d is H 1 2 .  The long-hand reaction is N 7 14 + H 1 2 O 8 15 + n 0 1 .
    1. It makes sense that “d” represents deuterium, H 1 2 , an isotope of hydrogen.

    21.82  Because of the relationship ΔE = Δmc 2 , the mass defect (Δm) is directly related to the binding energy (ΔE) of the nucleus.

    7 Be: 4p, 3n; 4(1.0072765) + 3(1.0086649) = 7.05510 amu

    Total mass defect = 7.0551 – 7.0147 = 0.0404 amu

    0.0404 amu/7 nucleons = 5.77 × 10 –3 amu/nucleon

    Δ E = Δ m × c 2 = 5.77 × 10 3 amu nucleon × 1 g 6 .022 × 10 23 amu × 1 kg 1 × 10 3 g × 8.988 × 10 16 m 2 s 2

    = 5.77 × 10 3 amu nucleon × 1.4925 × 10 10 J 1 amu = 8.612 × 10 13 = 8.61 × 10 13 J/nucleon

    9 Be: 4p, 5n; 4(1.0072765) + 5(1.0086649) = 9.07243 amu

    Total mass defect = 9.0724 – 9.0100 = 0.06243 = 0.0624 amu

    0.0624 amu/9 nucleons = 6.937 × 10 –3 = 6.94 × 10 –3 amu/nucleon

    6.937 × 10 –3 amu/nucleon × 1.4925 × 10 –10 J/amu = 1.035 × 10 –12 = 1.04 × 10 –12 J/nucleon

    10 Be: 4p, 6n; 4(1.0072765) + 6(1.0086649) = 10.0811 amu

    Total mass defect = 10.0811 – 10.0113 = 0.0698 amu

    0.0698 amu/10 nucleons = 6.98 × 10 –3 amu/nucleon

    6.98 × 10 –3 amu/nucleon × 1.4925 × 10 –10 J/amu = 1.042 × 10 –12 = 1.04 × 10 –12 J/nucleon

    The binding energies/nucleon for 9 Be and 10 Be are very similar; that for 10 Be is slightly higher.

    21.83  First, calculate k in s –1

    k = 0.693 12.3 yr × 1 yr 365 d × 1 d 24 h × 1 h 3600 s = 1.7866 × 10 9 = 1.79 × 10 9 s 1

    From Equation 21.18, 1.50 × 10 3 s –1 = (1.7866 × 10 –9 s –1 )(N);

    N = 8.396 × 10 11 = 8.40 × 10 11 . In 26.00 g of water, there are

    26.00 g H 2 O × 1 mol H 2 O 18.02 g H 2 O × 6.022 × 10 23 H 2 O 1 mol H 2 O × 2 H 1 H 2 O = 1.738 × 10 24 H atoms

    The mole fraction of 1 3 H atoms in the sample is thus

    8.396 × 10 11 /1.738 × 10 24 = 4.831 × 10 –13 = 4.83 × 10 –13


    21.84

    1. Δ m = Δ E/c 2 ; Δ m = 3.9 × 10 26 J/s ( 2.9979246 × 10 8 m/s) 2 × 1 kg-m 2 /s 2 1 J = 4.3 × 10 9 kg/s

      The rate of mass loss is 4.3 × 10 9 kg/s. (Fewer sig figs in the value of c produce the same result.)

    2. The mass loss arises from fusion reactions that produce more stable nuclei from less stable ones, e.g., Equations 21.26–21.29.
    3. Express the mass lost by the sun in terms of protons per second consumed in fusion reactions like Equations 21.26, 21.27, and 21.29.

      4.3 × 10 9 kg s × 1 proton 1.673 × 10 24 g × 1000 g 1 kg = 2.594 × 10 36 = 3 × 10 36 protons/s

    21.85 1000 Mwatts × 1 × 10 6 watts 1 Mwatt × 1 J 1 watt-s × 1 235 U atom 3 × 10 11 J × 1 mol U 6 .02214 × 10 23 atoms × 235 g U 1 mol × 3600 s 1 h × 24 h 1 d × 365 d 1 yr × 100 40 ( efficiency) = 1.03 × 10 6 = 1 × 10 6 g U/yr

    21.86 2 × 10 12 curies × 3 .7 × 10 10 dis/s 1 curie = 7.4 × 10 2 = 7 × 10 2 dis/s

    7.4 × 10 2 dis/s 75 kg × 8 × 10 13 J dis × 1 rad 1 × 10 2 J/g × 3600 s h × 24 h 1 d × 365 d 1 yr = 2.49 × 10 6 = 2 × 10 6 rad/yr

    Recall that there are 10 rem/rad for alpha particles.

    2.49 × 10 6 rad 1 yr × 10 rem 1 rad = 2.49 × 10 5 = 2 × 10 5 rem/yr

    Integrative Exercises

    21.87  Calculate the molar mass of NaClO 4 that contains 29.6% 36 Cl. Atomic mass of the enhanced Cl is 0.296(36.0) + 0.704(35.453) = 35.615 = 35.6. The molar mass of NaClO 4 is then (22.99 + 35.615 + 64.00) = 122.605 = 122.6. Calculate N, the number of 36 Cl nuclei, the value of k in s –1 , and the activity in dis/s.

    53.8 mg NaClO 4 × 1 g 1000 mg × 1 mol NaClO 4 122.605 g NaClO 4 × 1 mol Cl 1 mol NaClO 4 × 6.022 × 10 23 Cl atoms mol Cl × 29.6 36 Cl atoms 100 Cl atoms = 7.822 × 10 19 = 7.82 × 10 19 36 Cl atoms

    k = 0.693 / t 1/2 = 0.693 3.0 × 10 5 yr × 1 yr 365 × 24 × 3600 s = 7.32 × 10 14 = 7.3 × 10 14 s 1

    rate = kN = (7.32 × 10 –14 s –1 )(7.822 × 10 19 nuclei) = 5.729× 10 6 = 5.7 × 10 6 dis/s


    21.88  Calculate the amount of energy produced by the nuclear fusion reaction, the enthalpy of combustion, ΔH°, of C 8 H 18 , and then the mass of C 8 H 18 required.

    Δ m for the reaction 4 1 1 H 2 4 He + 2 + 1 0 e is:

    4 ( 1.00782 ) 4.00260 amu 2 ( 5.4858 × 10 4 amu ) = 0.027583 = 0.02758 amu

    Δ E = Δ mc 2 = 0.027583 amu × 1 g 6 .02214 × 10 23 amu × 1 kg 1000 g × ( 2.9979246 × 10 8 m/s) 2 = 4.11654 × 10 –12 = 4.117 × 10 –12 J/4 1 H nuclei

    1.0 g 1 H × 1 1 H nucleus 1 .00782 amu × 6.02214 × 10 23 amu g × 4.11654 × 10 12 J 4 1 H nuclei = 6.1495 × 10 11 J = 6.1 × 10 8 kJ produced by the fusion of 1.0 g 1 H.

    C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g)

    ΔH° = 8(–393.5 kJ) + 9(–241.82 kJ) – (–250.1 kJ) = –5074.3 kJ

    6.1495 × 10 8 kJ × 1 mol C 8 H 18 (l) 5074.3 kJ × 114.231 g C 8 H 18 mol C 8 H 18 = 1.384 × 10 7 g = 1.4 × 10 4 kg C 8 H 18

    14,000 kg C 8 H 18 (l) would have to be burned to produce the same amount of energy as fusion of 1.0 g 1 H.

    21.89  Refer to “Chemistry Put to Work: Gas Separations” in Section 10.8.

    1. The atomic weight of naturally occurring uranium is 238.02891 g/mol. The molar mass of UF 6 is then [238.02891 + 6(18.998403)] = 352.01933 g/mol.

      g = MM × RT VP = 352 .02 g 30 .0 L × 0.082058 L-atm mol-K × 350 K (695/760) atm = 368.522 = 369 g

      Check . This seems like a large mass, but 30.0 L is more than the molar volume of a gas at STP, so it is reasonable that more than one mole of UF 6 is in the flask.

    2. Of the 369 g sample, 0.720% is 235 UF 6 .

      (368.522 × 0.00720) = 2.65336 = 2.65 g 235 UF 6 .

      The mass of 235 U in 2.65 g 235 UF 6 is

      2 .65336 g 235 UF 6 × 235.044 g 235 U 349.034 g 235 UF 6 = 1.78681 = 1.79 g 235 U

    3. According to information in “Chemistry Put to Work: Gas Separations,” the ratio of effusion rates is 1.0043. That is, 235 UF 6 effuses (diffuses) 1.0043 times faster than 238 UF 6 . That is, the mass of 235 UF 6 in the diffused sample is 1.0043 times greater than in the initial sample. The mass of 235 UF 6 in the diffused sample is then (2.65336 g 235 UF 6 initial × 1.0043) = 2.66477 = 2.66 g 235 UF 6 .

      2 .66477 g 235 UF 6 × 235.044 g 235 U 349.034 g 235 UF 6 = 1.79449 = 1.79 g 235 U


    1. One more round of enrichment yields (0.266477 g 235 UF 6 × 1.0043) = 2.67623 = 2.68 g 235 UF 6 . The mass % of 235 UF 6 in the sample is then

      2.67623 g 235 UF 6 368.522 g sample × 100 = 0.726206 = 0.726 % 235 UF 6

      [This is the same result as twice applying the 1.0043 times enrichment to the natural abundance of 235 U. (0.720% × 1.0043 × 1.0043) = 0.726205 = 0.726%]

    21.90

    1. 0.18 Ci × 3.7 × 10 10 dis/s Ci × 3600 s h × 24 h d × 245 d = 1.41 × 10 17 = 1.4 × 10 17 alpha particles
    2. P = nRT/V = 1.41 × 10 17 He atoms × 1 mol He 6 .022 × 10 23 atoms × 295 K 0 .0250 L × 0.08206 L-atm mol-K

      = 2.27 × 10 –4 = 2.3 × 10 –4 atm = 0.17 torr

    21.91  Calculate N t in dis/min/g C from 1.5 × 10 –2 dis/0.788 g CaCO 3 . N o = 15.3 dis/min/g C. Calculate k from t 1/2 , calculate t from ln (N t / N o ) = –kt.

    C(s) + O 2 (g) → CO 2 (g) + Ca(OH 2 )(aq) → CaCO 3 (s) + H 2 O(l)

    1 C atom → 1 CaCO 3 molecule

    1.5 × 10 2 Bq 0 .788 g CaCO 3 × 1 dis/s 1 Bq × 60 s 1 min × 100.1 g CaCO 3 12.01 g C = 9.52 = 9.5 dis/min/g C

    k = 0.693/t 1/2 = 0.693/5.700 × 10 3 yr = 1.216 × 10 –4 = 1.22 × 10 –4 yr –1

    t = 1 k ln N t N o = 1 1.216 × 10 4 yr 1 ln 9 .52 dis/min/g C 15 .3 dis/min/g C = 3.90 × 10 3 yr

    21.92

    1. Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaNO 3 (aq)
    2. 1.25 mmol Ba 2+ + 1.25 mmol SO 4 2– → 1.25 mmol BaSO 4

      Neither reactant is in excess, so the activity of the filtrate is due entirely to [SO 4 2– ] from dissociation of BaSO 4 (s). Calculate [SO 4 2– ] in the filtrate by comparing the activity of the filtrate to the activity of the reactant.

      0.050 M SO 4 2 1.22 × 10 6 Bq/mL = x M filtrate 250 Bq/mL

      [SO 4 2– ] in the filtrate = 1.0246 × 10 –5 = 1.0 × 10 –5 M

      K sp = [Ba 2+ ][SO 4 2– ]; [SO 4 2– ] = [Ba 2+ ]

      K sp = (1.0246 × 10 –5 ) 2 = 1.0498 × 10 –10 = 1.0 × 10 –10


     

    22 Chemistry of the Nonmetals

    Visualizing Concepts

    22.1   Statement (c) is correct. C 2 H 4 , the structure on the left, is the stable compound. Carbon, with a relatively small covalent radius owing to its location in the second row of the periodic chart, is able to closely approach other atoms. This close approach enables significant π overlap, so carbon can form strong multiple bonds to satisfy the octet rule. Silicon, in the third row of the periodic table, has a covalent radius too large for significant π overlap. Si does not form stable multiple bonds and Si 2 H 4 is unstable.

    22.2

    1. Acid–base (Brønsted)
    2. Charges on species from left to right in the reaction: 0, 0, 1+, 1–
    3. NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH (aq)

    22.3 Analyze. The structure is a trigonal bipyramid where one of the five positions about the central atom is occupied by a lone pair, often called a see-saw.

    Plan A: Count the valence electrons in each molecule, draw a correct Lewis structure, and count the electron domains about the central atom.

    Plan B: Molecules (a)–(d) each contain four F atoms bound to a central atom through a single bond (F is unlikely to form multiple bonds because of its high electronegativity). This represents 16 electron pairs; the fifth position is occupied by a lone pair, for a total of 17 e pairs. A valence e count for (a)–(d) will tell us which molecules are likely to have the designated structure. Molecule (e), HClO 4 , is not exactly of the type AX 4 , so a Lewis structure will be required. Solve.

    1. XeF 4 36 e , 16 e pairs. Plan B predicts that this molecule will not adopt the see-saw structure.
      image
      6 e domains about the Xe octahedral domain geometry square planar structure
    2. BrF 4 + 34 e , 17 e pairs; structure will be see-saw
      image
      5 e domains about Br trigonal bipyramidal domain geometry see-saw structure

    1. SiF 4 32 e , 16 e pairs; structure will not be see-saw
      image
      4 e domains about Si tetrahedral domain geometry and structure
    1. TeCl 4 34 e , 17 e pairs; structure will be see-saw
      image
      5 e domains about Te trigonal bipyramidal domain geometry see-saw structure
    1. HClO 4 32 e , 16 e pairs; structure will not be see-saw
      image
      (HClO 4 is an oxyacid, so H is bound to O, not Cl. Other Lewis structures that optimize formal charges are possible; structure predictions are the same.)

    22.4  Both gases are colorless and odorless, so they cannot be distinguished by inspection. (In practice, for safety reasons, one should never work in a lab with unlabeled bottles.) One big difference in the chemical properties of the two gases is their ability to support combustion. Oxygen supports combustion, whereas nitrogen does not. Heat a small piece of steel wool, place it at the mouth of one of the bottles. If the metal flares or glows more strongly, the gas is oxygen. If not, the gas is nitrogen.

    22.5 Analyze. Given: space-filling models of molecules containing nitrogen and oxygen atoms. Find: molecular formulas and Lewis structures.

    Plan. Nitrogen atoms are blue, and oxygen atoms are red. Count the number of spheres of each color to determine the molecular formula. From each molecular formula, count the valence electrons (N = 5, O = 6) and draw a correct Lewis structure. Resonance structures are likely.

    Solve .  (We list the formulas and Lewis structures from left to right across each row of space-filling models.)

    image

    Many other resonance structures are possible. Those with double bonds to the central oxygen (like the right-hand structure above) do not minimize formal charge and are less significant in the net bonding model.

    image

    image

    Other equivalent resonance structures with different arrangement of the double bonds are possible.

    image

    We place the odd electron on N because of electronegativity arguments.

    image

    We place the odd electron on N because of electronegativity arguments.

    image

    The right-most structure above does not minimize formal charge and makes smaller contribution to the net bonding model.

    22.6  The graph is applicable only to (c) density. Density depends on both atomic mass and volume (radius). Both increase going down a family, but atomic mass increases to a greater extent. Density, the ratio of mass to volume, increases going down the family; this trend is consistent with the data in the figure.

    According to periodic trends, (a) electronegativity and (b) first ionization energy both decrease rather than increase going down the family. According to Table 22.5, both (d) X–X single bond enthalpy and (e) electron affinity are somewhat erratic, with the trends decreasing from S to Po, and anomalous values for the properties of O, probably owing to its small covalent radius.

    22.7  Statements (a) and (c) are true.

    Statement (b) is false. Although nuclear charge does increase going down the group, this effect, taken by itself, would decrease atomic radii. The increase in principle quantum number ( n ) of the valence electrons dominates and atomic radii increase going down the group. Statement (d) is false because the anion that is the strongest base in water is the conjugate base of the weakest conjugate acid. According to trends in binary hydrides, the acid with the longest X–H bond will be the most readily ionized and the strongest acid. The strongest acid is SeH and the weakest is OH . Therefore, O 2– is the strongest base in water.

    22.8 Analyze/Plan .  Evaluate the graph, describe the trend in data, recall the general trend for each of the properties listed, and use details of the data to discriminate between possibilities.

    Solve .  The property depicted is (a), first ionization energy. The general trend is an increase in value moving from left to right across the period, with a small discontinuity at S. Considering just this overall feature, both (a) first ionization energy and (c) electronegativity increase moving from left to right, so these are possibilities. (b), Atomic radius, decreases, and can be eliminated. Because Si is a solid and Cl and Ar are gases at room temperature, melting points must decrease across the row; (d), melting point, can be eliminated. According to data in Tables 22.2, 22.5, 22.7, and 22.8, (e), X–X single bond enthalpies, show no consistent trend. Furthermore, there is no known Ar–Ar single bond, so no value for this property can be known; (e) can be eliminated.


    Now let’s examine trends in (a), first ionization energy, and (c), electronegativity, more closely. From electronegativity values in Chapter 8 , we see a continuous increase with no discontinuity at S, and no value for Ar. Values for (a), first ionization energy, from Chapter 7 do match the pattern in the figure. The slightly lower value of I 1 for S is the result of a decrease in repulsion when an electron is removed from a fully occupied orbital. In summary, only (a), first ionization energy, fits the property depicted in the graph.

    22.9  The compound on the left, with the strained three-membered ring, will be the most generally reactive. For central atoms with four electron domains*, idealized bond angles are 109 o . From left to right, the bond angles in the three molecules pictured are 60 o , 90 o , and 108 o . The larger the deviation from ideal bond angles, the more strain in the molecule and the more generally reactive it is.

    *For the stick structures shown in the exercise, each line represents a C–C single bond and the intersection of two lines is a C atom. To determine the number of electron domains about each atom, visualize or draw the hydrogen atoms and nonbonded electron pairs in each molecule. Alternatively, note that both C and O atoms form only single bonds, so hybridization must be sp 3 and idealized bond angles are 109 o .

    22.10 Analyze/Plan .  The structure shown is a diatomic molecule or ion, depending on the value of n. Each species has 10 valence electrons and 5 electron pairs. Solve.

    1. Only second row elements are possible, because of the small covalent radius required for multiple bonding. Likely candidates are CO, N 2 , NO + , CN , and C 2 2– .
      image
    2. Because C 2 2– has the highest negative charge, it is likely to be the strongest H + acceptor and strongest Brønsted base. This is confirmed in Section 22.9 under “Carbides.”

    Periodic Trends and Chemical Reactions (Section 22.1)

    22.11 Analyze/Plan .  Use the color-coded periodic chart on the front-inside cover of the text to classify the given elements. Solve.

    Metals: (b) Sr, (c) Mn, (e) Na; nonmetals: (a) P, (d) Se, (f) Kr; metalloids: none

    22.12  Metals: (a) Ga, (b) Mo, (f) Ru nonmetals: (e) Xe  metalloid: (c) Te, (d) As

    22.13 Analyze/Plan .  Follow the logic in Sample Exercise 22.1. Solve.

    1. O
    2. Br
    3. Ba
    4. O
    5. Co
    6. Br

    22.14

    1. Cl
    2. K
    3. K in the gas phase (lowest ionization energy), Li in aqueous solution (most positive E° value)
    4. Ne; Ne and Ar are difficult to compare to the other elements because they do not form compounds and their radii are not measured in the same way as other elements. However, Ne is several rows to the right of C and surely has a smaller atomic radius. The next smallest is C.
    5. C
    6. C (graphite, diamond, fullerenes, carbon nanotubes, and graphene)

    22.15  Statements (b) and (d) are true.

    Statement (a) is false because a nitrogen atom is too small to accommodate five fluorine atoms around it. Statement (c) is false because the reduction potential of Cl 2 is larger than the reduction potential of I 2 . The substance with the smaller reduction potential is easier to oxidize.

    22.16  Statements (a) and (d) are true.

    Statement (b) is false because Si does not readily form π bonds (to itself or other atoms), so Si 2 H 4 and Si 2 H 2 are not known as stable compounds. Statement (c) is false; the oxidation number of N and P in the two compounds is +5.

    22.17 Analyze/Plan .  Follow the logic in Sample Exercise 22.2. Solve.

    1. NaOCH 3 (s) + H 2 O(l) → NaOH(aq) + CH 3 OH(aq)
    2. CuO(s) + 2 HNO 3 (aq) → Cu(NO 3 ) 2 (aq) + H 2 O(l)
    3. WO 3 (s) + 3 H 2 (g) Δ W(s) + 3 H 2 O(g)
    4. 4 NH 2 OH(l) + O 2 (g) → 6 H 2 O(l) + 2 N 2 (g)
    5. Al 4 C 3 (s) + 12 H 2 O(l) → 4 Al(OH) 3 (s) + 3 CH 4 (g)

    22.18

    1. Mg 3 N 2 (s) + 6 H 2 O(l) → 2 NH 3 (g) + 3 Mg(OH) 2 (s)

      Because H 2 O(l) is a reactant, the state of NH 3 in the products could be expressed as NH 3 (aq).

    2. 2 C 3 H 7 OH(l) + 9 O 2 (g) → 6 CO 2 (g) + 8 H 2 O(l)
    3. MnO 2 (s) + C(s) Δ CO(g) + MnO(s) or

      MnO 2 (s) + 2 C(s) Δ 2 CO(g) + Mn(s) or

      MnO 2 (s) + C(s) Δ CO 2 (g) + Mn(s)

    4. AlP(s) + 3 H 2 O(l) → PH 3 (g) + Al(OH) 3 (s)
    5. Na 2 S(s) + 2 HCl(aq) → H 2 S(g) + 2 NaCl(aq)

    Hydrogen, the Noble Gases, and the Halogens (Sections 22.2, 22.3, and 22.4)

    22.19 Analyze/Plan .  Use information on the isotopes of hydrogen in Section 22.2 to list their symbols, names, and relative abundances. Solve.

    1. 1 1 H-protium; 1 2 H-deuterium; 1 3 H-tritium
    2. The order of abundance is proteum > deuterium > tritium.

    1. 1 3 H-tritium is radioactive.
    1. 1 3 H 2 3 He + 1 0 e

    22.20  Statement (d) is correct. Because deuterium is almost twice as heavy as protium, we expect physical properties influenced by mass to be different for the two isotopes of hydrogen. In fact, D 2 O has higher melting and boiling points, and a greater density than H 2 O.

    22.21 Analyze/Plan .  Consider the electron configuration of hydrogen and the Group 1A elements. Solve.

    Like other elements in group 1A, hydrogen has only one valence electron and its most common oxidation number is +1.

    22.22  In its standard state, hydrogen is a gas, and thus a nonmetal, like the halogens. Hydrogen can gain an electron to form an anion with a 1– charge. Chemically, hydrogen can combine with group 1A metals to form ionic compounds, where H is the anion.

    22.23 Analyze/Plan .  Use information on the descriptive chemistry of hydrogen given in Section 22.2 to complete and balance the equations. Solve.

    1. NaH(s) + H 2 O(l) → NaOH(aq) + H 2 (g)
    2. Fe(s) + H 2 SO 4 (aq) → Fe 2+ (aq) + H 2 (g) + SO 4 2– (aq)
    3. H 2 (g) + Br 2 (g) → 2 HBr(g)
    4. 2 Na(l) + H 2 (g) → 2 NaH(s)
    5. PbO(s) + H 2 (g) Δ Pb(s) + H 2 O(g)

    22.24

    1. 2 Al(s) + 6 H + (aq) → 2 Al 3+ (aq) + 3 H 2 (g)
    2. Mg(s) + H 2 O(g) → MgO(s) + H 2 (g)
    3. MnO 2 (s) + H 2 (g) → MnO(s) + H 2 O(g)
    4. CaH 2 (s) + 2 H 2 O(l) → Ca(OH) 2 (aq) + 2 H 2 (g)

    22.25 Analyze/Plan .  If the element bound to H is a nonmetal, the hydride is molecular. If H is bound to a metal with integer stoichiometry, the hydride is ionic; with noninteger stoichiometry, the hydride is metallic. Solve.

    1. ionic (metal hydride)
    2. molecular (nonmetal hydride)
    3. metallic (nonstoichiometric transition metal hydride)

    22.26

    1. molecular
    2. ionic
    3. metallic

    22.27  Vehicle fuels produce energy via combustion reactions. The reaction H 2 (g) + 1/2 O 2 (g) → H 2 O(g) is very exothermic, producing 242 kJ per mole of H 2 burned. The only product of combustion is H 2 O, a nonpollutant (but like CO 2 , a greenhouse gas).

    22.28  Electrolysis of water is the cleanest way to produce hydrogen, but it is energy intensive. To make this process sustainable, the energy must come from renewable sources, such as hydroelectric or nuclear power plants, wind generators, or solar cells. The biomass production of fuel for electric power generation would also be a sustainable energy source.


    22.29 Analyze/Plan .  Consider the periodic properties of Xe and Ar. Solve.

    Xenon is larger, and can more readily accommodate an expanded octet. More important is the lower ionization energy of xenon; because the valence electrons are a greater average distance from the nucleus, they are more readily promoted to a state in which the Xe atom can form bonds with fluorine.

    22.30  Your friend cannot be correct. In general, noble gas elements have very stable electron configurations with complete s and p subshells. They have very large positive ionization energies; they do not lose electrons easily. They have positive electron affinities; they do not attract electrons to themselves. They do not easily gain, lose or share electrons, so they do not readily form the chemical bonds required to create compounds. To date, the only known compounds of noble gases involve Xe and Kr bound to other nonmetals. Specifically, there are no known compounds of Ne.

    22.31 Analyze/Plan .  Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve.

    1. Ca(O Br ) 2 , Br, +1
    2. H Br O 3 , Br, +5
    3. Xe O 3 , Xe, +6
    4. Cl O 4 , Cl, +7
    5. H I O 2 , I, +3
    6. IF 5 ; I, +5; F, –1

    22.32

    1. Cl O 3 , Cl +5
    2. H I , I, –1
    3. ICl 3 ; I, +3; Cl, –1
    4. NaO Cl , Cl, +1
    5. H Cl O 4 , Cl, +7
    6. XeF 4 ; Xe, +4; F, –1

    22.33 Analyze/Plan .  Review the nomenclature rules and ion names in Section 2.8, as well as the rules for assigning oxidation numbers in Section 4.4. Solve.

    1. iron(III) chlorate , Cl, +5
    2. chlorous acid, Cl, +3
    3. xenon hexafluoride, F, –1
    4. bromine pentafluoride; Br, +5; F, –1
    5. xenon oxide tetrafluoride, F, –1
    6. iodic acid, I, +5

    22.34

    1. potassium chlorate, Cl, +5
    2. calcium iodate, I, +5
    3. aluminum chloride, Cl, –1
    4. bromic acid, Br, +5
    5. paraperiodic acid, I, +7
    6. xenon tetrafluoride, F, –1

    22.35 Analyze/Plan .  Consider intermolecular forces and periodic properties, including oxidizing power, of the listed substances. Solve.

    1. Van der Waals intermolecular attractive forces increase with increasing numbers of electrons in the atoms.
    2. F 2 reacts with water: F 2 (g) + H 2 O(l) → 2 HF(aq) + 1/2 O 2 (g). That is, fluorine is too strong an oxidizing agent to exist in water.
    3. HF has extensive hydrogen bonding.
    4. Oxidizing power is related to electronegativity. Electronegativity decreases in the order given.

    22.36

    1. The more electronegative the central atom, the greater the extent to which it withdraws charge from oxygen, in turn making the O–H bond more polar, and enhancing ionization of H + .
    2. HF reacts with the silica which is a major component of glass:

      6 HF(aq) + SiO 2 (s) → SiF 6 2– (aq) + 2 H 2 O(l) + 2 H + (aq)


    1. If the reaction of NaI and H 2 SO 4 did produce HI, it would be a metathesis reaction. If HI is not the product of this reaction, then I is probably oxidized by H 2 SO 4 to produce I 2 .
    1. The major factor is size; there is not room about Br for the three chlorides plus the two unshared electron pairs that would occupy the bromine valence shell orbitals.

    Oxygen and the Other Group 6A Elements (Sections 22.5 and 22.6)

    22.37 Analyze/Plan .  Use information on the descriptive chemistry of oxygen given in Section 22.5 to complete and balance the equations. Solve.

    1. 2 HgO(s) Δ 2 Hg(l) + O 2 (g)
    2. 2 Cu(NO 3 ) 2 (s) Δ 2 CuO(s) + 4 NO 2 (g) + O 2 (g)
    3. PbS(s) + 4 O 3 (g) → PbSO 4 (s) + 4 O 2 (g)
    4. 2 ZnS(s) + 3 O 2 (g) Δ 2 ZnO(s) + 2 SO 2 (g)
    5. 2 K 2 O 2 (s) + 2 CO 2 (g) → 2 K 2 CO 3 (s) + O 2 (g)
    6. 3 O 2 (g) h ν 2 O 3 (g)

    22.38

    1. CaO(s) + H 2 O(l) → Ca 2+ (aq) + 2 OH (aq)
    2. Al 2 O 3 (s) + 6 H + (aq) → 2 Al 3+ (aq) + 3 H 2 O(l)
    3. Na 2 O 2 (s) + 2 H 2 O(l) → 2 Na + (aq) + 2 OH (aq) + H 2 O 2 (aq)
    4. N 2 O 3 (g) + H 2 O(l) → 2 HNO 2 (aq)
    5. 2 KO 2 (s) + 2 H 2 O(l) → 2 K + (aq) + 2 OH (aq) + O 2 (g) + H 2 O 2 (aq)
    6. NO(g) + O 3 (g) → NO 2 (g) + O 2 (g)

    22.39 Analyze/Plan .  Oxides of metals are bases, oxides of nonmetals are acids, oxides that act as both acids and bases are amphoteric, and oxides that act as neither acids nor bases are neutral. Solve.

    1. acidic (oxide of a nonmetal)
    2. acidic (oxide of a nonmetal)
    3. amphoteric
    4. basic (oxide of a metal)

    22.40

    1. Mn 2 O 7 (higher oxidation state of Mn)
    2. SnO 2 (higher oxidation state of Sn)
    3. SO 3 (higher oxidation state of S)
    4. SO 2 (more nonmetallic character of S)
    5. Ga 2 O 3 (more nonmetallic character of Ga)
    6. SO 2 (more nonmetallic character of S)

    22.41 Analyze/Plan .  Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve.

    1. H 2 Se O 3 , +4
    2. KH S O 3 , +4
    3. H 2 Te , –2
    4. C S 2 , –2
    5. Ca S O 4 , +6
    6. Cd S , –2
    7. Zn Te , –2

    Oxygen (a group 6A element) is in the –2 oxidation state in compounds (a), (b), and (e).

    22.42

    1. S Cl 4 , +4
    2. Se O 3 , +6
    3. Na 2 S 2 O 3 , +2
    4. H 2 S , –2
    5. H 2 S O 4 , +6
    6. S O 2 , +4
    7. Hg Te , –2

    Oxygen (a group 6A element) is in the –2 oxidation state in compounds (b), (c), (e) and (f).

    22.43 Analyze/Plan .  The half-reaction for oxidation in all these cases is:

    H 2 S(aq) → S(s) + 2 H + + 2 e [The product could be written as S 8 (s), but this is not necessary. In fact it is not necessarily the case that S 8 would be formed, rather than some other allotropic form of the element.] Combine this half-reaction with the given reductions to write complete equations. The reduction in (c) happens only in acid solution. The reactants in (d) are acids, so the medium is acidic. Solve.

    1. 2 Fe 3+ (aq) + H 2 S(aq) → 2 Fe 2+ (aq) + S(s) + 2 H + (aq)
    2. Br 2 (l) + H 2 S(aq) → 2 Br–(aq) + S(s) + 2 H + (aq)
    3. 2 MnO 4 (aq) + 6 H + (aq) + 5 H 2 S(aq) → 2 Mn 2+ (aq) + 5 S(s) + 8 H 2 O(l)
    4. 2 NO 3 (aq) + H 2 S(aq) + 2 H + (aq) → 2 NO 2 (aq) + S(s) + 2 H 2 O(l)

    22.44  An aqueous solution of SO 2 contains H 2 SO 3 and is acidic. Use H 2 SO 3 as the reducing agent and balance assuming acid conditions.

    1. 2 [ MnO 4 (aq) + 8 H + (aq) + 5 e Mn 2 + (aq) + 4 H 2 O(l)] 5[H 2 SO 3 (aq) + H 2 O(l) SO 4 2 (aq) + 4 H + (aq) + 2 e ] 2 MnO 4 (aq) + 5 H 2 SO 3 (aq) 2 MnSO 4 (aq) + 3 SO 4 2 (aq) + 3 H 2 O(l) + 4 H + (aq)
    2. Cr 2 O 7 2 (aq) + 14 H + (aq) + 6 e 2 Cr 3 + (aq) + 7 H 2 O(l) 3 [H 2 SO 3 (aq) + H 2 O(l) SO 4 2 (aq) + 4 H + (aq) + 2 e ] Cr 2 O 7 2 (aq) + 3 H 2 SO 3 (aq) + 2 H + (aq) 2 Cr 3 + (aq) + 3 SO 4 2 (aq) + 4 H 2 O(l)
    3. Hg 2 2 + (aq) + 2 e 2 Hg(l) H 2 SO 3 (aq) + H 2 O(l) SO 4 2 (aq) + 4 H + (aq) + 2 e Hg 2 2 + (aq) + H 2 SO 3 (aq) + H 2 O(l) 2 Hg(l) + SO 4 2 (aq) + 4 H + (aq)

    22.45 Analyze/Plan .  For each substance, count valence electrons, draw the correct Lewis structure, and apply the rules of VSEPR to decide electron domain geometry and geometric structure. Solve.

    1. image
    2. image
    3. image

    22.46

    image

    22.47 Analyze/Plan .  Use information on the descriptive chemistry of sulfur given in Section 22.6 to complete and balance the equations. Solve.

    1. SO 2 (s) + H 2 O(l) → H 2 SO 3 (aq) ⇌ H + (aq) + HSO 3 (aq)
    2. ZnS(s) + 2 HCl(aq) → ZnCl 2 (aq) + H 2 S(g)
    3. 8 SO 3 2– (aq) + S 8 (s) → 8 S 2 O 3 2– (aq)
    4. SO 3 (aq) + H 2 SO 4 (l) → H 2 S 2 O 7 (l)

    22.48

    1. Al 2 Se 3 (s) + 6 H + (aq) → 2 Al 3+ (aq) + 3 H 2 Se(g)
    2. Cl 2 (aq) + S 2 O 3 2– (aq) + H 2 O(l) → 2 Cl (aq) + S(s) + SO 4 2– (aq) + 2 H + (aq)

    Nitrogen and the Other Group 5A Elements (Sections 22.7 and 22.8)

    22.49 Analyze/Plan .  Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve.

    1. Na N O 2 , +3
    2. N H 3 , –3
    3. N 2 O, +1
    4. NaC N , –3
    5. H N O 3 , +5
    6. N O 2 , +4
    7. N 2 , 0
    8. B N , –3

    22.50

    1. N O, +2
    2. N 2 H 4 , –2
    3. KC N , –3
    4. Na N O 2 , +3
    5. N H 4 Cl, –3
    6. Li 3 N , –3

    22.51 Analyze/Plan .  For each substance, count valence electrons, draw the correct Lewis structure, and apply the rules of VSEPR to decide electron domain geometry and geometric structure. Solve.

    1. image

      The molecule is bent around the central oxygen and nitrogen atoms; the four atoms need not lie in a plane. The right-most form does not minimize formal charges and is less important in the actual bonding model. The oxidation state of N is +3.

    2. image

      The molecule is linear. The oxidation state of N is –1/3.


    1. image

      The geometry is tetrahedral around the left nitrogen, trigonal pyramidal around the right. The oxidation state of N is –2.

    1. image

      (three equivalent resonance forms) The ion is trigonal planar. The oxidation state of N is +5.

    22.52

    1. image

      The ion is tetrahedral. The oxidation state of N is –3.

    2. image

      The ion is bent with a 120° O–N–O angle. The oxidation state of N is +3.

    3. image

      The molecule is linear. Again, the third resonance form makes less contribution to the structure because of the high formal charges involved. The oxidation state of N is +1.

    4. image

      The molecule is bent (nonlinear). The odd electron resides on N because it is less electronegative than O. The oxidation state of N is +4.

    22.53 Analyze/Plan .  Use information on the descriptive chemistry of nitrogen given in Section 22.7 to complete and balance the equations. Solve.

    1. Mg 3 N 2 (s) + 6 H 2 O(l) → 2 NH 3 (g) + 3 Mg(OH) 2 (s)

      Because H 2 O(l) is a reactant, the state of NH 3 in the products could be expressed as NH 3 (aq).

    2. 2 NO(g) + O 2 (g) → 2 NO 2 (g), redox reaction
    3. N 2 O 5 (g) + H 2 O(l) → 2 H + (aq) + 2 NO 3 (aq)
    4. NH 3 (aq) + H + (aq) → NH 4 + (aq)
    5. N 2 H 4 (l) + O 2 (g) → N 2 (g) + 2 H 2 O(g), redox reaction

    22.54

    1. 4 Zn(s) + 2 NO 3 (aq) + 10 H + (aq) → 4 Zn 2+ (aq) + N 2 O(g) + 5 H 2 O(l)
    2. 4 NO 3 (aq) + S(s) + 4 H + (aq) → 4 NO 2 (g) + SO 2 (g) + 2 H 2 O(l) (or 6 NO 3 (aq) + S(s) + 4 H + (aq) → 6 NO 2 (g) + SO 4 2– (aq) + 2 H 2 O(l)
    3. 2 NO 3 (aq) + 3 SO 2 (g) + 2 H 2 O(l) → 2 NO(g) + 3 SO 4 2– (aq) + 4 H + (aq)

    1. N 2 H 4 (g) + 5 F 2 (g) → 2 NF 3 (g) + 4 HF(g)
    1. 4 CrO 4 2– (aq) + 3 N 2 H 4 (aq) + 4 H 2 O(l) → 4 Cr(OH) 4 (aq) + 4 OH (aq) + 3 N 2 (g)

    22.55 Analyze/Plan .  Follow the method for writing balanced half-reactions given in Section 20.2 and Sample Exercises 20.2. Solve.

    1. HNO 2 (aq) + H 2 O(l) → NO 3 (aq) + 3 H + (aq) + 2 e , E red ° = 0.96 V
    2. N 2 (g) + H 2 O(l) → N 2 O(g) + 2 H + (aq) + 2 e , E red ° = 1.77 V

    22.56

    1. NO 3 (aq) + 4 H + (aq) + 3 e → NO(g) + 2 H 2 O(l)
    2. HNO 2 (aq) → NO 2 (g) + H + (aq) + 1 e

    22.57 Analyze/Plan .  Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve.

    1. H 3 P O 3 , +3
    2. H 4 P 2 O 7 , +5
    3. Sb Cl 3 , +3
    4. Mg 3 As 2 , –3
    5. P 2 O 5 , +5
    6. Na 3 P O 4 , +5

    22.58

    1. P O 4 3– , +5
    2. H 3 As O 3 , +3
    3. Sb 2 S 3 , +3
    4. Ca(H 2 P O 4 ) 2 , +5
    5. K 3 P , –3
    6. Ga As , –3

    22.59 Analyze/Plan .  Consider the structures of the compounds of interest when explaining the observations. Solve.

    1. Phosphorus is a larger atom and can more easily accommodate five surrounding atoms and an expanded octet of electrons than nitrogen can. Also, P has energetically “available” 3d orbitals that participate in the bonding, but nitrogen does not.
    2. Only one of the three hydrogens in H 3 PO 2 is bonded to oxygen. The other two are bonded directly to phosphorus and are not easily ionized because the P–H bond is not very polar.
    3. PH 3 is a weaker base than H 2 O (PH 4 + is a stronger acid than H 3 O + ). Any attempt to add H + to PH 3 in the presence of H 2 O merely causes protonation of H 2 O.
    4. Refer to the structures of white and red phosphorus in Figure 22.26. White phosphorus consists of P 4 molecules, with P–P–P bond angles of 60°. Each P atom has four VSEPR pairs of electrons, so the predicted electron pair geometry is tetrahedral and the preferred bond angle is 109°. Because of the severely strained bond angles in P 4 molecules, white phosphorus is highly reactive. Red phosphorus is a chain of groups of four P atoms. It has fewer severely strained P–P–P bond angles and is less reactive than white phosphorus.

    22.60

    1. Only two of the hydrogens in H 3 PO 3 are bound to oxygen. The third is attached directly to phosphorus, and not readily ionized, because the H–P bond is not very polar.
    2. The smaller, more electronegative nitrogen withdraws more electron density from the O–H bond, making it more polar and more likely to ionize.
    3. Phosphate rock consists of Ca 3 (PO 4 ) 2 , which is only slightly soluble in water. The phosphorus is unavailable for plant use.

    1. N 2 can form stable π bonds to complete the octet of both N atoms. Because phosphorus atoms are larger than nitrogen atoms, they do not form stable π bonds with themselves and must form σ bonds with several other phosphorus atoms (producing P 4 tetrahedral or chain structures) to complete their octets.
    1. In solution Na 3 PO 4 is completely dissociated into Na + and PO 4 3– . PO 4 3– , the conjugate base of the very weak acid HPO 4 2– , has a K b of 2.4 × 10 –2 and produces a considerable amount of OH by hydrolysis of H 2 O.

    22.61 Analyze/Plan .  Use information on the descriptive chemistry of phosphorus given in Section 22.8 to complete and balance the equations. Solve.

    1. 2 Ca 3 (PO 4 ) 2 (s) + 6 SiO 2 (s) + 10 C(s) Δ P 4 (g) + 6 CaSiO 3 (l) + 10 CO(g)
    2. PBr 3 (l) + 3 H 2 O(l) → H 3 PO 3 (aq) + 3 HBr(aq)
    3. 4 PBr 3 (g) + 6 H 2 (g) → P 4 (g) + 12 HBr(g)

    22.62

    1. PCl 5 (l) + 4 H 2 O(l) → H 3 PO 4 (aq) + 5 HCl(aq)
    2. 2 H 3 PO 4 (aq) Δ H 4 P 2 O 7 (aq) + H 2 O(l)
    3. P 4 O 10 (s) + 6 H 2 O(l) → 4 H 3 PO 4 (aq)

    Carbon, the Other Group 4A Elements, and Boron (Sections 22.9, 22.10, and 22.11)

    22.63 Analyze/Plan .  Review the nomenclature rules and ion names in Section 2.8. Solve.

    1. HCN
    2. Ni(CO) 4
    3. Ba(HCO 3 ) 2
    4. CaC 2
    5. K 2 CO 3

    22.64

    1. H 2 CO 3
    2. NaCN
    3. KHCO 3
    4. C 2 H 2
    5. Fe(CO) 5

    22.65 Analyze/Plan .  Use information on the descriptive chemistry of carbon given in Section 22.9 to complete and balance the equations. Solve.

    1. ZnCO 3 (s) Δ ZnO(s) + CO 2 (g)
    2. BaC 2 (s) + 2 H 2 O(l) → Ba 2+ (aq) + 2 OH (aq) + C 2 H 2 (g)
    3. 2 C 2 H 2 (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O(g)
    4. CS 2 (g) + 3 O 2 (g) → CO 2 (g) + 2 SO 2 (g)
    5. Ca(CN) 2 (s) + 2 HBr(aq) → CaBr 2 (aq) + 2 HCN(aq)

    22.66

    1. CO 2 (g) + OH (aq) → HCO 3 (aq)
    2. NaHCO 3 (s) + H + (aq) → Na + (aq) + H 2 O(l) + CO 2 (g)
    3. 2 CaO(s) + 5 C(s) Δ 2 CaC 2 (s) + CO 2 (g)
    4. C(s) + H 2 O(g) Δ H 2 (g) + CO(g)
    5. CuO(s) + CO(g) → Cu(s) + CO 2 (g)

    22.67 Analyze/Plan .  Use information on the descriptive chemistry of carbon given in Section 22.9 to complete and balance the equations. Solve.

    1. 2 CH 4 (g) + 2 NH 3 (g) + 3 O 2 (g) cat 800 ° C 2 HCN(g) + 6 H 2 O(g)
    2. NaHCO 3 (s) + H + (aq) → CO 2 (g) + H 2 O(l) + Na + (aq)
    3. 2 BaCO 3 (s) + O 2 (g) + 2 SO 2 (g) → 2 BaSO 4 (s) + 2 CO 2 (g)

    22.68

    1. 2 Mg(s) + CO 2 (g) → 2 MgO(s) + C(s)
    2. 6 CO 2 (g) + 6 H 2 O(l) h ν C 6 H 12 O 6 (aq) + 6 O 2 (g)
    3. CO 3 2– (aq) + H 2 O(l) → HCO 3 (aq) + OH (aq)

    22.69 Analyze/Plan .  Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample 4.8. Solve.

    1. H 3 B O 3 , +3
    2. Si Br 4 , +4
    3. Pb Cl 2 , +2
    4. Na 2 B 4 O 7 ⋅ 10H 2 O, +3
    5. B 2 O 3 , +3
    6. Ge O 2 , +4

    22.70

    1. Si O 2 , +4
    2. Ge Cl 4 , +4
    3. Na B H 4 , +3
    4. Sn Cl 2 , +2
    5. B 2 H 6 , +3
    6. B Cl 3 , +3

    22.71 Analyze/Plan .  Consider periodic trends within a family, particularly metallic character, as well as descriptive chemistry in Sections 22.9 and 22.10. Solve.

    1. Tin; see Table 22.8. The filling of the 4f subshell at the beginning of the sixth row of the periodic table increases Z and Z eff for later elements. This causes the ionization energy of Pb to be greater than that of Sn.
    2. Carbon, silicon, and germanium; these are the nonmetal and metalloids in group 4A. They form compounds ranging from XH 4 (–4) to XO 2 (+4). The metals tin and lead are not found in negative oxidation states.
    3. Silicon; silicates are the main component of sand.

    22.72

    1. carbon
    2. lead
    3. germanium

    22.73 Analyze/Plan .  Consider the structural chemistry of silicates discussed in Section 22.10 and shown in Figure 22.32. Solve.

    1. Tetrahedral
    2. Metasilicic acid will probably adopt the single-strand silicate chain structure shown in Figure 22.32(b). The empirical formula shows 3 O and 2 H atoms per Si atom. The chain has the same Si to O ratio as metasilicic acid. Furthermore, in the chain structure, there are two terminal (not bridging) O atoms on each Si. These can accommodate the 2 H atoms associated with each Si atom of the acid. The sheet structure does not fulfill these requirements.

    22.74  Carbon forms carbonates rather than silicates to take advantage of its ability to form π bonds. Because of its relatively compact 2p valence orbitals, carbon can engage in effective π-type overlap and form multiple bonds with itself and other elements. Carbonate, CO 3 2– , the anion present in carbonates, takes advantage of this ability to form stable π bonds, and is additionally stabilized by resonance. In silicates, silicon atoms form only single bonds and are tetrahedral.


    22.75 Analyze/Plan . In silicate anions, the oxidation number of silicon is +4 and that of oxygen is –2 (Section 22.10). Determine the charge on the silicate anion, then balance the charges of the cations and anions in the minerals. Solve.

    1. x = 2. The charge on Si 2 O 7 6– anion is 6–, that on Zn 2+ cation is 2+. Two Ca 2+ cations are required to balance charge.
    2. x = 2. The charge on Si 2 O 5 2– anion is 2–, that on Al 3+ cation is 3+. Two OH anions are required to balance charge.

    22.76

    1. x = 1. The charge on Si 3 O 8 4– anion is 4–, that on Al 3+ cation is 3+. One Na + cation is required to balance charge.
    2. x = 2. The charge on Si 4 O 11 6– anion is 6–, that on the cations is 2+. Two OH anions are required to balance charge.

    22.77

    1. Diborane (Figure 22.34 and below) has bridging H atoms linking the two B atoms. The structure of ethane shown below has the C atoms bound directly, with no bridging atoms.
      image
    2. B 2 H 6 is an electron deficient molecule. It has 12 valence electrons, whereas C 2 H 6 has 14 valence electrons. The 6 valence electron pairs in B 2 H 6 are all involved in B–H σ bonding, so the only way to satisfy the octet rule at B is to have the bridging H atoms shown in Figure 22.34.
    3. A hydride ion, H–, has two electrons, whereas an H atom has one. The term hydridic indicates that the H atoms in B 2 H 6 have more than the usual amount of electron density for a covalently bound H atom.

    22.78

    1. B 2 H 6 (g) + 6 H 2 O(l) → 2 H 3 BO 3 (aq) + 6 H 2 (g)
    2. 4 H 3 BO 3 (s) Δ H 2 B 4 O 7 (s) + 5 H 2 O(g)
    3. B 2 O 3 (s) + 3 H 2 O(l) → 2 H 3 BO 3 (aq)

    Additional Exercises

    22.79

    1. False. H 2 (g) and D 2 (g) are composed of different isotopes of hydrogen. Allotropes are composed of atoms of a single element bound into different structures.
    2. True. An interhalogen is a compound formed from atoms of two or more halogens.
    3. False. MgO(s) is a basic anhydride; it is the oxide of a metal.
    4. True. SO 2 (g) is the oxide of a nonmetal.
    5. True. A condensation reaction is the combination of two molecules to form a large molecule and a small one such as H 2 O or HCl.

    1. True. The nucleus of tritium contains one proton and two neutrons.
    1. False. Disproportionation is an oxidation–reduction process where the same element is both oxidized and reduced. In this reaction, sulfur is oxidized and oxygen is reduced.

    22.80

    1. BrO 3 (aq) + XeF 2 (aq) + H 2 O(l) → Xe(g) + 2 HF(aq) + BrO 4 (aq)
    2. BrO 3 , +5; BrO 4 , +7

    22.81

    1. SO 2 (g) + H 2 O(l) ⇌ H 2 SO 3 (aq)
    2. Cl 2 O 7 (g) + H 2 O(l) ⇌ 2 HClO 4 (aq)
    3. Na 2 O 2 (s) + 2 H 2 O(l) → H 2 O 2 (aq) + 2 NaOH(aq)
    4. BaC 2 (s) + 2 H 2 O(l) → Ba 2+ (aq) + 2 OH (aq) + C 2 H 2 (g)
    5. 2 RbO 2 (s) + 2 H 2 O(l) → 2 Rb + (aq) + 2 OH (aq) + O 2 (g) + H 2 O 2 (aq)
    6. Mg 3 N 2 (s) + 6 H 2 O(l) → 3 Mg(OH) 2 (s) + 2 NH 3 (g)
    7. NaH(s) + H 2 O(l) → NaOH(aq) + H 2 (g)

    22.82

    1. H 2 SO 4 – H 2 O → SO 3
    2. 2 HClO 3 – H 2 O → Cl 2 O 5
    3. 2 HNO 2 – H 2 O → N 2 O 3
    4. H 2 CO 3 – H 2 O → CO 2
    5. 2 H 3 PO 4 – 3H 2 O → P 2 O 5

    22.83  Assume that the reactions occur in acidic solution. The half-reaction for reduction of H 2 O 2 is in all cases H 2 O 2 (aq) + 2 H + (aq) + 2 e → 2 H 2 O(aq).

    1. N 2 H 4 (aq) + 2 H 2 O 2 (aq) → N 2 (g) + 4 H 2 O(l)
    2. SO 2 (g) + H 2 O 2 (aq) → SO 4 2– (aq) + 2 H + (aq)
    3. NO 2– (aq) + H 2 O 2 (aq) → NO 3 (aq) + H 2 O(l)
    4. H 2 S(g) + H 2 O 2 (aq) → S(s) + 2 H 2 O(l)
    5. 2 H + (aq) + H 2 O 2 (aq) + 2 e 2 H 2 O(l) 2[Fe 2 + (aq) Fe 3 + (aq) + e ] 2 Fe 2 + (aq) + H 2 O 2 (aq) + 2 H + (aq) 2 Fe 3 + (aq) + 2 H 2 O(l)

    22.84 S(g) + O 2 (g) SO 2 (g) ΔH = 296.9 kJ SO 2 (g) + 1/2 O 2 (g) SO 3 (g) ΔH = 98.3 kJ SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) ΔH = 130 kJ S(g) + 3/2 O 2 (g) + H 2 O(l) H 2 SO 4 (aq) ΔH = 525.2 = 5.3 × 10 2 kJ ( a ) ( b ) ( c )

    5000 lb H 2 SO 4 × 453.6 g 1 lb × 1 mol H 2 SO 4 98.09 g × 525 kJ mol H 2 SO 4 = 1.214 × 10 7 = 1 × 10 7 kJ

    One mole of H 2 SO 4 produces 5.3 × 10 2 kJ of heat, 5000 lb of H 2 SO 4 produces 1 × 10 7 kJ. (If trailing zeros are not significant, 130 kJ limits the first result to 2 sig figs and 5000 lb limits the second result to 1 sig fig.)


    22.85

    1. PO 4 3– , + 5; NO 3 , + 5
    2. The Lewis structure for NO 4 3– would be:
      image

      The formal charge on N is +1 and on each O atom is –1. The four electronegative oxygen atoms withdraw electron density, leaving the nitrogen deficient. Because N can form a maximum of four bonds, it cannot form a π bond with one or more of the O atoms to regain electron density, as the P atom in PO 4 3– does. Also, the short N–O distance would lead to a tight tetrahedron of O atoms subject to steric repulsion.

    22.86

    1. Although P 4 , P 4 O 6 , and P 4 O 10 all have four P atoms in a tetrahedral arrangement, the bonding between P atoms and by P atoms is not the same in the three molecules. In P 4 , the 4 P atoms are bound only to each other by P–P single bonds with strained bond angles of approximately 60°. In the two oxides, the 4 P atoms are directly bound to oxygen atoms, not to each other. Bonding by P atoms in P 4 O 6 and P 4 O 10 is very similar. Each contains the P 6 O 6 cage, formed by four P 3 O 3 rings that share a P–O–P edge. Phosphorus bonding to oxygen maintains the overall P 4 tetrahedron but allows the P atoms to move away from each other so that the angle strain is relieved relative to molecular P 4 . The P–O–P and O–P–O angles in both oxides are near the ideal 109°. In P 4 O 6 , each P is bound to 3 O atoms and has a lone pair completing its octet. In P 4 O 10 , the lone pair is replaced by a terminal O atom and each P is bound to 3 bridging and 1 terminal O atom.
    2. image

      In both structures there are unshared pairs on all oxygens to give octets and the geometry around each P is approximately tetrahedral.

    22.87  3 H 3 PO 4 → H 5 P 3 O 10 + 2 H 2 O

    22.88 GeO 2 (s) + C(s) Δ Ge(l) + CO 2 (g)

    Ge(l) + 2 Cl 2 (g) → GeCl 4 (l)

    GeCl 4 (l) + 2 H 2 O(l) → GeO 2 (s) + 4 HCl(g)

    GeO 2 (s) + 2 H 2 (g) → Ge(s) + 2 H 2 O(l)

    22.89  KAlSi 3 O 8 . The oxidation number of Si is +4, while the oxidation number of Al is +3. If K is present for charge balance, each Al requires one K. That is, (K+Al) replaces Si. The ratio of Si to O in the mineral is 1 : 2. Consider four SiO 2 units, Si 4 O 8 . If 25% of Si are replaced, that is one out of four Si atoms. Replace one Si with (K+Al). The formula becomes KAlSi 3 O 8 .


    22.90

    1. The charge on the aluminosilicate ion shown (AlSi 3 O 10 ) is 5– (+3 from Al, +12 from Si, –20 from O).
    2. In the silicate ions pictured in Figure 22.32, Si is central and O is either bridging or terminal. We want to select a structure from Figure 22.32 that has the same ratio of central atoms and O atoms as AlSi 3 O 10 5– . The ratio of central to O atoms in AlSi 3 O 10 5– is 4 to 10, or 2 to 5. The analogous ion in Figure 22.32 is Si 2 O 5 2– . The AlSi 3 O 10 5– ion will have a sheet structure similar to the one shown in Figure 22.32(c), with ¼ of the Si central atoms replaced by Al central atoms.

    Integrative Exercises

    22.91

    1. 100.0 × 10 3 g FeTi × 1 mol FeTi 103 .7 g FeTi × 1 mol H 2 1 mol FeTi × 2.016 g H 2 1 mol H 2 = 1944.1 = 1.94 × 10 3 g H 2
    2. V = 1944.1 g H 2 2.016 g/mol H 2 × 0.08206 L-atm mol-K × 273 K 1 atm = 21 , 603 = 2.16 × 10 4 L H 2
    3. 2 H 2 (g) + O 2 (g) → 2 H 2 O(l)

      Δ H o = 2 Δ H f o H 2 O(l) – 2 Δ H f o H 2 (g) – Δ H f o O 2 (g)

      Δ H o = 2(–285.83) – 2(0) – (0) = –571.66 kJ

      1944.1 g H 2 × 1 mol H 2 2.016 g H 2 × 571.66 kJ 2 mol H 2 = 275 , 636 = 2.76 × 10 5 kJ

      The minus sign indicates that energy is produced.

    22.92  From Appendix C, we need only Δ H f o for F(g), so that we can estimate ΔH for the process:

    F 2 (g) F(g) + F(g); ΔH° = 160 kJ XeF 2 (g) Xe(g) + F 2 (g) ΔH f o = 109 kJ XeF 2 (g) Xe(g) + 2 F(g) ΔH° = 269 kJ

    The average Xe–F bond enthalpy is thus 269/2 = 134 kJ. Similarly,

    XeF 4 (g) Xe(g) + 2 F 2 (g) ΔH f o = 218 kJ 2 F 2 (g) 4 F(g) ΔH° = 320 kJ XeF 4 (g) Xe(g) + 4 F(g) ΔH° = 538 kJ

    Average Xe–F bond energy = 538/4 = 134 kJ

    XeF 6 (g) Xe(g) + 3 F 2 (g) ΔH f o = 298 kJ 3 F 2 (g) 6 F(g) ΔH° = 480 kJ XeF 6 (g) Xe(g) + 6 F(g) ΔH° = 778 kJ

    Average Xe–F bond energy = 778/6 = 130 kJ

    The average bond enthalpies are: XeF 2 , 134 kJ; XeF 4 , 134 kJ; XeF 6 , 130 kJ. They are remarkably constant in the series.


    22.93

    1. H 2 (g) + 1/2 O 2 (g) → H 2 O(l); ΔH = –285.83 kJ/mol H 2

      CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)

      ΔH = 2(–285.83) – 393.5 – (–74.8) = –890.4 kJ/ mol CH 4

    2. for H 2 : 285.83 kJ 1 mol H 2 × 1 mol H 2 2.0159 g H 2 = 141.79 kJ/g H 2

      for CH 4 : 890.4 kJ 1 mol CH 4 × 1 mol CH 4 16.043 g CH 4 = 55.50 kJ/g CH 4

    3. Find the number of moles of gas that occupy 1 m 3 at STP:

      n = 1 atm × 1 m 3 273 K × 1 mol-K 0 .08206 L-atm × [ 100 cm 1 m ] 3 × 1 L 10 3 cm 3 = 44.64 mol

      for H 2 : 285.83 kJ 1 mol H 2 × 44.64 mol H 2 1 m 3 H 2 = 1.276 × 10 4 kJ/m 3 H 2

      for CH 4 : 890.4 kJ 1 mol CH 4 × 44.64 mol CH 4 1 m 3 CH 4 = 3.975 × 10 4 kJ/m 3 CH 4

    22.94 Analyze/Plan . Δ G o = –RT ln K. Use Δ G o for ozone from Appendix C to calculate K for the reaction at 290.0 K, assuming no electrical input. Solve.

    The reaction under consideration is: 3 O 2 (g) → 2 O 3 (g)

    Δ G o for ozone at 298 K = 163.4 kJ/mol

    Δ G o for the reaction is then 2(163.4 kJ) – 3(0 kJ) = 326.8 kJ.

    l n K = Δ G o RT = ( 326.8 × 10 3 ) J 8.314 J/K × 298.0 K = 131.903 = 131.9 ; K = 5 × 10 58

    22.95

    1. First, calculate the molar solubility of Cl 2 in water.

      n = 1 atm (0 .310 L) 0.08206 L-atm 1 mol-K × 273 K = 0.01384 = 0.0138 mol Cl 2 M = 0.01384 mol 0 .100 L = 0.1384 = 0.138 M

      [ Cl ] = [ HOCl] = [ H + ] Let this quantity = x . Then, x 3 ( 0.1384 x) = 4.7 × 10 4

      Assuming that x is small compared with 0.1384:

      x 3 = (0.1384)(4.7 × 10 –4 ) = 6.504 × 10 –5 ; x = 0.0402 = 0.040 M

      We can correct the denominator using this value, to get a better estimate of x:

      x 3 0.1384 0 .0402 = 4.7 × 10 4 ; x = 0.0359 = 0.036 M

      One more round of approximation gives x = 0.0364 = 0.036 M . This is the equilibrium concentration of HClO.


    1. From the equilibrium reaction in part (a), [H + ] = 0.036 M . pH = –log[H + ] = 1.4

      The HOCl produced by this equilibrium will ionize slightly to produce additional H + (aq). However, the K a value for HOCl is small, 3.0 × 10 –8 , and the acid ionization will be suppressed by the presence of H + (aq) from the solubility equilibrium. [H + ] from ionization of HOCl will be small compared to 0.036 M and will not significantly impact the pH.

    22.96

    1. 2 NH 4 ClO 4 (s) Δ N 2 (g) + 2 HCl(g) + 3 H 2 O(g) + 5/2 O 2 ( g )

      NH 4 ClO 4 (s) Δ 1/2 N 2 (g) + HCl(g) + 3/2 H 2 O(g) + 5/4 O 2 ( g )

    2. Δ H ° = Σ Δ H f ° prod Σ Δ H react

      Δ H ° = Δ H f o HCl(g) + 3/2 Δ H f o H 2 O(g) + 1/2 Δ H f o N 2 (g) + 5/4 Δ H f o O 2 ( g ) Δ H f o NH 4 ClO 4 Δ H ° = 92.30 kJ + 3/2( 241.82 kJ) + 1/2 (0 kJ) + 5/4 (0 kJ) ( 295.8 kJ) = 159.2 kJ/mol NH 4 ClO 4

    3. The aluminum reacts exothermically with O 2 (g) and HCl(g) produced in the decomposition, providing additional heat and thrust.
    4. There are (1/2 + 1 + 3/2 + 5/4) = 4.25 mol gas per mol NH 4 ClO 4 decomposed

      1 lb NH 4 ClO 4 × 453.6 g 1 lb 1 mol NH 4 ClO 4 117.49 g NH 4 ClO 4 × 4.25 mol gas 1 mol NH 4 ClO 4 = 16.408 = 16.4 mol gas

      V = nRT P = 16.408 mol gas × 0.08206 L-atm mol-K × 2 73 K 1 atm = 367.57 = 368 L

    22.97

    1. N 2 H 4 (g) + O 2 (g) → N 2 (g) + 2 H 2 O(l)
    2. Δ H ° = Δ H f o N 2 (g) + 2 Δ H f o H 2 O(l) Δ H f o N 2 H 4 (aq) Δ H f o O 2 (g)

      = 0 + 2(–285.83) – 95.40 – 0 = –667.06 kJ

    3. 9.1 g O 2 1 × 10 6 g H 2 O × 1.0 g H 2 O 1 mL H 2 O × 1000 mL 1 L × 3.0 × 10 4 L = 273 = 2.7 × 10 2 g O 2

      2.73 × 10 2 g O 2 × 1 mol O 2 32.00 g O 2 × 1 mol N 2 H 4 1 mol O 2 × 32.05 g N 2 H 4 1 mol N 2 H 4 = 2.7 × 10 2 g N 2 H 4

    22.98

    1. SO 2 (g) + 2 H 2 S(aq) → 3 S(s) + 2 H 2 O(g) or, if we assume S 8 is the product, 8 SO 2 (g) + 16 H 2 S(aq) → 3 S 8 (s) + 16 H 2 O(g).
    2. Assume that all S in the coal becomes SO 2 upon combustion, so that 1 mol S (coal) = 1 mol SO 2 ; 1 ton = 2000 lb; 760 torr = 1.00 atm

      4000 lb coal × 0 .035 lb S 1 lb coal × 453.6 g S 1 lb S × 1 mol S (coal) 32.07 g S × 1 mol SO 2 1 mol S (coal) × 2 mol H 2 S 1 mol SO 2 = 3960 = 4.0 × 10 3 mol H 2 S

      V = 3960 mol × ( 0 .08206 L-atm/mol-K) × 300 K 1.00 atm = 97 , 496 = 9.7 × 10 4 L


    1. 3960 mol H 2 S × 3 mol S 2 mol H 2 S × 32.07 g S 1 mol S = 1.9 × 10 5 g S

      This is about 210 lb S per ton of coal combusted. (However, two-thirds of this comes from the H 2 S, which presumably at some point was also obtained from coal.)

    22.99 Plan .  Vol air → kg air → g H 2 S → g FeS. Use the ideal-gas equation to change volume of air to mass of air, (assuming 1.00 atm, 298 K and an average molar mass (MM) for air of 29.0 g/mol. Use (20 mg H 2 S/kg air) to find the mass of H 2 S in the given mass of air. Solve .

    V air = 12 ft × 2 0 ft × 8 ft × 12 3 in 3 ft 3 × 2 .54 3 cm 3 1 3 in 3 × 1 L 1000 cm 3 = 5.4368 × 10 4 = 5 × 10 4 L

    g air = PV MM RT ; assume P = 1.00 atm, T = 298 K, M M air = 29.0 g/mol

    g air = 1.00 atm × 5 .4368 × 10 4 L × 29 .0 g/mol 298 K × mol-K 0.08206 L-atm = 64 , 476 = 6 × 10 4 g air

    6.4476 × 10 4 g air × 1 kg 1000 g × 20 mg H 2 S 1 kg air × 1 g 1000 mg = 1.2895 = 1 g H 2 S

    FeS(s) + 2 HCl(aq) → FeCl 2 (aq) + H 2 S(g)

    1.2895 g H 2 × 1 mol H 2 34.08 g H 2 S × 1 mol FeS 1 mol H 2 S × 87.91 g FeS 1 mol FeS = 3.3263 = 3 g FeS

    22.100  The reactions can be written as follows:

    H 2 (g) + X(std state) H 2 X(g) ΔH f o 2 H(g) H 2 (g) ΔH f o (H H) X(g) X(std state) ΔH 3 Add: 2 H(g) + X(g) H 2 X(g) ΔH = ΔH f o + ΔH f o ( H H) + ΔH 3

    These are all the necessary ΔH values. Thus,

    Compound ΔH D H–X
    H 2 O ΔH = –241.8 kJ – 436 kJ – 248 kJ = –926 kJ 463 kJ
    H 2 S ΔH =   –20.17 kJ – 436 kJ – 277 kJ = –733 kJ 367 kJ
    H 2 Se ΔH =   +29.7 kJ – 436 kJ – 227 kJ = –633 kJ 317 kJ
    H 2 Te ΔH = +99.6 kJ – 436 kJ – 197 kJ = –533 kJ 267 kJ

    The average H–X bond energy in each case is just half of ΔH. The H–X bond energy decreases steadily in the series. The origin of this effect is probably the increasing size of the orbital from X with which the hydrogen 1s orbital must overlap.

    22.101

    1. MnSi: more than one element, so not metallic; high melting, so not molecular; insoluble in water, so not ionic; therefore covalent network.
    2. MnSi(s) + HF(aq) → SiH 4 (g) + MnF 4 (s)

      Reduction of Mn(IV) to Mn(II) is unlikely, because F is an extremely weak reducing agent. E red o for F 2 (g) + 2 e → 2 F (aq) = 2.87 V


    22.102  N 2 H 5 + (aq) → N 2 (g) + 5 H + (aq) + 4 e E red o = 0.23 V

    Reduction of the metal should occur when E red o of the metal ion is more positive than about –0.15 V. This is the case for (b) Sn 2+ (marginal), (c) Cu 2+ , (d) Ag + and (f) Co 3+

    22.103  First write the balanced equation to give the number of moles of gaseous products per mole of hydrazine.

    1. (CH 3 ) 2 NNH 2 + 2 N 2 O 4 → 3 N 2 (g) + 4 H 2 O(g) + 2 CO 2 (g)
    2. (CH 3 )HNNH 2 + 5/4 N 2 O 4 → 9/4 N 2 (g) + 3 H 2 O(g) + CO 2 (g)

      In case (A) there are nine moles gas per one mole (CH 3 ) 2 NNH 2 plus two moles N 2 O 4 . The total mass of reactants is 60 + 2(92) = 244 g. Thus, there are

      9 mol gas 244 g reactants = 0.0369 mol gas 1 g reactants

      In case (B) there are 6.25 moles of gaseous product per one mole (CH 3 )HNNH 2 plus 1.25 moles N 2 O 4 . The total mass of this amount of reactants is 46.0 + 1.25(92.0) = 161 g.

      6.25 mol gas 161 g reactants = 0.0388 mol gas 1 g reactants

    Thus the methylhydrazine (B) has marginally greater thrust.

    22.104

    1. 3 B 2 H 6 (g) + 6 NH 3 (g) → 2 (BH) 3 (NH) 3 (l) + 12 H 2 (g)

      3 LiBH 4 (s) + 3 NH 4 Cl(s) → 2 (BH) 3 (NH) 3 (l) + 9 H 2 (g) + 3 LiCl(s)

    2. 30 valence e , 15 e pairs
      image

      The structure with nonbonded pairs minimizes formal charge, but these electrons are almost certainly delocalized about the six-membered ring, mimicking the bonding in benzene.

    3. n = PV RT = 1.00 atm × 2.00 L 273 K × mol-K 0.08206 L-atm = 0.08929 = 8.93 × 10 2 mol NH 3

      0.08929 mol NH 3 × 2 mol (BH ) 3 (NH ) 3 6 mol NH 3 × 80.50 g (BH ) 3 (NH ) 3 1 mol (BH ) 3 (NH ) 3 = 2.3956 = 2.40 g (BH ) 3 (NH ) 3


     

    23 Transition Metals and Coordination Chemistry

    Visualizing Concepts

    23.1 Analyze/Plan .  Given graphs of three properties moving across the fourth period of the chart, match each graph to the atomic property it represents. Solve.

    Graph (a) is the trend in maximum oxidation state. This value corresponds to the maximum number of (4s + 3d) electrons that can be removed from a neutral atom. It increases up to a maximum of +7 for Mn, then decreases. The decrease is partly because of an increase in the attraction of 3d electrons for the nucleus as Z eff increases.

    Graph (b) is the trend in effective nuclear charge, Z eff . Moving from left to right in a period, Z eff increases because the increase in Z is not offset by a significant increase in shielding.

    Graph (c) is the trend in radius. Increasing Z eff leads to decreasing atomic radius.

    23.2 Analyze .  Given the formula of a coordination compound, draw the structure, determine the coordination number, coordination geometry, oxidation state of the metal, and number of unpaired electrons.

    Plan .  From the formula, determine the identity of the ligands and the number of coordination sites they occupy. From the total coordination number, decide on a likely geometry. Use ligand and overall complex charges to calculate the oxidation number of the metal. Refer to the d-orbital energy level diagram that corresponds to the structure of the compound and the field strength of the ligands to determine the number of unpaired electrons.

    Solve . The ligands are 2Cl , one coordination site each, and en, ethylenediamine, two coordination sites, for a coordination number of 4. This coordination number has two possible geometries, tetrahedral and square planar. Pt is one of the metals known to adopt square planar geometry when CN = 4.

    image
    1. Coordination number is 4
    2. Coordination geometry is square planar
    3. The oxidation state of Pt is +2. Pt(en)Cl 2 is a neutral compound, the en ligand is neutral, and the 2Cl ligands are each –1, so the oxidation state of Pt must be +2, Pt(II).
    4. There are no unpaired electrons. Square planar d 8 complexes are usually low spin, especially with heavier metals like Pt.

    23.3 Analyze .  Given a ball-and-stick figure of a ligand, write the Lewis structure and answer questions about the ligand.

    Plan. Assume that each atom in the Lewis structure obeys the octet rule. Complete each octet with unshared electron pairs or multiple bonds, depending on the bond angles in the ball-and-stick model. Black = C, blue = N, red = O, gray = H.

    image

    There is a second resonance structure with the double bond drawn to the other O atom.

    Check .  Write the molecular formula, count the valence electron pairs and see if it matches your structure. [C 4 H 9 N 2 O 2 ] (16 + 9 + 10 + 12 + 1) = 48 valence e , 24 e pair Our Lewis structure also has 24 e pairs.

    1. Donor atoms have unshared electron pairs. The potential donors in this structure are the two N and two O atoms.

      The ligand is polydentate. If bound to a single metal center, it is likely to be tridentate. Even though there are four possible donor atoms, the structure would be strained if all four were bound to one metal center. It is likely that only one of the two O atoms binds to the same metal as the two N atoms. If it is bound to more than one metal, it could use all four donor atoms.

    2. An octahedral complex has 6 coordination sites. A single ligand has 3 likely donors, so two ligands are needed. From a steric perspective, the likely donors would be the 2 N atoms and 1 of the carbonyl oxygen atoms. The chelate bite of a carboxyl group is relatively small and would require an O M O angle of less than 90°.

    23.4 Analyze/Plan . Given the molecular formula, name the compound. Use the ball-and-stick structures to decide the number of geometric isomers for each molecular geometry. Consider the d-electron configuration of platinum(II) and the appropriate crystal field splitting diagrams (Figures 23.33 and 23.34) to determine magnetic properties for the two molecular geometries. Solve .

    1. Diaminodichloroplatinum(II). If the geometry is square planar, the prefix cis-would be added.
    2. No, the tetrahedral molecule would not have a geometric isomer.
    3. The tetrahedral molecule would be paramagnetic. Platinum(II) is a d 8 transition metal ion. In a tetrahedral crystal field (Figure 23.33 and Sample Exercise 23.8), there would be two unpaired d-electrons and the molecule would be paramagnetic.
    4. Yes, the square planar molecule would have a geometric isomer, the trans isomer.

    1. The square planar molecule would be diamagnetic. A d 8 transition metal ion in a square planar crystal field (Figure 23.34 and Sample Exercise 23.8) has no unpaired electrons.
    1. Yes, you would be able to distinguish the two geometries by determining the number of geometric isomers. The square planar geometry has two possible isomers, cis and trans, while the tetrahedral geometry has one unique structure.
    1. Yes, you would be able to distinguish the two geometries by measuring the molecule’s response to a magnetic field. The tetrahedral molecule is attracted to a magnetic field, while the square planar molecule is not.

    23.5 Analyze .  Given 5 structures, visualize which are identical to (1) and which are geometric isomers of (1).

    Plan .  There are two possible ways to arrange MA 3 X 3 . The first has bond angles of 90° between all similar ligands; this is structure (1). The second has one 180° angle between similar ligands. Visualize which description fits each of the five structures.

    Solve .  (1) has all 90° angles between similar ligands.

    (2) has a 180° angle between similar ligands (see the blue ligands in the equatorial plane of the octahedron)

    (3) has all 90° angles between similar ligands

    (4) has all 90° angles between similar ligands

    (5) has a 180° angle between similar ligands (see the blue axial ligands)

    Structures (3) and (4) are identical to (1); (2) and (5) are geometric isomers.

    23.6 Analyze .  Given four structures, decide which are chiral.

    Plan .  Chiral molecules have nonsuperimposable mirror images. Draw the mirror image of each molecule and visualize whether it can be rotated into the original molecule. If so, the complex is not chiral. If the original orientation cannot be regenerated by rotation, the complex is chiral. Solve .

    image

    The two orientations are not superimposable and molecule (1) is chiral.

    image

    The two orientations are superimposable. Rotate the right-most structure 90° counterclockwise about the B-M-B axis to align the G’s; the bidentate ligands then also overlap. Molecule (2) is not chiral.

    image

    The two orientations are not superimposable and molecule (3) is chiral.

    image

    The two orientations are not superimposable and molecule (4) is chiral.

    23.7 Analyze .  Given the visible colors of two solutions, determine the colors of light absorbed by each solution.

    Plan .  Apparent color is transmitted or reflected light, absorbed color is basically the complement of apparent color. Use the color wheel in Figure 23.25 to obtain the complementary absorbed color for the solutions.

    Solve .  Moving from left to right, the solutions appear blue-green (cyan), yellow, green, and red. The solutions absorb red-orange, violet, red, and green.

    23.8 Analyze. Fit the crystal field splitting diagram to the complex description in each part.

    Plan .  Determine the number of d-electrons in each transition metal. On the splitting diagrams match the d-orbital splitting patterns to complex geometry and electron pairing to the definition of high-spin and low-spin.

    Solve .  Octahedral complexes have the 3 lower, 2 higher splitting pattern, whereas tetrahedral complexes have the opposite 2 lower, 3 higher pattern. Low spin complexes favor electron pairing because of large d-orbital splitting. High-spin complexes have maximum occupancy because of small orbital splitting.

    1. Fe 3+ , 5 d-electrons; weak field: spins unpaired; octahedral: 3 lower, 2 higher d-splitting ∴ diagram (4)
    2. Fe 3+ , 5 d-electrons; strong field: spins paired; octahedral: 3 lower, 2 higher d-splitting ∴ diagram (1)
    3. Fe 3+ , 5 d-electrons; tetrahedral: 2 lower, 3 higher d-splitting ∴ diagram (3)
    4. Ni 2+ , 8 d-electrons; tetrahedral: 2 lower, 3 higher d-splitting ∴ diagram (2)

    Check .  Diagram (2) was the remaining choice for (d) and it fits the description.

    23.9 Analyze/Plan .  Given the linear diagram and axial labels, answer the questions and predict crystal field splitting. Orbitals with lobes nearest ligand charges (or partial charges) will be highest in energy; orbitals with lobes away from charges are lowest in energy.


    Solve .  Diagram (c) is the best choice. The d z 2 orbital has lobes nearest the charges and is at the highest energy. The d x 2 y 2 and d xy orbitals have lobes in the xy-plane farthest from the charges and are lowest in energy. The d xz and d yz orbitals point between the respective axes and are intermediate in energy.

    image

    23.10 Analyze .  Given the colors of two low spin Fe(II) complexes, determine which complex contains the stronger field ligand. Plan . We can make this direct comparison because both solutions contain low spin d 6 ions. A solution that appears one color absorbs visible light of the complementary color. Use the color wheel in Figure 23.25 to decide which color and approximate wavelength of visible light is absorbed by the two solutions. The stronger-field ligand causes a larger d-orbital splitting and absorbs light with the shorter wavelength.

    Solve . The green solution absorbs red light in the 650 to 750 nm range. The red solution absorbs green light in the 490 to 560 nm range. The complex that produces the red solution has the larger d-orbital splitting and the stronger-field ligand.

    The Transition Metals (Section 23.1)

    23.11   Trend (c). The lanthanide contraction is the name given to the decrease in atomic size because of the build-up in effective nuclear charge as we move through the lanthanides (elements 57–71) and beyond them. This effect offsets the expected increase in atomic size going from period 5 to period 6 transition elements.

    23.12   Trend (b) explains the peak in maximum oxidation state of the transition-metal elements near groups 7B and 8B. As effective nuclear charge increases, d-electrons are more strongly attracted to the nucleus and more difficult to remove from the atom.

    23.13

    1. Ti 2+ , [Ar]3d 2
    2. Ti 4+ , [Ar]
    3. Ni 2+ , [Ar]3d 8
    4. Zn 2+ , [Ar]3d 10

    23.14   Refer to Figure 23.5. Among the period 4 transition metals, only Sc and Zn do not have at least one oxidation state with partially filled 3d orbitals. Sc 3+ has zero 3d electrons and Zn 2+ has ten 3d electrons. In Sc 3+ the 3d orbitals are unoccupied; in Zn 2+ the 3d orbitals are filled.

    23.15

    1. Ti 3+ , [Ar]3d 1
    2. Ru 2+ , [Kr]4d 6
    3. Au 3+ , [Xe]4f 14 5d 8
    4. Mn 4+ , [Ar]3d 3

    23.16

    1. Co 3+ , [Ar]3d 6 ; 6 valence d-electrons
    2. Cu + ; [Ar]3d 10 ; 10 valence d-electrons
    3. Cd 2+ , [Kr]4d 10 ; 10 valence d-electrons
    4. Os 3+ : [Xe] 4f 14 5d 5 ; 5 valence d-electrons

    23.17 Analyze/Plan .  Consider the definitions of paramagnetic and diamagnetic. Solve .

    The unpaired electrons in a paramagnetic substance cause it to be weakly attracted into a magnetic field. (A diamagnetic material, where all electrons are paired, is very weakly repelled by a magnetic field.)

    23.18   Antiferromagnetic materials cannot be used to make permanent magnets. In an antiferromagnetic material, coupled spins are aligned in opposite directions and the opposing spins exactly cancel.

    23.19 Analyze/Plan .  Consider the orientation of spins in various types of magnetic materials as shown in Figure 23.6.

    The diagram shows a material with misaligned spins that become aligned in the direction of an applied magnetic field. This is a paramagnetic material.

    23.20

    1. Fe 2 O 3 has all Fe atoms in the +3 oxidation state, whereas Fe 3 O 4 contains Fe atoms in both the +2 and +3 states. Each Fe 3 O 4 formula unit has one Fe(II) and two Fe(III).
    2. In an antiferromagnetic material, spins on coupled atoms are oppositely aligned, producing a net spin of zero. This is only possible for Fe 2 O 3 , where all Fe atoms have the same oxidation state, d-electron configuration and number of unpaired electrons. In Fe 3 O 4 , Fe(II) and Fe(III) atoms have different d-electron configurations and different numbers of unpaired electrons. Assuming an Fe(II) is coupled to an Fe(III), even if spins on coupled centers are oppositely aligned, their spins do not fully cancel and the material is ferromagnetic.

    Transition-Metal Complexes (Section 23.2)

    23.21

    1. Primary valence. In Werner’s theory, primary valence is the charge of the metal cation at the center of the complex. “Oxidation state” is a broader term than “ionic charge,” but Werner’s complexes contain metal ions where cation charge and oxidation state are equal.
    2. Coordination number. In Werner’s theory, secondary valence is the number of atoms bound or coordinated to the central metal ion.
    3. NH 3 can serve as a ligand because it has an unshared electron pair, whereas BH 3 does not. Ligands act as a Lewis base in metal-ligand interactions. As such, they must possess at least one unshared electron pair. BH 3 , with fewer than 8 electrons about B, has no unshared electron pair and cannot act as a ligand. In fact, BH 3 acts as a Lewis acid, an electron pair acceptor, because it is electron-deficient.

    23.22

    1. Negatively charged ions are more likely to act as ligands, to take advantage of the electrostatic attraction between a positively charged metal cation and a negatively charged ligand.
    2. Polar molecules are more likely to act as ligands to take advantage of ion-dipole interactions between a positively charged metal cation and a polar ligand.

    23.23 Analyze/Plan .  Follow the logic in Sample Exercises 23.1 and 23.2. Solve .

    1. This compound is electrically neutral, and the NH 3 ligands carry no charge, so the charge on Ni must balance the –2 charge of the 2 Br ions. The charge and oxidation state of Ni is +2.
    2. Because there are 6 NH 3 molecules in the complex, the likely coordination number is 6. In some cases Br acts as a ligand, so the coordination number could be other than 6.
    3. Assuming that the 6 NH 3 molecules are the ligands, 2 Br ions are not coordinated to the Ni 2+ , so 2 mol AgBr(s) will precipitate.

    23.24

    1. [Cr(H 2 O) 6 ] 3+
    2. Three. Because the Cl ions are not in the coordination sphere, all 3 anions react with Ag + to form 3 moles of AgCl(s).
    3. If the empirical formula of the compound is CrCl 3 and Cr 3+ has a coordination number of six, some or all of the Cl ions must be shared by more than one Cr 3+ ion. This produces a network solid that is very difficult to dissolve or break down.

    23.25 Analyze/Plan .  Count the number of donor atoms in each complex, taking the identity of polydentate ligands into account. Follow the logic in Sample Exercise 23.2 to obtain oxidation numbers of the metals.

    1. Coordination number = 4, oxidation number = +2
    2. 5, +4
    3. 6, +3
    4. 5, +2
    5. 6, +3
    6. 4, +2

    23.26

    1. Coordination number = 6, oxidation number = +3
    2. 4, +2
    3. 6, +4
    4. 6, +3
    5. 6, +3
    6. 5, +2

    Common Ligands in Coordination Chemistry (Section 23.3)

    23.27

    1. CH 3 CH 2 NH 2 , 20 e , 10 e pr
      image

      monodentate ligand, only N atom has a nonbonded pair of electrons


    1. P(CH 3 ) 3 , 26 e , 13 e pr
      image

      monodentate ligand, only P atom has a nonbonded pair of electrons

    1. CO 3 2− , 24 e , 12 e pr
      image

      either monodentate or bidentate

      [All three O atoms are possible bonding sites, but it is not geometrically possible for all three O atoms to be bound to the same metal ion.]

    1. C 2 H 6 , 14 e , 7 e pr
      image

      unlikely to act as a ligand, no nonbonded pairs of electrons

    23.28

    1. 2 coordination sites, 2 N donor atoms
    2. 2 coordination sites, 2 N donor atoms
    3. 2 coordination sites, 2 O donor atoms (Although there are four potential O donor atoms in C 2 O 4 2– , it is geometrically impossible for more than two of these to be bound to a single metal ion.)
    4. 4 coordination sites, 4 N donor atoms
    5. 6 coordination sites, 2 N and 4 O donor atoms

    23.29 Analyze/Plan .  Given the formula of a coordination compound, determine the number of coordination sites occupied by the polydentate ligand. The coordination number of the complexes is probably 4 or 6. Note the number of monodentate ligands and determine the number of coordination sites occupied by the polydentate ligands. Solve .

    1. ortho -Phenanthroline, o -phen, is bidentate. The complex is 6-coordinate, there are 4 monodentate NH 3 ligands, so o -phen occupies 2 sites.
    2. Oxalate, C 2 O 4 2– , is bidentate. The complex is 6-coordinate, there are 4 monodentate H 2 O ligands, so oxalate occupies 2 coordination sites.

    1. Ethylenediaminetetraacetate, EDTA, is hexadentate. The complex is probably 6-coordinate octahedral with EDTA occupying all six coordination sites.
    1. Ethylenediamine, en, is bidentate. The complex is 4-coordinate and each en ligand occupies 2 coordination sites.

    23.30

    1. 6
    2. 6
    3. 6
    4. 6

    23.31 Analyze/Plan . Anions and polar molecules (with nonbonded electron pairs) are most likely to act as ligands in a metal complex (Solution 23.22). Solve .

    1. CH 3 CN, polar molecule with nonbonded electron pair
    2. H , anion
    3. CO, polar molecule with a nonbonded electron pair

    23.32

    1. Pyridine is a mono dentate ligand because it has one N donor atom and therefore occupies one coordination site in a metal complex.
    2. K for this reaction will be smaller than one. Two free pyridine molecules are replaced by one free bipy molecule. There are more moles of particles in the reactants than products, so ΔS is predicted to be negative. Processes with a net decrease in entropy are usually nonspontaneous, have positive ΔG, and values of K less than one. This equilibrium is likely to be spontaneous in the reverse direction.

    23.33   False. The ligand shown in the figure does not typically act as a bidentate ligand for a single metal center. The entire molecule is planar; there is no “bend” in the central 6-membered ring that includes the two N atoms. The benzene rings on either side of the two N atoms inhibit their approach in the correct orientation for chelation. It might act as a bridging ligand between two metal centers, but again, the benzene rings would create a significant amount of steric hindrance.

    23.34

    1. The complex shown in the exercise has tetrahedral geometry about the silver.
    2. The ligands are neutral molecules and the metal is Ag(I), so the complex will have a 1+ charge.
    3. Yes, one nitrate ion, NO 3– , will be present in the crystal to provide charge balance for the complex cation.
    4. [Ag( o -phen) 2 ]NO 3
    5. bis( ortho -phenanthroline)silver(I) nitrate

    Nomenclature and Isomerism in Coordination Chemistry (Section 23.4)

    23.35 Analyze/Plan .  Given the name of a coordination compound, write the chemical formula. Refer to Tables 23.4 and 23.5 to find ligand formulas. Place the metal complex (metal ion + ligands) inside square brackets and the counterion (if there is one) outside the brackets. Solve .

    1. [Cr(NH 3 ) 6 ](NO 3 ) 3
    2. [Co(NH 3 ) 4 CO 3 ] 2 SO 4
    3. [Pt(en) 2 Cl 2 ]Br 2
    4. K[V(H 2 O) 2 Br 4 ]
    5. [Zn(en) 2 ][HgI 4 ]

    23.36

    1. [Mn(H 2 O) 4 Br 2 ]ClO 4
    2. [Cd(bipy) 2 ]Cl 2
    3. K[Co( o -phen)Br 4 ]
    4. Cs[Cr(NH 3 ) 2 (CN) 4 ]
    5. [Rh(en) 3 ][Co(ox) 3 ]

    23.37 Analyze/Plan .  Follow the logic in Sample Exercise 23.3, paying attention to naming rules in Section 23.4. Solve .

    1. tetraamminedichlororhodium(III) chloride
    2. potassium hexachlorotitanate(IV)
    3. tetrachlorooxomolybdenum(VI)
    4. tetraaqua(oxalato)platinum(IV) bromide

    23.38

    1. dichloroethylenediamminecadmium(II)
    2. potassium hexacyanomanganate(II)
    3. pentaamminecarbonatochromium(III) chloride
    4. tetraamminediaquairidium(III) nitrate

    23.39 Analyze/Plan . Consider the coordination number and geometry of each of the complexes, along with the definitions of the various types of isomerism. Use this information to decide which of the complexes could exhibit isomerism of the specified type. Solve.

    Complex 1 has a coordination number of 6 and octahedral geometry about the metal. There are 4 monodentate ligands of one kind and two of another.

    Complex 2 has a coordination number of 4 and square planar geometry. There are two monodentate ligands of one kind and two of another.

    Complex 3 has a coordination number of 6 and octahedral geometry about the metal. There are 2 bidentate ligands and two monodentate ligands.

    1. Complexes 1, 2, and 3 can have geometric isomers. These are different arrangements of the same set of ligands. All three complexes have cis-trans isomers, where a pair of ligands is either opposite or adjacent to each other.
    2. Only complex 2 can have linkage isomers. Nitrite ion, NO 2 , can coordinate through either N or O. It is the only ligand in the three complexes that has this ability.
    3. Only the cis geometric isomer of complex 3 can have optical isomers. These are isomers, with the same arrangement of bonds, that are mirror images of each other and cannot be superimposed.
    4. Only complex 1 can have coordination sphere isomers. This is where an anion can either be a ligand or a counterion. Complex 1 is the only example with a counterion that can also be a ligand.

    23.40   Complex 1 has a coordination number of 6 and octahedral geometry about the metal. There are 5 monodentate ligands of one kind and one of another.

    Complex 2 has a coordination number of 6 and octahedral geometry. There are three monodentate ligands of one kind and three of another.

    Complex 3 has a coordination number of 6 and octahedral geometry about the metal. There are 5 monodentate ligands of one kind and one of another.

    1. Only complex 2 has geometric isomers. It is the only complex with different possible arrangements of the same ligands. There is only one unique way to arrange 5 ligands of one kind and one of another, as in complex 1 and 3.
    2. Only complex 1 can have linkage isomers. Thiocyanate ion, SCN–, can coordinate through either N or S. It is the only ligand in the three complexes that has this ability.
    3. None of the complexes has optical isomers. None of the complexes has bidentate ligands, which are usually present in optically active octahedral complexes.
    4. Only complex 3 can have coordination sphere isomers. This is where an anion can either be a ligand or a counterion. Complex 3 is the only example with anions that are not specifically written as ligands.

    23.41   Yes. A tetrahedral complex of the form MA 2 B 2 would have neither structural isomers nor stereoisomers. For a tetrahedral complex, no differences in connectivity are possible for a single central atom, so the terms cis and trans do not apply. No optical isomers with tetrahedral geometry are possible because M is not bound to four different groups. The complex must be square planar with cis and trans geometric isomers.

    23.42   Two geometric isomers are possible for an octahedral MA 3 B 3 complex (see below). All other arrangements, including mirror images, can be rotated into these two structures. Neither isomer is optically active.

    image

    23.43 Analyze/Plan .  Follow the logic in Sample Exercise 23.4 and 23.5. Solve .

    1. No geometric isomers
      image
    2. Two geometric isomers, cis and trans
      image

    1. Three geometric isomers: cis and trans; the cis isomer has enantiomers
      image

      (The three isomeric complex ions in part (c) each have a 1+ charge.)

    23.44

    1. Two geometric isomers (enantiomers)
      image
    2. Two geometric isomers. In one, the three ammonia ligands are all cis to each other; in the other, one pair of ammonia ligands are trans.
      image
    3. No geometric isomers
      image

    23.45 Analyze/Plan . Consider the geometry and arrangement of ligands in each complex. Decide whether the structure has a nonsuperimposable mirror image. Solve .

    1. Not chiral. The tetrahedral Zn 2+ is not bonded to four different ligands.
    2. Not chiral. One mirror image can be superimposed on the other. [See similar trans structure in Solution 23.43 (c)]
    3. Chiral, has an optical isomer. [See similar cis structures in Solution 23.43 (c)]

    23.46

    1. Not chiral. A square planar complex cannot be chiral.
    2. Not chiral. One mirror image can be superimposed on the other.
    3. Chiral. One mirror image cannot be superimposed on the other.

    Color and Magnetism in Coordination Chemistry; Crystal-Field Theory (Sections 23.5 and 23.6)

    23.47

    1. Visible light with a wavelength of 610 nm is orange. If the complex absorbs orange light, it will appear blue.
    2. E(J/photon) = hν = hc/ λ.

      E = 6.626 × 10 34 J-s 610 nm × 2.998 × 10 8 m s × 1 nm 1 × 10 9 m = 3.257 × 10 19 = 3.26 × 10 19 J

    3. Change J/photon to kJ/mol.

      3.259 × 10 19 J photon × 1 kJ 1000 J × 6.022 × 10 23 photons mol = 196 kJ/mol

    23.48

    1. E(J/photon) = hc/λ. λ = hc/E.

      λ = 6.626 × 10 34 J-s 4.51 × 10 19 J/photon × 2.998 × 10 8 m s × 1 nm 1 × 10 9 m = 440.46 = 440 nm

    2. If the complex absorbs only 441 nm visible light, it absorbs violet and appears yellow.

    23.49 Analyze/Plan .  Given the formula of a coordination complex, determine the oxidation state and electron configuration of the central metal ion. If necessary, use a d-orbital energy level diagram appropriate for the geometry of the complex to decide if the metal ion has unpaired electrons. If so, it is paramagnetic.

    1. Zn 2+ , [Ar]3d 10 . There are no unpaired electrons, so the complex is diamagnetic.
    2. Pd 2+ , [Kr]4d 8 . The complex is probably square planar. Square planar complexes with 8 d-electrons are usually diamagnetic, especially with a heavy metal center like Pd.
    3. V 3+ , [Ar]3d 2 . There are 2 d-electrons, so the complex is paramagnetic. The complex is octahedral, but the two electrons would be unpaired in any of the d-orbital energy level diagrams.
    4. Ni 2+ , [Ar]3d 8 . The complex is paramagnetic. The geometry is octahedral and there are two unpaired electrons.

    23.50

    1. Ag + , [Kr]3d 10 . The complex is diamagnetic; all electrons are paired.
    2. Cu 2+ , [Ar]3d 9 . The complex is paramagnetic. The geometry is square planar, but there would be one unpaired electron in any of the d-orbital energy level diagrams.
    3. Ru 2+ , [Kr]3d 6 . The complex is octahedral and bipy is a strong field ligand. The 6 d-electrons are paired in the 3 lower energy d orbitals. The complex is diamagnetic.
    4. Co 2+ , [Ar]3d 7 . The complex is paramagnetic. The complex is either tetrahedral or square planar, but in either diagram there will be at least one unpaired electron.

    23.51   An electron in a d orbital with lobes that point directly at the ligands will have higher energy than an electron in a d orbital with lobes that do not point directly at the ligands.

    23.52

    1. The six ligands in an octahedral arrangement are oriented along the x, y, and z axes of the metal. The d xy , d xz , and d yz metal orbitals point between the x, y, and z axes, and also between the ligands in this arrangement.
    2. The four ligands in a tetrahedral arrangement are oriented between the x, y, and z axes of the metal. The d orbitals along the axes, d x 2 y 2 and d z 2 , point between the ligands.

    23.53

    1. image
    2. The magnitude of Δ and the energy of the d-d transition for a d 1 complex are equal.
    3. 6.626 × 10 34 J-s 545 nm × 2.998 × 10 8 m s × 1 nm 1 × 10 9 m × 1 kJ 1000 J × 6.022 × 10 23 photons mol = 220 kJ/mol

    23.54

    1. Δ E = hc/ λ = 6.626 × 10 34 J-s × 2 .998 × 10 8 m/s 500 × 10 9 m photon = 3.973 × 10 19 = 3.97 × 10 19 J/photon Δ = 3.973 × 10 19 J/photon × 6 .022 × 10 23 photons 1 mol × 1 kJ 1000 J = 239.25 = 239 kJ/mol
    2. NH 3 is higher than H 2 O in the spectrochemical series, an ordering of ligands according to their ability to increase the energy gap, Δ. If H 2 O is replaced by NH 3 in the complex, the magnitude of Δ would increase because NH 3 is higher in the spectrochemical series and creates a stronger ligand field.

    23.55 Analyze/Plan .  Determine the oxidation state of the copper ions in each mineral from their molecular formulas. Write the appropriate electron configuration(s). Consider the relationship between d-orbital electron configuration, the color of a complex, the wavelength of absorbed light, and the magnitude of the crystal field splitting Δ. Solve .

    1. Both minerals contain Cu 2+ ions. The electron configuration of Cu 2+ is [Ar]3d 9 .
    2. Azurite will probably have the larger ∆. Malachite appears green and absorbs red. Azurite appears blue and absorbs orange. Orange light has a wavelength range of 580 to 650 nm, whereas red light has wavelengths between 650 and 750 nm. The shorter wavelengths of the orange light absorbed by azurite correspond to higher-energy electron transitions and larger ∆ values.

    23.56

    1. Red. The [Ni(bipy) 2 ] 2+ ion absorbs 520 nm green light and appears as the complementary color, red. This fits the absorbed wavelength vs observed color trend shown in Figure 23.30.
    2. The shorter the wavelength of light absorbed, the greater the value of Δ, and the stronger the ligand field. The order of increasing ligand field strength is the order of decreasing wavelength absorbed.

      H 2 O < NH 3 < en < bipy

    23.57 Analyze/Plan .  Determine the charge on the metal ion, subtract it from the row number (3-12) of the transition metal, and the remainder is the number of d-electrons. Solve .

    1. Ti 3+ , d 1
    2. Co 3+ , d 6
    3. Ru 3+ , d 5
    4. Mo 5+ , d 1
    5. Re 3+ , d 4

    23.58

    1. Fe 3+ , d 5
    2. Mn 2+ , d 5
    3. Ag + , d 10
    4. Cr 3+ , d 3
    5. Sr 2+ , d 0

    23.59   Yes. A weak-field ligand leads to a small ∆ value and a small d-orbital splitting energy. If the splitting energy of a complex is smaller than the energy required to pair electrons in an orbital, the complex is high-spin.

    23.60   Octahedral. In an octahedral crystal field, there are more ligand point charges to interact with d orbitals on the metal. And, the point charges point directly at the metal d orbitals so the interaction is greater.

    23.61 Analyze/Plan .  Follow the logic in Sample Exercise 23.7. Solve .

    1. Mn: [Ar]4s 2 3d 5

      Mn 2+ : [Ar]3d 5

      image

      1 unpaired electron

    2. Ru: [Kr]5s 1 4d 7

      Ru 2+ : [Kr]4d 6

      image

      0 unpaired electrons

    3. Rh: [Kr]5s 1 4d 8

      Rh 2+ : [Kr]4d 7

      image

      1 unpaired electron


    23.62

    1. Fe: [Ar]4s 2 3d 6

      Fe 3+ : [Ar]3d 5

      image

      5 unpaired electrons

    2. Mo: [Kr]5s 1 4d 5

      Mo 3+ : [Kr]4d 3

      image

      3 unpaired electrons

    3. Co: [Ar]4s 2 3d 7

      Co 3+ : [Ar]3d 6

      image

      4 unpaired electrons

    23.63 Analyze/Plan .  All complexes in this exercise are six-coordinate octahedral. Use the definitions of high-spin and low-spin along with the orbital diagram from Sample Exercise 23.7 to place electrons for the various complexes. Solve .

    1. image

      d 4 , high spin

    2. image

      d 5 , high spin

    3. image

      d 6 , low spin

    4. image

      d 5 , low spin

    5. image

      d 3

    6. image

      d 8

    23.64

    1. image

      d 4 , high spin

    2. image

      d 5 , high spin

    3. image

      d 6 , low spin

    4. image

      d 5 , low spin

    5. image

      d 3

    6. image

      d 8

    23.65 Analyze/Plan .  Follow the ideas but reverse the logic in Sample Exercise 23.7. Solve .

    image

    high spin


    23.66

    image

    Both complexes contain Fe 3+ , a d 5 ion. CN–, a strong field ligand, produces such a large Δ that the splitting energy is greater than the pairing energy, and the complex is low spin. NCS– produces a smaller Δ, so it is energetically favorable for d-electrons to be unpaired in the higher energy d-orbitals. NCS– is a much weaker-field ligand than CN–. It is probably weaker than NH 3 and near H 2 O in the spectrochemical series.

    Additional Exercises

    23.67   The paper clip must contain a significant amount of Ni, a ferromagnetic metal. At ambient temperature, the paper clip is below its Curie temperature, behaves ferromagnetically, and is strongly attracted to the permanent magnet. The lighter heats the left paperclip above its Curie temperature (354 o C), and it switches from from ferromagnetic to paramagnetic behavior. That is, below its Curie temperature, the spins of the unpaired electrons in Ni are perfectly aligned and the clip is strongly attracted to the permanent magnet. Above the Curie temperature, the unpaired spins become randomly aligned, and the paper clip loses most of its attraction for the permanent magnet.

    23.68   We expect radii in a group to increase moving down the periodic table as principle quantum number increases. However, the nuclear build-up associated with filling of the 4f subshell at the beginning of period 6 counteracts this trend. The increased nuclear charge for transition metals of period 6 means that the valence electrons experience a Z eff large enough to offset the increase in principle quantum number. The increased Z eff causes the radii of the metals in group 6 to be smaller than expected, and period 5 and 6 metals in the same group to have very similar radii. This phenomenon is called the lanthanide contraction.

    23.69   [Pt(NH 3 ) 6 ]Cl 4 ; [Pt(NH 3 ) 4 Cl 2 ]Cl 2 ; [Pt(NH 3 ) 3 Cl 3 ]Cl; [Pt(NH 3 ) 2 Cl 4 ]; K[Pt(NH 3 )Cl 5 ]

    23.70

    image

    23.71

    1. image
    2. image
    3. image
    4. image

    octahedral octahedral

    1. cis -tetraamminediaquacobalt(II) nitrate
    2. sodium aquapentachlororuthenate(III)
    3. ammonium trans -diaquabisoxalatocobaltate(III)
    4. cis -dichlorobisethylenediammineruthenium(II)

    23.72   Only the complex in 23.69(d) has optical isomers. The chelating ethylenediamine ligands in (d) prevent its mirror images (enantiomers) from being superimposable.

    23.73

    1. Valence electrons: 2P + 6C + 16H = 10 + 24 + 16 = 50 e , 25 e pr
      image

      Both dmpe and en are bidentate ligands. The dmpe ligand binds through P, whereas en binds through N. Phosphorus is less electronegative than N, so dmpe is a stronger electron pair donor and Lewis base than en. Dmpe creates a stronger ligand field and is higher on the spectrochemical series.

      Structurally, P has a larger covalent radius than N, so M–P bonds are longer than M–N bonds. This is convenient because the two –CH 3 groups on each P atom in dmpe create more steric hindrance (bumping with adjacent atoms) than the H atoms on N in en.

    2. CO and dmpe are neutral, 2 CN– = 2–, 2 Na + = 2+. The ion charges balance, so the oxidation state of Mo is zero.

    1. The symbol image represents the bidentate dmpe ligand.
      image

    23.74   The trans isomer is not observed. In a square planar complex such as [Pt(en)Cl 2 ], if one pair of ligands is trans, the remaining two coordination sites are also trans to each other. Ethylenediamine is a relatively short bidentate ligand that cannot occupy trans coordination sites, so the trans isomer is unknown.

    23.75   We will represent the end of the bidentate ligand containing the CF 3 group by a shaded oval, the other end by an open ova702l:

    image

    23.76

    1. Iron. Hemoglobin is the iron-containing protein that transports O 2 in human blood.
    2. Magnesium. Chlorophylls are magnesium-containing porphyrins in plants. They are the key components in the conversion of solar energy into chemical energy that can be used by living organisms.
    3. Iron. Siderophores are iron-binding compounds or ligands produced by a microorganism. They compete on a molecular level for iron in the medium outside the organism and carry needed iron into the cells of the organism.
    4. Copper. Hemocyanine is a copper-containing protein responsible for oxygen transport in the blue blood of certain marine animals.

    23.77

    1. Zero. The CO ligand is a neutral molecule and the charge on the complex is zero, so nickel must be present as Ni(0).
    2. The electron configuration of a Ni atom in the absence of a ligand field is [Ar]4s 2 3d 8 . A tetrahedral complex with 8 d-electrons would have two unpaired electrons and be paramagnetic. Because the compound is diamagnetic, the Ni atom must have 10 d-electrons. The electron configuration of Ni in the complex is [Ar]3d 10 .
    3. tetracarbonylnickel(0)

    23.78

    1. pentacarbonyliron(0)
    2. Because CO is a neutral molecule, the oxidation state of iron must be zero.
    3. [Fe(CO) 4 CN]– has two geometric isomers. In a trigonal bipyramid, the axial and equatorial positions are not equivalent and not superimposable. One isomer has CN in an axial position and the other has it in an equatorial position.
      image

    23.79

    1. left shoe
    2. wood screw
    3. a typical golf club

    23.80

    1. image

      d 2

    2. These complexes are colored because the crystal-field splitting energy, Δ, is in the visible portion of the electromagnetic spectrum. Visible light with λ = hc/Δ is absorbed, promoting one of the d-electrons into a higher energy d-orbital. The remaining wavelengths of visible light are reflected or transmitted; the combination of these wavelengths is the color we see.
    3. [V(H 2 O) 6 ] 3+ will absorb light with higher energy. H 2 O is in the middle of the spectrochemical series, and causes a larger Δ than F–, a weak-field ligand. Because Δ and λ are inversely related, larger Δ corresponds to higher energy and shorter λ.

    23.81

    1. Formally, the two Ru centers have different oxidation states; one is +2 and the other is +3.
    2. image
    3. There is extensive bonding-electron delocalization in the isolated pyrazine molecule. When pyrazine acts as a bridging ligand, its delocalized molecular orbitals provide a pathway for delocalization of the “odd” d-electron in the Creutz-Taube ion. The two metal ions appear equivalent because the odd d-electron is delocalized across the pyrazine bridge.

    23.82   According to the spectrochemical series, the order of increasing Δ for the ligands is Cl < H 2 O < NH 3 . (The tetrahedral Cl complex will have an even smaller Δ than an octahedral one.) The smaller the value of Δ, the longer the wavelength of visible light absorbed. The color of light absorbed is the complement of the observed color. A blue complex absorbs orange light (580 to 650 nm), a pink complex absorbs green light (490 to 560 nm), and a yellow complex absorbs violet light (400 to 430 nm). Because [CoCl 4 ] 2– absorbs the longest wavelength, it appears blue. [Co(H 2 O) 6 ] 2+ absorbs green and appears pink, and [Co(NH 3 ) 6 ] 3+ absorbs violet and appears yellow.

    23.83

    1. image
    2. In deoxyhemoglobin, H 2 O is bound to Fe in place of O 2 .
    3. The two forms of hemoglobin have different colors because they absorb different wavelengths of visible light. They differ by just the H 2 O or O 2 ligand, which means that the two ligands have slightly different ligand fields. Oxyhemoglobin appears red and absorbs green light, whereas deoxyhemoglobin appears bluish and absorbs longer wavelength yellow-green light. O 2 has a stronger ligand field than H 2 O.

    1. According to Table 18.1, air is 20.948 mole percent O 2 . This translates to 209,480 ppm O 2 . This is approximately 500 times the 400 ppm concentration of CO in the experiment. If air with a CO concentration 1/500th that of O 2 converts 1/10 of the oxyhemoglobin to carboxyhemoglobin, the equilibrium constant for binding CO is much larger than that for binding O 2 .
    1. If CO is a stronger field ligand than O 2 , carboxyhemoglobin will absorb shorter wavelengths than hemoglobin. It will absorb blue-green light and appear orange-red.

    23.84

    1. The term isoelectronic means that the three ions have the same number of electrons.
    2. In each ion, the metal is in its maximum oxidation state and has a d 0 electron configuration. That is, the metal ions have no d-electrons, so there should be no d-d transitions.
    3. A ligand-metal charge transfer transition occurs when an electron in a filled ligand orbital is excited to an empty d-orbital of the metal.
    4. Absorption of 565 nm yellow light by MnO 4– causes the compound to appear violet, the complementary color. CrO 4 2– appears yellow, so it is absorbing violet light of approximately 420 nm. The wavelength of the LMCT transition for chromate, 420 nm, is shorter than the wavelength of LCMT transition in permanganate, 565 nm. This means that there is a larger energy difference between filled ligand and empty metal orbitals in chromate than in permanganate.
    5. UV. A white compound indicates that no visible light is absorbed. Going left on the periodic chart from Mn to Cr, the absorbed wavelength got shorter and the energy difference between ligand and metal orbitals increased. The 420 nm absorption by CrO 4– is at the short wavelength edge of the visible spectrum. It is not surprising that the ion containing V, further left on the chart, absorbs at a still shorter wavelength in the ultraviolet region and that VO 4 3– appears white.

    23.85   The higher the oxidation state of the metal, the smaller the energy separation between the ligand orbitals and the empty d-orbitals on the metal. The oxidation states of the metals in the tetrahedral oxoanions are: Mn, +7; Cr, +6; V, +5. From Solution 23.84, the energy separation between the ligand orbitals and the empty d-orbitals on the metals increases in the order Mn < Cr < V.

    23.86   Application of pressure would result in shorter metal ion–oxide distances. This would have the effect of increasing the ligand-electron repulsions, and would result in a larger splitting in the d-orbital energies. Thus, application of pressure should result in a shift in the absorption to a higher energy and shorter wavelength.

    image

    23.87

    1. image
    2. sodium dicarbonyltetracyanoferrate(II)
    3. +2, 6 d-electrons
    4. We expect the complex to be low spin. Cyanide (and carbonyl) are high on the spectrochemical series, which means the complex will have a large Δ splitting characteristic of low spin complexes.

    23.88

    image

    23.89   A large part of the metal-ligand interaction is electrostatic attraction between the positively charged metal and the fully or partially negatively charged ligand. For the same ligand, the greater the charge on the metal or the shorter the M–L separation, the stronger the interaction and the more stable the complex. The greater positive charge and smaller ionic radius of a metal in the 3+ oxidation state means that, for the same ligand, complexes with metals in the 3+ state are more stable than those with metals in the 2+ state.

    23.90

    1. Only one
    2. Two
      image
    3. Four; two are geometric, the other two are stereoisomers of each of these.
      image

    23.91

    1. Zero. The CO ligand is a neutral molecule and the charge on the complex is zero, so chromium must be present as Cr(0).
    2. The electron configuration of a Cr atom in the absence of a ligand field is [Ar]4s 2 3d 4 . An octahedral complex with 4 d-electrons would have unpaired electrons and be paramagnetic. In order for the complex to be diamagnetic, all valence electrons must be paired. The electron configuration of Cr in the complex is [Ar]3d 6 . In a strong field complex, the 6 d-electrons will be paired in the three lower energy d orbitals.

    1. A colorless complex has no d-d electron transitions. This indicates a large Δ for the complex, which means CO is a strong-field ligand.
    1. hexacarbonylchromium(0)

    Integrative Exercises

    23.92

    1. image
    2. The pK a of pure water is 14, that of carbonic anhydrase is 7.5. The active site of carbonic anhydrase is much more acidic than the bulk water. In carbonic anhydrase, the Zn 2+ ion withdraws electron density from the O atom of water. The electronegative oxygen atom compensates by withdrawing electron-density from the O H bond. The O H bond is polarized and H becomes more ionizable, more acidic than in the bulk solvent. This is similar to the effect of an electronegative central atom in an oxyacid such as H 2 SO 4 .
    3. When the water molecule is deprotonated, the ligand coordinated to water becomes hydroxide ion, OH . The three N atoms are unaffected.
    4. In [Zn(H 2 O) 6 ] 2+ , the Zn 2+ ion has six bound O atoms from which to withdraw electron density. Each O atom donates less electron density than the single O atom in carbonic anhydrase and each O atom withdraws less electron density from its O H bonds. The O H bonds in [Zn(H 2 O) 6 ] 2+ are less polarized and less acidic than those in carbonic anhydrase. [Zn(H 2 O) 6 ] 2+ is a weaker acid and has a higher pK a than carbonic anhydrase.
    5. No, we do not expect carbonic anhydrase to have a deep color like hemoglobin. The Zn 2+ ion is a d 10 metal center. Its d-orbitals are completely occupied and there is no possibility for the d-d transitions that lead to colored complexes.

    23.93

    1. Both compounds have the same general formulation, so Co is in the same (+3) oxidation state in both complexes.
    2. Cobalt(III) complexes are generally inert; that is, they do not rapidly exchange ligands inside the coordination sphere. Therefore, the ions that form precipitates in these two cases are probably outside the coordination sphere. The dark violet compound A forms a precipitate with BaCl 2 (aq) but not AgNO 3 (aq), so it has SO 4 2– outside the coordination sphere and coordinated Br , [Co(NH 3 ) 5 Br]SO 4 . The red-violet compound B forms a precipitate with AgNO 3 (aq) but not BaCl 2 (aq) so it has Br outside the coordination sphere and coordinated SO 4 2– , [Co(NH 3 ) 5 SO 4 ]Br.

    image
    1. Compounds A and B have the same formula but different properties (color, chemical reactivity), so they are isomers. They vary by which ion is inside the coordination sphere, so they are coordination sphere isomers .
    1. Compound A is an ionic sulfate and compound B is an ionic bromide, so both are strong electrolytes. According to the solubility guidelines in Table 4.1, both should be water-soluble.

    23.94   Determine the empirical formula of the complex, assuming a 100 g sample.

    10.0 g Mn × 1 mol Mn 54 .94 g Mn = 0.1820 mol Mn; 0 .182/0 .182 = 1

    28.6 g K × 1 mol K 39 .10 g K = 0.7315 mol K; 0.732 / 0 .182 = 4

    8.8 g C × 1 mol C 12 .0 g C = 0.7327 mol C; 0.733 /0 .182 = 4

    29.2 g Br × 1 mol Br 79 .904 g Br = 0.3654 mol Br; 0.365 /0 .182 = 2

    23.4 g O × 1 mol O 16.00 g O = 1.463 mol O; 1.46 /0 .182 = 8

    There are 2 C and 4 O per oxalate ion, for a total of two oxalate ligands in the complex. To match the conductivity of K 4 [Fe(CN) 6 ], the oxalate and bromide ions must be in the coordination sphere of the complex anion. Thus, the compound is K 4 [Mn(ox) 2 Br 2 ].

    23.95

    1. ΔG° = –nFE°. The positive E° values for both sets of complexes correspond to –ΔG° values. Negative values of ΔG° mean that both processes are spontaneous. For both o -phen and CN ligands, the Fe(II) complex is more thermodynamically favorable than the Fe(III) complex.
    2. The CN complex, with the smaller positive E° value, is more difficult to reduce.
    3. That both the Fe(II) complexes are low spin means that both CN and o -phen are strong-field ligands. The negatively charged CN has a stronger electrostatic interaction with Fe 3+ than the neutral o -phen has. This stabilizes the Fe(III) complex of CN relative to the Fe(III) complex of o -phen, which reduces the driving force for reduction of [Fe(CN) 6 ] 3– relative to reduction of [Fe( o -phen) 3 ] 3+ . The E° value and magnitude of ΔG° for the reduction of the CN complex are thus smaller than those values for the o -phen complex.

    23.96   First determine the empirical formula, assuming that the remaining mass of complex is Pd.

    37.6 g Br × 1 mol Br 79 .904 g Br = 0.4706 mol Br; 0.4706 /0 .2361 = 2

    28.3 g C × 1 mol C 12 .01 g C = 2.356 mol C; 2.356 /0 .2361 = 10

    6.60 g N × 1 mol N 14 .01 g N = 0.4711 mol N; 0.4711 /0 .2361 = 2

    2.37 g H × 1 mol H 1 .008 g H = 2.351 mol H; 2.351 / 0.2361 = 10

    25.13 g Pd × 1 mol Pd 106 .42 g Pd = 0.2361 mol Pd; 0.2361 /0 .2361 = 1

    The chemical formula is [Pd(NC 5 H 5 ) 2 Br 2 ]. This should be a neutral square-planar complex of Pd(II), a nonelectrolyte. Because the dipole moment is zero, we can infer that it must be the trans isomer.

    image

    23.97

    1. The reaction that occurs increases the conductivity of the solution by producing a greater number of charged particles, particles with higher charges, or both. It is likely that H 2 O from the bulk solvent exchanges with a coordinated Br according to the reaction below. This reaction would convert the 1:1 electrolyte, [Co(NH 3 ) 4 Br 2 ]Br, to a 1:2 electrolyte, [Co(NH 3 ) 3 (H 2 O)Br]Br 2 .
    2. [Co(NH 3 ) 4 Br 2 ] + (aq) + H 2 O(l) → [Co(NH 3 ) 4 (H 2 O)Br] 2+ (aq) + Br (aq)
    3. Before the exchange reaction, there is one mole of free Br per mole of complex. mol Br = mol Ag +

      M = mol/L; L AgNO 3 = mol AgNO 3 / M AgNO 3

      3.87 g complex 0.500 L soln × 1 mol complex 366 .77 g complex × 0.02500 L soln used = 5.276 × 10 4 = 5.28 × 10 4 mol complex

      5.276 × 10 4 mol complex × 1 mol Br 1 mol complex × 1 mol Ag + 1 mol Br × 1 L Ag + (aq) 0.0100 mol Ag + (aq) = 0.05276 L = 52.8 mL AgNO 3 (aq)


    1. After the exchange reaction, there are 2 mol free Br per mol of complex. Because M AgNO 3 (aq) and volume of complex solution are the same for the second experiment, the titration after conductivity changes will require twice the volume calculated in part (c), 105.52 = 106 mL of 0.0100 M AgNO 3 (aq).

    23.98   Calculate the concentration of Mg 2+ alone, and then the concentration of Ca 2+ by difference. M × L = mol

    0.0104 mol EDTA 1 L × 0.0187 L × 1 mol Mg 2 + 1 mol EDTA × 24.31 g Mg 2 + 1 mol Mg 2 + × 1000 mg g × 1 0.100 L H 2 O = 47.28 = 47.3 mg Mg 2 + / L

    0.0104 M EDTA × 0 .0315 L = mol (Ca 2 + + Mg 2 + ) 0.0104 M EDTA × 0 .0187 L = mol Mg 2 + 0 .0104 M EDTA × 0 .0128 L = mol Ca 2 +

    0.0104 M EDTA × 0 .0128 L × 1 mol Ca 2 + 1 mol EDTA × 40.08 g Ca 2 + 1 mol Ca 2 + × 1000 mg g × 1 0.100 L H 2 O = 53.35 = 53.4 mg Ca 2 + / L

    23.99   Use Hess’ law to calculate Δ G o for the desired equilibrium. Then Δ G o = –RT ln K to calculate K.

    Hb + CO HbCO HbO 2 Hb + O 2 HbO 2 + Hb + CO HbCO + Hb + O 2 HbO 2 + CO HbCO + O 2 Δ G o = 80 kJ Δ G o = 70 kJ Δ G o = 10 kJ

    Δ G ° = RT ln K, ln K = Δ G o RT = ( 10 kJ) 8.314 J/K-mol × 298 K × 1000 J kJ = 4.036 = 4.04

    K = e 4.04 = 56.61 = 57

    23.100 182 × 10 3 J 1 mol × 1 mol 6 .022 × 10 23 molecules = 3.022 × 10 19 = 3.02 × 10 19 J/photon

    ΔE = hν = 3.02 × 10– 19 J; ν = ΔE/h

    ν = 3.022 × 10– 19 J/6.626 × 10– 34 J - s = 4.561 × 10 14 = 4.56 × 10 14 s– 1

    λ = 2.998 × 10 8 m/s 4.561 × 10 14 s 1 = 6.57 × 10 7 m = 657 nm

    We expect that this complex will absorb in the visible, at around 660 nm. It will thus exhibit a blue-green color (Figure 23.25).


    23.101   The process can be written:

    H 2 (g) + 2 e 2 H + (aq) Cu(s) Cu 2 + + 2 e Cu 2 + (aq) + 4 NH 3 (aq) [Cu(NH 3 ) 4 ] 2 + (aq) H 2 (g) + Cu(s) + 4 NH 3 (aq) 2 H + (aq) + [Cu(NH 3 ) 4 ] 2 + (aq) E red o = 0.0 V E red o = 0.337 V "E f o " = ? E = 0.08 V

    E = E o RT ln K; K = [ H + ] 2 [ Cu(NH 3 ) 4 2 + ] P H 2 [ NH 3 ] 4

    P H 2 = 1 atm, [H + ] = 1 M , [NH 3 ] = 1 M , [Cu(NH 3 ) 4 ] 2+ = 1 M , Q = 1

    E = E o RT ln ( 1 ) ; E = E o RT ( 0 ) ; E = E o = 0.08 V

    Because we know E o values for two steps and the overall reaction, we can calculate “E°” for the formation reaction and then K f , using E o = 0.0592 n log K f for the step.

    E cell = 0.08 V = 0.0 V 0.337 V + "E f o " "E f o " = 0.08 V + 0 .337 V = 0.417 V = 0.42 V

    "E f o " = 0.0592 n log K f ; log K f = n(E f o ) 0.0592 = 2 ( 0.417 ) 0.0592 = 14.0878 = 14

    K f = 10 14.0878 = 1.2 × 10 14 = 10 14


     

    24 The Chemistry of Life: Organic and Biological Chemistry

    Visualizing Concepts

    24.1 Analyze/Plan. Follow the logic in Sample Exercise 24.1 to name each compound. Decide which structures are the same compound. Solve .

    1. 2,2,4-trimethylpentane
    2. 3-ethyl-2-methylpentane
    3. 2,3,4-trimethylpentane
    4. 2,3,4-trimethylpentane

    Structures (c) and (d) are the same molecule.

    24.2 Analyze/Plan. Given structural formulas, specify which molecules are unsaturated . Consider the definition of unsaturated and apply it to the molecules in the exercise. Solve .

    Unsaturated molecules contain one or more multiple bonds. Saturated molecules contain only single bonds. Molecules (c) and (d) are unsaturated.

    24.3 Analyze/Plan. Given structural formulas, decide which molecule will undergo addition. Consider which functional groups are present in the molecules, and which are most susceptible to addition. Solve .

    1. Molecule (iii), an alkene, will readily undergo addition. Addition reactions are characteristic of alkenes. [Molecule (i) will not typically undergo addition, because its delocalized electron cloud is too difficult to disrupt. Molecules (b) and (d) contain carbonyl groups (actually carboxylic acid groups) that do not typically undergo addition, except under special conditions.]
    2. Molecule (i) is an aromatic hydrocarbon.
    3. Molecule (i) most readily undergoes a substitution reaction.

    24.4 Analyze/Plan. Given condensed structural formulas, predict which molecule will have the highest boiling point. Boiling point is determined by strength of intermolecular forces; for neutral molecules with similar molar masses, the strongest intermolecular force is hydrogen bonding.

    We are also asked to identify various functional groups, given condensed structural formulas. Refer to Table 24.6 for information on functional groups. Solve .

    1. Molecule (ii), an alcohol, forms hydrogen bonds with like molecules; it has the highest boiling point.
    2. Molecule (iv) is most oxidized; it has the most oxygen atoms.

    1. None of the compounds is an ether. Ethers are characterized by an oxygen atom bound to two carbon atoms. Compound (iv) is an ester, which has an ether linkage.
    1. Compound (iv) is an ester.
    1. None of the compounds is a ketone. Ketones are characterized by a carbonyl group bonded to two alkyl groups. Compound (i) is an aldehyde; the carbonyl group is bonded to one alkyl group and one hydrogen atom.

    24.5 Analyze. Given structural formulas, name the compound, identify the functional groups present, and determine if the molecule is chiral. Plan. Apply principles of organic nomenclature in Section 24.2. Identify functional groups using Table 24.6. Assess chirality by looking for one or more carbon atoms that are bonded to four different groups. Solve .

    1. Valine. This molecule is an amino acid shown in the zwitterion form. Use Figure 24.16 to name the amino acid. In the neutral amino acid, there is a carboxlyic acid group and an amine group. (In the zwitterion, these become a carboxyl group and an ammonium group.) The amino acid is chiral. The C atom to which the NH 3 + group is bound is a chiral center, because it is bound to four different groups. All amino acids except glycine are chiral.
    2. 3-chlorobenzoic acid. The functional groups are a carboxylic acid group and an (aryl) halide. The molecule is not chiral; no C atom is bound to four different groups.
    3. 2-pentene. The molecule contains an alkene functional group and is not chiral.
    4. Propane. The molecule is an alkane, there are no functional groups, and it is not chiral.

    24.6 Analyze/Plan. Given ball-and-stick models, select the molecule that fits the description given. From the models, decide the type of molecule or functional group represented.

    Solve. Molecule (i) is a sugar, (ii) is an ester with a long hydrocarbon chain, (iii) is an organic base and a component of nucleic acids, (iv) is an amino acid, and (v) is an alcohol.

    1. Molecule (i) is a disaccharide composed of galactose (left) and glucose (right); it can be hydrolyzed to form a solution containing glucose. Because it is the only sugar molecule depicted, it was not necessary to know the exact structure of glucose to answer the question.
    2. Amino acids form zwitterions, so the choice is molecule (iv).
    3. Molecule (iii) is an organic base present in DNA (again, the only possible choice).
    4. Molecule (v) because alcohols react with carboxylic acids to form esters.
    5. Molecule (ii), because it has a long hydrocarbon chain and an ester functional group.

    Introduction to Organic Compounds; Hydrocarbons (Sections 24.1 and 24.2)

    24.7

    1. False. Butane is an alkane; it contains carbon atoms that are sp 3 hybridized.
    2. False. Cyclohexane is a saturated hydrocarbon, whereas benzene is aromatic.
    3. True.
    4. False. Olefin is another name for alk ene .

    24.8

    1. False. Hexane contains six carbon atoms and pentane contains only five.
    2. True. For molecules with similar structures, the larger the molecule, the stronger their dispersion forces and the higher the boiling point.
    3. True. The carbon atoms involved in an alkyne group are sp hybridized.
    4. False. There is only one way to arrange the three carbon atoms in propane.

    24.9 Analyze/Plan. Given a condensed structural formula, determine the bond angles and hybridization about each carbon atom in the molecule. Visualize the number of electron domains about each carbon. State the bond angle and hybridization based on electron domain geometry. Solve .

    image

    C1 has trigonal planar electron domain geometry, 120° bond angles, and sp 2 hybridization. C3 and C4 have linear electron domain geometry, 180° bond angles, and sp hybridization. C2 and C5 both have tetra-hedral electron domain geometry, 109° bond angles, and sp 3 hybridization.

    24.10

    image
    1. C2, C5, and C6 have sp 3 hybridization (4 e domains around C)
    2. C7 has sp hybridization (2 e domains around C)
    3. C1, C3, and C4 have sp 2 hybridization (3 e domains around C)

    24.11 Analyze/Plan. For each molecule, count the number of carbon atoms in the root name and in the substituent group(s). Solve .

    1. One. The root “meth” indicates one carbon atom. Methane is the simplest alkane.
    2. Ten. The root “dec” indicates 10 carbon atoms.
    3. Seven. The root “hex” indicates 6 carbon atoms, plus 1 in the “methyl” substituent.
    4. Five. Although this is a common name (not an IUPAC name), the root “pent” still indicates 5 total carbon atoms.
    5. Two. Again a common name, acetylene is the simplest two-carbon alkyne. The IUPAC name is ethyne.

    24.12   True. Note that we are comparing enthalpies for only single bonds in the molecule.

    24.13

    1. True.
    2. True.
    3. False. Alkenes contain carbon-carbon double bonds.
    4. False. Alkynes contain carbon-carbon triple bonds.
    5. True.
    6. False. Cyclohexane is a saturated hydrocarbon; benzene is aromatic.
    7. True.

    24.14   All the classifications listed are hydrocarbons; they contain only the elements hydrogen and carbon.

    1. Alkanes are hydrocarbons that contain only single bonds.
    2. Cycloalkanes contain at least one ring of three or more carbon atoms joined by single bonds. Because it is a type of alkane, all bonds in a cycloalkane are single bonds.
    3. Alkenes contain at least one C=C double bond.
    4. Alkynes contain at least one C≡C triple bond.
    5. A saturated hydrocarbon contains only single bonds. Alkanes and cycloalkanes fit this definition.
    6. An aromatic hydrocarbon contains one or more planar, six-membered rings of carbon atoms with delocalized π-bonding throughout the ring.

    24.15 Analyze/Plan. Follow the rules for naming alkanes given in Section 24.2 and illustrated in Sample Exercise 24.1. Solve .

    1. 2-methylhexane
    2. 4-ethyl-2,4-dimethyldecane
    3. image
    4. image
    5. image

    24.16

    1. 3,3,5-trimethylheptane
    2. 3,4,4-trimethylheptane
    3. image
    4. image
    5. image

    24.17 Analyze/Plan. Follow the rules for naming alkanes given in Section 24.2 and illustrated in Sample Exercise 24.1. Solve .

    1. 2,3-dimethylheptane
    2. image
    3. image
    4. 2,2,5-trimethylhexane
    5. 3-ethylheptane

    24.18

    1. image

    1. image
    1. image
    1. 2,4-dimethylhexane
    1. methylcyclobutane

    24.19  Assuming that each component retains its effective octane number in the mixture (and this isn’t always the case), we obtain: octane number = 0.35(0) + 0.65(100) = 65.

    24.20  Octane number can be increased by increasing the fraction of branched-chain alkanes or aromatics, because these have high octane numbers. This can be done by cracking. The octane number also can be increased by adding an anti-knock agent such as tetraethyl lead, Pb(C 2 H 5 ) 4 (no longer legal); methyl t-butyl ether (MTBE); or an alcohol, methanol, or ethanol.

    Alkenes, Alkynes, and Aromatic Hydrocarbons (Section 24.3)

    24.21

    1. C 4 H 6 is an unsaturated hydrocarbon. The maximum number of hydrogen atoms for 4 C atoms in a saturated alkane is [(2 × 4) + 2] = 10. C 4 H 6 does not contain the maximum possible hydrogen atoms and is unsaturated.
    2. Yes, all alkynes are unsaturated. The presence of a triple bond means that the alkyne carbon atoms are not bound to the maximum possible number of hydrogen atoms.

    24.22

    1. The molecule CH 3 CH=CH 2 is unsaturated because it contains a double bond. It is possible to add more hydrogen to the molecule.
    2. The formula CH 3 CH 2 CH=CH 3 has too many H atoms bound to the right-most C atom; the formula as it stands implies 5 bonds to this atom. A correct formula is CH 3 CH 2 CH=CH 2 , with 2 H atoms on the right-most C atom.

    24.23 Analyze/Plan. Consider the definition of the stated classification and apply it to a compound containing five C atoms. Solve .

    1. CH 3 CH 2 CH 2 CH 2 CH 3 , C 5 H 12
    2. image
    3. CH 2 =CHCH 2 CH 2 CH 3 , C 5 H 10
    4. HC≡CCH 2 CH 2 CH 3 , C 5 H 8

    24.24   cyclic alkane, image

    cyclic alkene, image

    alkyne, CH 3 –CH 2 –C≡C–CH 2 –CH 3 , C 6 H 10

    aromatic hydrocarbon, image

    24.25 Analyze/Plan. We are given the class of compounds “enediyne.” Based on organic nomenclature, determine the structural features of an enediyne. Construct a molecule with 6 C atoms in a row that has these features. Solve .

    The term “enediyne” contains the suffixes –ene and –yne. The suffix –ene is used to name alkenes, molecules with one double bond. The suffix –yne is used to name alkynes, molecules with one triple bond. An enediyne then features one double and two triple bonds. Possible arrangements of these bonds involving 6 C atoms in a row are:

    CH 2 =CH–C≡C–C≡CH  CH≡C≡CH=CH–C≡CH

    Check. The formula of a saturated alkane is C n H 2n+2 . For each double bond subtract 2 H atoms, for each triple bond subtract 4 H atoms. A saturated 6 C alkane has 14 H atoms. For the enediyne, subtract (2 + 4 + 4 =) 10 H atoms. The molecular formula is C 6 H 4 . That is the molecular formula of each structure above.

    24.26   C n H 2n–2

    24.27 Analyze/Plan. Follow the logic in Sample Exercise 24.3.

    Solve. There are many correct structures that are alkenes or alkynes and have the molecular formula C 6 H 10 . The molecule will have two points of unsaturation. Molecules can include various arrangements of one alkyne group, two alkene groups, or one cyclic mono-alkene. A few of the possibilities are shown below.

    image

    24.28

    image

    24.29 Analyze/Plan. Follow the logic in Sample Exercises 24.1 and 24.4. Solve .

    1. image
    2. image
    3. cis -6-methyl-3-octene
    4. para -dibromobenzene (or 1,4-dibromobenzene)
    5. 4,4-dimethyl-1-hexyne

    24.30

    1. image
    2. image
    3. image
    4. 1-butyne
    5. trans -2-heptene

    24.31

    1. True
    2. True. (Alkenes can have cis and trans isomers, but whether they do have them depends on the groups bonded to the sp 2 C atoms of the alkene.)
    3. False. The geometry of the alkyne functional group is linear.

    24.32   Butene is an alkene, C 4 H 8 . There are two possible placements for the double bond:

    CH 2 = CHCH 2 CH 3 or  CH 3 CH = = CHCH 3 1 - butene                          2 - butene

    These two compounds are structural isomers .


    For 2-butene, there are two different, noninterchangeable ways to construct the carbon skeleton (owing to the absence of free rotation around the double bond). These two compounds are geometric isomers .

    image

    24.33 Analyze/Plan. In order for geometrical isomerism to be possible, the molecule must be an alkene with two different groups bound to each of the alkene C atoms. Solve .

    1. image
    2. image
    3. no, not an alkene
    4. no, not an alkene

    24.34

    image

    24.35

    1. True
    2. Plan. Draw the condensed structural formula of 2-pentene. Consider the part of the molecule that is likely to react with Br 2 . Solve .

      CH 3 CH 2 CH = CH CH 3 + Br 2 CH 3 CH 2 CH(Br)CH(Br)CH 3 2-pentene 2,3-dibromopentane

      This is an addition reaction. The π bond is broken and a Br atom adds to each of the C atoms involved in the π bond.

    3. Plan. Draw the structure of benzene. The term para means the Cl atoms will be opposite each other across the benzene ring in the product. Solve .
      image

      This is a substitution reaction. None of the double bonds in the benzene ring are broken. Each Cl atom has replaced an H atom on the the ring. The Cl atoms have substituted for two of the H atoms that are opposite each other across the ring.


    24.36

    1. image
    2. image
    3. image

    24.37

    1. Plan. Consider the structures of cyclopropane, cyclopentane, and cyclohexane. Solve .

      The small 60° C–C–C angles in the cyclopropane ring cause strain that provides a driving force for reactions that result in ring opening. There is no comparable strain in the five- or six-membered rings.

    2. Plan. First form an alkyl halide: C 2 H 4 (g) + HBr(g) → CH 3 CH 2 Br(l); then carry out a Friedel-Crafts reaction. Solve .
      image

    24.38

    1. The reaction of Br 2 with an alkene to form a colorless halogenated alkane is an addition reaction. Aromatic hydrocarbons do not readily undergo addition reactions, because their π-electrons are stabilized by delocalization.
    2. Plan. Use a Friedel-Crafts reaction to substitute a CH 2 CH 3 onto benzene. Do a second substitution reaction to get para-bromoethylbenzene. Solve .
      image

      It appears that ortho, meta, and para geometric isomers of bromoethylbenzene would be possible. However, because of electronic effects beyond the scope of this chapter, the ethyl group favors formation of ortho and para isomers, but not the meta. The ortho and para products must be separated by distillation or some other technique.

    24.39   Yes, this information suggests (but does not prove) that the reactions proceed in the same manner. That the rate laws are both first order in both reactants and second order overall indicates that the activated complex in the rate-determining step in each mechanism is bimolecular and contains one molecule of each reactant. This is usually an indication that the mechanisms are the same, but it does not rule out the possibility of different fast steps, or a different order of elementary steps.


    24.40   The partially positive end of the hydrogen halide, H δ + X, δ is attached to the π electron cloud of the alkene cyclohexene. The electrons that formed the π bond in cyclohexene form a sigma bond to the H atom of HX, leaving a halide ion, X . The intermediate is a carbocation; one of the C atoms formerly involved in the π bond is now bound to a second H atom. The other C atom formerly involved in the π bond carries a full positive charge and forms only three sigma bonds, two to adjacent C atoms and one to H.

    image

    24.41 Analyze/Plan. Both combustion reactions produce CO 2 and H 2 O:

    C 3 H 6 (g) + 9/2 O 2 (g) → 3 CO 2 (g) + 3 H 2 O(l)

    C 5 H 10 (g) + 15/2 O 2 (g) → 5 CO 2 (g) + 5 H 2 O(l)

    Thus, we can calculate the ΔH comb / CH 2 group for each compound. Solve .

    Δ H comb CH 2 group = 2089 kJ/mol C 3 H 6 3 CH 2 groups = 696.3 kJ mol CH 2 ; 3317 kJ/mol C 5 H 10 5 CH 2 groups = 663.4 kJ/mol CH 2

    ΔH comb /CH 2 group for cyclopropane is greater because C 3 H 6 contains a strained ring. When combustion occurs, the strain is relieved and the stored energy is released during the reaction.

    24.42

    Δ H _ C 10 H 8 (l) + 12 O 2 (g) 10 CO 2 (g) + 4 H 2 O(l) 5157 kJ [ C 10 H 18 (l) + 29/2 O 2 (g) 10 CO 2 (g) + 9 H 2 O(l) ( 6286 ) kJ C 10 H 8 (l) + 5 H 2 O(l) C 10 H 18 (l) + 5/2 O 2 (g) + 1129 kJ 5 / 2 O 2 (g) + 5 H 2 (g) 5 H 2 O(l) 5 ( 285.8 ) kJ C 10 H 8 (l) + 5 H 2 (g) C 10 H 18 (l) 300 kJ

    Compare this with the heat of hydrogenation of ethylene:

    C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g); ΔH = –84.7 – (52.3) = –137 kJ. This value applies to just one double bond. For five double bonds, we would expect about –685 kJ. The fact that hydrogenation of napthalene yields only –300 kJ indicates that the overall energy of the napthalene molecule is lower than expected for five isolated double bonds. The resonance energy is then [–685 kJ – (–300 kJ)] or –385 kJ. Resonance has stabilized or lowered the overall energy of naphthalene by 385 kJ.

    Functional Groups and Chirality (Sections 24.4 and 24.5)

    24.43 Analyze/Plan. Match the structural features of various functional groups shown in Table 24.6 to the molecular structures in this exercise. Solve .

    1. (iii)
    2. (i)

    1. (ii) Amines are organic bases; they are H + acceptors because of the the lone pair of electrons on the N atom.
    1. (iv)
    1. (v)

    24.44

    1. image ester
    2. –Cl, halocarbon; –OH, alcohol (aromatic alcohols are phenols)
    3. image , amide
    4. alkane
    5. C=C , alkene; image , aldehyde
    6. image , ketone

    24.45 Analyze/Plan. Given the name of a compound, write its molecular formula. Identify the structure of the functional group present in the isomer. Finally, draw the structural formula of a molecule that contains the new functional group and has the same molecular formula as the parent compound. Solve .

    1. The formula of acetone is C 3 H 6 O. An aldehyde contains the group image

      An aldehyde that is an isomer of acetone is propionaldehyde (or propanal):

      image
    2. The formula of 1-propanol is C 3 H 8 O. An ether contains the group –O–. An ether that is an isomer of 1-propanol is ethylmethyl ether:
      image

    24.46

    1. image
    2. image

      CH 2 = = CH 2 CH 2 CH 2 OH,     CH 3 CH = = CHCH 2 OH,    (cis and trans) CH 2 = = CHCH(OH)CH 3 (enantiomers)

      (Structures with the —OH group attached to an alkene carbon atom are not included. These molecules are called “vinyl alcohols” and are not the major form at equilibrium.)


    24.47 Analyze/Plan. From the hydrocarbon name, deduce the number of C atoms in the acid; one carbon atom is in the carboxyl group. Solve .

    1. meth = 1 C atom
      image
    2. pent = 5 C atoms
      image
    3. dec = 10 C atoms in backbone
      image

    24.48

    1. image
    2. image
    3. image
    4. image

    24.49 Analyze/Plan. In a condensation reaction between an alcohol and a carboxylic acid, the alcohol loses its —OH hydrogen atom and the acid loses its —OH group. The alkyl group from the acid is attached to the carbonyl group and the alkyl group from alcohol is attached to the ether oxygen of the ester. The name of the ester is the alkyl group from the alcohol plus the alkyl group from the acid plus the suffix - oate . Solve .

    1. image
    2. image
    3. image

    24.50

    1. image
    2. image
    3. image

    24.51 Analyze/Plan. Follow the logic in Sample Exercise 24.6. Solve .

    1. image
    2. image

    24.52

    1. image
    2. image

    24.53   High melting and boiling points are indicators of strong intermolecular forces in the bulk substance. The strongest intermolecular force among neutral covalent molecules is hydrogen bonding. The carboxyl group of acetic acid has —OH, which acts as a donor, and —C=O, which acts as an acceptor in hydrogen bonding. We expect acetic acid to be a strongly hydrogen-bonded substance, as shown by its physical properties. The melting and boiling points of acetic acid are somewhat higher than those of water, another substance that experiences strong hydrogen bonding.

    24.54

    image

    24.55 Analyze/Plan. Follow the logic in Sample Exercise 24.2, incorporating functional group information from Table 24.6. Solve .

    1. CH 3 CH 2 CH 2 CH(OH)CH 3
    2. CH 3 CH(OH)CH 2 OH
    3. image
    4. image
    5. CH 3 OCH 2 CH 3

    24.56

    1. CH 3 CH 2 CH 2 CH 2 CH(C 2 H 5 )CH 2 OH
    2. image
    3. image
    4. CH 3 CH 2 OCH 2 CH 2 CH 2 CH 3
    5. image

    24.57 Analyze/Plan. Review the rules for naming alkanes and haloalkanes; draw the structure. That is, draw the carbon chain indicated by the root name, place substituents, fill remaining positions with H atoms. Each C atom attached to four different groups is chiral. Solve .

    image

    The correct choice is (c); there are two chiral carbon atoms in the molecule. C2 is obviously attached to four different groups. C3 is chiral because the substituents on C2 render the C1-C2 group different than the C4-C5 group.

    24.58

    image

    Yes, the molecule is chiral. The chiral carbon atom is attached to chloro, methyl, ethyl, and propyl groups. (If the root was a 5-carbon chain, the molecule would not have optical isomers because two of the groups would be ethyl groups.)


    Introduction to Biochemistry; Proteins (Sections 24.6 and 24.7)

    24.59

    1. image
    2. In forming a protein, amino acids undergo a condensation reaction between the amino group and carboxylic acid:
      image
    3. The bond that links amino acids in proteins is called the peptide bond.
      image

    24.60

    1. True.
    2. True. Lysine is in the group of basic amino acids. Near pH 7, these amino acids have more positively charged functional groups than negatively charged groups.
    3. False. There is one amide group, but other N-containing group is an amine; it is separated from the carbonyl group by one C atom.
    4. False. Leucine and isoleucine are structural isomers, but not enantiomers.
    5. False. The R group of valine is hydrophobic; it is a nonpolar hydrocarbon. The R group of arginine is hydrophilic; although it is larger than that of valine, it has one imine and two amine groups capable of hydrogen-bonding interactions with water molecules.

    24.61 Analyze/Plan. Two dipeptides are possible. Either peptide can have the terminal carboxyl group or the terminal amino group. Solve .

    histadylaspartic acid, His-Asp or HD

    image

    aspartylhistidine, Asp-His or DH

    image

    24.62

    image

    24.63 Analyze/Plan. Follow the logic in Sample Exercise 24.7. Solve .

    1. image
    2. Three tripeptides are possible: Gly-Gly-His, GGH; Gly-His-Gly, GHG; His-Gly-Gly, HGG

    24.64

    1. Valine, serine, glutamic acid
    2. Six (assuming the tripeptide contains all three amino acids):

      Gly-Ser-Glu, GSE; Gly-Glu-Ser, GES; Ser-Gly-Glu, SGE; Ser-Glu-Gly, SEG; Glu-Ser-Gly, ESG; Glu-Gly-Ser, EGS

    24.65

    1. True.
    2. False. Alpha helix and beta sheet are examples of secondary structure. They are different configurations of the protein chain.
    3. False.

    24.66

    1. False. In an alpha helix, hydrogen bonding occurs between carbonyl groups and amide hydrogen atoms along the protein backbone.
    2. False. In a beta sheet, hydrogen bonding occurs between carbonyl groups and amide hydrogen atoms along two different protein chains. This interaction zips the two chains together into a beta sheet.

    Carbohydrates and Lipids (Sections 24.8 and 24.9)

    24.67

    1. True. A disaccharide is composed of two sugar units. Sugars are carbohydrates.
    2. False. Sucrose is a disaccharide.
    3. True. Well, most carbohydrates have this general formula.

    24.68

    1. No. Carbon 1 in the cyclic form of glucose is chiral. The configuration at C1 in alpha glucose is the opposite of that in beta glucose. However, there are four other chiral centers in each of the two forms that do not have opposite configurations. In order for molecules to be enantiomers, they must be mirror images of each other; all analogous chiral centers must have opposite configurations.

    1. image
    1. image

    24.69

    1. The empirical formula of cellulose is C 6 H 10 O 5 .
    2. The six-membered ring form of glucose forms the monomer unit that is the basis of the polymer cellulose. In cellulose, glucose monomer units are joined by β linkages.
    3. Ether linkages connect the glucose monomer units in cellulose.

    24.70

    1. The empirical formula of starch is C 6 H 10 O 5 .
    2. The six-membered ring form of glucose is the unit that forms the basis of starch. The glucose monomer units are joined by α linkages.
    3. Ether linkages connect the glucose monomer units in starch.

    24.71

    1. Yes, D-mannose is a sugar; it is a polyhydroxy aldehyde.
    2. In the linear form of mannose, there are four chiral carbon atoms, C2, C3, C4, and C5. The two terminal carbon atoms, C1 and C6, are not chiral.
    3. Both the α (left) and β (right) forms are possible.
      image

    24.72

    1. Galactose is a sugar; it is a polyhydroxy aldehyde.
    2. In the linear form of galactose, there are four chiral carbon atoms, C2, C3, C4, and C5. The two terminal carbon atoms, C1 and C6, are not chiral.
    3. The structure is best deduced by comparing galactose with glucose, and inverting the configurations at the appropriate carbon atoms. Recall from Solution 25.71 that both the β-form (shown here) and the α-form (OH on carbon 1 on the opposite side of ring as the CH 2 OH on carbon 5) are possible.
      image

    24.73

    1. False. Fat molecules contain ester functional groups.
    2. True. See Figure 24.24.
    3. True. See Figure 24.24

    24.74

    1. True. Consider the fuels ethane, C 2 H 6 , and ethanol, C 2 H 5 OH, where one C–H bond has been replaced by C–O–H, a C–O and an O–H bond. Combustion reactions for the two fuels follow.

      C 2 H 6 + 7/2 O 2 → 2 CO 2 + 3 H 2 O; C 2 H 5 OH + 3 O 2 → 2 CO 2 + 3 H 2 O

      More energy is required to break bonds in the combustion of one mole of ethanol. The reaction is less exothermic overall than the combustion of ethane. As ethane has the more exothermic combustion reaction, we say that more energy is “stored” in C 2 H 6 than in C 2 H 5 OH.

    1. False. Trans fats contain at least one double bond, with substituents in the trans orientation.
    2. True.
    3. False. Monounsaturated fatty acids have one carbon-carbon double bond in the chain, whereas the rest are single bonds.

    Nucleic Acids (Section 24.10)

    24.75   Dispersion forces increase as molecular size (and molar mass) increases. The larger purines (2 rings vs. 1 ring for pyrimidines) have larger dispersion forces.

    24.76

    image

    24.77 Analyze/Plan. Consider the structures of the organic bases in Section 24.10. The first base in the sequence is attached to the sugar with the free phosphate group in the 5′ position. The last base is attached to the sugar with a free –OH group in the 3′ position. Solve .

    The DNA sequence is 5′–TACG–3′.

    24.78 Analyze/Plan. Refer to Figures 24.26 and 24.27 for details of DNA structure. In the DNA double strand, each adenine is hydrogen-bonded to a thymine and each cytosine is hydrogen-bonded to a guanine. Each base in a DNA strand is associated with a phosphate group with a 1– charge.

    1. One adenine to one thymine
    2. One cytosine to one guanine
    3. 24 sodium ions per dodecamer. The double-stranded dodecamer has 24 bases, 12 in each strand, for an overall 24– charge. 24 sodium ions per dodecamer are required to achieve charge balance.

    24.79 Analyze/Plan. Recall that there is complimentary base pairing in nucleic acids because of hydrogen bond geometry. The DNA pairs are A–T and G–C. (The RNA pairs are A–U and G–C.)

    Solve. From the single strand sequence, formulate the complimentary strand. Note that 3′ of the complimentary strand aligns with 5′ of the parent strand.

    5′–GCATTGGC–3′

    3′–CGTAACCG–5′

    24.80   Statement (d) best explains the chemical differences between DNA and RNA. Statement (b) is true, but it does not prevent hydrogen bonding between complimentary base pairs.

    Additional Exercises

    24.81

    image

    Structures with the –OH group attached to an alkene carbon atom are not included. These molecules are called “vinyl alcohols” and are not the major form at equilibrium.

    24.82 Analyze/Plan. We are asked the number of structural isomers for two specified carbon chain lengths and a certain number of double bonds. Structural isomers have different connectivity. Since the chain length is specified, we can ignore structural isomers created by branching. We are not asked about geometrical isomers, so we ignore those as well. The resulting question is: How many ways are there to place the specified number of double bonds along the specified C chain? Solve .

    5 C chain with one double bond: 2 structural isomers

    C = = C C C C       C C = = C C C

    6 C chain with two double bonds: 6 structural isomers

    C=C—C=C—C—C C=C—C—C=C—C C=C—C—C—C=C C—C=C—C=C—C C=C=C—C—C—C C—C=C=C—C—C

    24.83

    1. image
    2. Cyclopentene does not show cis-trans isomerism because the existence of the ring demands that the C–C bonds be cis to one another.
    3. 1-pentyne does not have enantiomers because the geometry about the alkyne group is linear.

    24.84   The suffix –ene signifies an alkene, –one a ketone. The molecule has alkene and ketone functional groups.


    24.85

    1. image
    2. image
    3. image
    4. image

    24.86

    1. Quinine, molecule (b), and indigo, molecule (c), both contain amine functional groups and would produce basic solutions if dissolved in water.
    2. Acetaminophen, molecule (d), would produce an acidic solution if dissolved in water. None of the four molecules is a carboxylic acid. Acetaminophen is both an amide and a phenol. Both of these functional groups have ionizable H atoms and produce very weakly acidic aqueous solutions.
    3. Acetaminophen, molecule (d), is probably most water soluble. Both (c) and (d) can form hydrogen bonds with water, but acetaminophen has a lower molecular weight and smaller nonpolar portion to interfere with interactions between solute and Solve nt.

    24.87

    1. image
    2. image
    3. image
    4. image

    24.88

    1. nitroglycerin, 1,2,3-trinitroxypropane
      image
    2. Putrescine, 1,4-diaminobutane
      image
    3. Cyclohexanone, a precursor to Nylon
      image
    4. 1,1,2,2-tetrafluoroethene, precursor to Teflon
      image
    5. Oleic acid, cis-9-octanedecenoic acid
      image

    24.89   In order for indole to be planar, the N atom must be sp 2 hybridized. The nonbonded electron pair on N is in a pure p orbital perpendicular to the plane of the molecule. The electrons that form the π bonds in the molecule are also in pure p orbitals perpendicular to the plane of the molecule. Thus, each of these p orbitals is in the correct orientation for π overlap; the delocalized π system extends over the entire molecule and includes the “nonbonded” electron pair on N. The reason that indole is such a weak base (H + acceptor) is that the nonbonded electron pair is delocalized and an H + ion does not feel the attraction of a full localized electron pair.

    24.90

    1. The molecule has one ketone and two alcohol functional groups. There are no chiral centers, no carbon atoms attached to four different groups.
    2. The molecule has one ketone and three alcohol functional groups. There is one chiral center. The carbon bearing the secondary −OH has four different groups attached, and is thus chiral.
    3. The molecule has a carboxylic acid and an amine functional group. It is an amino acid, shown in its neutral (not zwitterion) form. There are two chiral centers. The carbon bearing the −NH 2 group and the carbon bearing the –CH 3 group are both chiral.

    24.91   In the zwitterion form of a tripeptide present in aqueous solution near pH 7, the terminal carboxyl group is deprotonated and the terminal amino group is protonated, resulting in a net zero charge. The molecule has a net charge only if a side (R) group contains a charged (protonated or deprotonated) group. The tripeptide is positively charged if a side group contains a protonated amine. According to Figure 24.16, the only amino acids with protonated amines in their side groups are histidine (His), lysine (Lys), and arginine (Arg). Of the tripeptides listed, only (a) Gly-Ser-Lys will have a net positive charge at pH 7. [Note that aspartic acid (Asp) has a deprotonated carboxyl in its side group, so (c) Phe-Tyr-Asp will have a net negative charge at pH 7.]

    24.92   Glu-Cys-Gly is the only possible order. Glutamic acid has two carboxyl groups that can form a peptide bond with cysteine, so there are two possible structures.

    image

    24.93   Both glucose and fructose contain six C atoms, so both are hexoses. Glucose contains an aldehyde group at C1, so it is an aldohexose. Fructose has a ketone at C2, so it is a ketohexose.

    24.94   DNA and RNA have the bases guanine, cytosine, and adenine in common, but DNA contains thymine, whereas RNA contains uracil. Thymine and uracil differ by a single methyl group, so both have a similar hydrogen bonding pattern with the complementary base adenine. This means that there is the possibility of a DNA strand binding to a “complementary” RNA strand.

    Given this possibility, RNA is not involved in DNA replication. However, during the process of transcription and in the presence of the enzyme RNA polymerase, a complementary strand of mRNA is assembled along a segment of the backbone of a single DNA strand. During transcription, when there is adenine (A) in the DNA strand, uracil (U) is added to the RNA strand.

    Integrative Exercises

    24.95

    image
    1. Methane boils at –128 °C. Methane is nonpolar and has the smallest molar mass. It experiences only weak dispersion forces and has the lowest boiling point.
    2. Difluoromethane boils at –52 °C. It is somewhat polar and experiences weak dipole-dipole and dispersion forces. Although difluoromethane has a larger molar mass than dimethyl ether, it has approximately spherical shape, which reduces the strength of its dispersion forces.

    1. Dimethyl ether boils at –25 °C. It is somewhat polar and experiences weak dipole-dipole and dispersion forces. Its shape is bent, which enables stronger dispersion forces than the distorted spherical shape of difluoromethane. [This explanation of the relative boiling points of dimethyl ether and difluoromethane is a rationalization based on the actual boiling points of the compounds. It is difficult to distinguish the two based on simple principles of relative strengths of intermolecular forces.]
    1. Ethanol boils at 78 °C. It is an alcohol that participates in hydrogen bonding. This is the strongest kind of intermolecular force and explains the highest boiling point.

    24.96   Determine the empirical formula of the unknown compound and its oxidation product. Use chemical properties to propose possible structures.

    68.1 g C × 1 mol C 12 .01 g C = 5.6703 ; 5.6703 / 1.1375 = 4.98 5

    13.7 g H × 1 mol H 1 .008 g H = 13.5913 ; 13.5913 / 1.1375 = 11.95 12

    18.2 g P × 1 mol O 16 .00 g O = 1.1375 ; 1.1375 / 1.1375 = 1

    The empirical formula of the unknown is C 5 H 12 O.

    69.7 g C × 1 mol C 12 .01 g C = 5.8035 ; 5.8035 / 1.1625 = 4.99 5

    11.7 g H × 1 mol H 1 .008 g H = 11.6071 ; 11.6071 / 1.1625 = 9.99 10

    18.6 g O × 1 mol O 16 .00 g O = 1.1625 ; 1.1625 / 1.1625 = 1

    The empirical formula of the oxidation product is C 5 H 10 O.

    The compound is clearly an alcohol. Its slight solubility in water is consistent with the properties expected of a secondary alcohol with a five-carbon chain. The fact that oxidation results in a ketone, rather than an aldehyde or a carboxylic acid, tells us that it is a secondary alcohol. Some reasonable structures for the unknown secondary alcohol are:

    image

    24.97   Determine the empirical formula, molar mass, and thus molecular formula of the compound. Confirm with physical data.

    66 .7 g C × 1 mol C 12 .01 g C = 5 .554 mol C; 5 .554/1 .381 = 4 .021 = 4

    11.2 g H × 1 mol H 1 .008 g H = 11.11 mol H; 11 .11/1 .381 = 8.043 = 8

    22.1 g O × 1 mol O 16 .00 g O = 1.381 mol O; 1 .381/1 .381 = 1


    The empirical formula is C 4 H 8 O. Using Equation 10.11 (MM = molar mass):

    MM = ( 2.28 g/L ) ( 0.08206 L-atm/mol-K ) ( 373 K ) 0.970 -atm = 71.9 g/mol

    The formula weight of C 4 H 8 O is 72, so the molecular formula is also C 4 H 8 O. Because the compound has a carbonyl group and cannot be oxidized to an acid, the only possibility is 2-butanone.

    image

    The boiling point of 2-butanone is 79.6 °C, confirming the identification.

    24.98   Determine the empirical formula, molar mass, and thus molecular formula of the compound. Confirm with physical data.

    85.7 g C × 1 mol C 12 .01 g C = 7.136 mol C; 7 .136/7 .136 = 1

    14.3 g H × 1 mol H 1 .008 g H = 14.19 mol H; 14.19 /7 .136 2

    Empirical formula is CH 2 . Using Equation 10.11 (MM = molar mass):

    MM = ( 2.21 g/L ) ( 0.08206 L-atm/mol-K ) ( 373 K ) ( 735 / 760 ) atm = 69.9 g/mol

    The molecular formula is thus C 5 H 10 . The absence of reaction with aqueous Br 2 indicates that the compound is not an alkene, so the compound is probably the cycloalkane cyclopentane. According to the Handbook of Chemistry and Physics , the boiling point of cyclopentane is 49 °C at 760 torr. This confirms the identity of the unknown.

    24.99   The reaction is: 2 NH 2 CH 2 COOH(aq) → NH 2 CH 2 CONHCH 2 COOH(aq) + H 2 O(l)

    ΔG° = (–488) + (–237.13) – 2(–369) = 12.87 = 13 kJ

    24.100

    1. image
    2. image

      K a = 1.8 × 10– 5 , pK a = –log(1.8 × 10– 5 ) = 4.74

      The conjugate acid of NH 3 is NH 4 + .

      NH 4 + (aq) ⇌ NH 3 (aq) + H + (aq)

      K a = K w /K b = 1.0 × 10– 14 /1. 8 × 10– 5 = 5.55 × 10– 10 = 5.6 × 10– 10

      pK a = –log(5.55 × 10– 10 ) = 9.26


      In general, a –COOH group is a stronger acid than a –NH 3 + group. The smaller pK a value for amino acids is for the ionization (deprotonation) of the –COOH group and the larger pK a is for the deprotonation of the –NH 3 + group.

    1. image

      By analogy to serine, the carboxyl group near the amine will have pK a ~2 and the amino group will have pK a ~9. By elimination, the carboxyl group in the side chain has pK a ~4.

    1. From the titration curve, the unknown amino acid has three pK a values: 2.0, 4.0, and 10.0. The unknown will be an acidic amino acid, because two of the pK a values are significantly less than 7. The two acidic amino acids are glutamic acid and aspartic acid. From part (c), the pK a values of glutamic acid are close to but not an exact match to the ones on the titration curve. The most likely candidate is aspartic acid.

      Asparagine and glutamine are the amide forms of aspartic and glutamic acid. The “middle” pK a value for these two amino acids represents ionization of the amide hydrogen and will be large than 4.0.

    24.101

    1. Because the native form is most stable, it has a lower, more negative free energy than the denatured form. Another way to say this is that ΔG for the process of denaturing the protein is positive.
    2. ΔS is negative in going from the denatured form to the folded (native) form; the native protein is more ordered.
    3. The four S–S linkages are strong covalent links holding the chain in place in the folded structure. A folded structure without these links would be less stable (higher G) and have more motional freedom (more positive entropy).
    4. After reduction, the eight S–S groups will form hydrogen-bond-like interactions with acceptors along the protein backbone, but these will be weaker and less specifically located than the S–S covalent bonds of the native protein. Overall, the tertiary structure of the reduced protein will be looser and less compact because of the loss of the S–S linkages; the entropy will be higher.
    5. The amino acid cysteine must be present in order for –SH bonds to be found in ribonuclease A. (Methionine contains S, but no –SH functional group.)

    24.102   AMPOH (aq) ⇌ AMPO 2– (aq) + H + (aq)

    pK a = 7.21 ; K a = 10 pK a = 6.17 × 10 8 = 6.2 × 10 8

    K a = [ AMPO 2 ] [ H + ] [ AMPOH ] = 6.2 × 10 8 . When pH = 7.40 , [ H + ] = 3.98 × 10 8 = 4 × 10 8 .

    Then [AMPOH ] [ AMPO 2 ] = 3.98 × 10 8 / 6.17 × 10 8 = 0.6457 = 0.6